Class 12 Physics (Part 1) Chapter 7 Moving Charges and Magnetic Field

Key Concept: Helical Motion

b) $\frac{m v \sin \theta}{eB}$
[Solution Description]
For helical motion, radius is determined by the perpendicular component of velocity: $v_{\perp} = v \sin \theta$.
The radius is $r = \frac{m v_{\perp}}{qB} = \frac{m v \sin \theta}{eB}$ (since charge of electron is $-e$).

Key Concept: Magnetic Force on a Moving Charge

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The magnetic force $\overrightarrow{F}$ on a charged particle moving with velocity $\overrightarrow{v}$ in a uniform magnetic field $\overrightarrow{B}$ is given by $\overrightarrow{F} = q (\overrightarrow{v} \times \overrightarrow{B})$. The cross product indicates that $\overrightarrow{F}$ is perpendicular to both $\overrightarrow{v}$ and $\overrightarrow{B}$. Since force is perpendicular to velocity, the work done $W = \overrightarrow{F} \cdot \overrightarrow{v} = 0$, meaning no energy is transferred. Hence, the speed of the particle remains constant while its direction changes. Therefore, both Assertion and Reason are true, and Reason correctly explains why the force is perpendicular to the velocity.

Key Concept: Ampere's Circuital Law, Magnetic Field

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that the line integral of $\vec{B}$ around a closed path is zero when no current passes through the enclosed area. This is true because according to Ampere’s circuital law, $\oint \vec{B} \cdot d\vec{l} = \mu_0 I$, and if $I = 0$, then $\oint \vec{B} \cdot d\vec{l} = 0$. The Reason provides the correct mathematical formulation of Ampere’s law, which directly explains why the Assertion is true. Thus, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Direction of Magnetic Field

a) Upwards (out of the plane)
[Solution Description]
Using the right-hand thumb rule, if the current flows from north to south, then the magnetic field at a point east of the wire will be directed upwards (out of the plane).

Key Concept: Cyclotron Kinetic Energy

c) 5.51 MeV
[Solution Description] The kinetic energy $K$ of a particle accelerated in a cyclotron is given by:
$K = \frac{q^2 B^2 R^2}{2m}$
Substitute the given values:
$K = \frac{(1.6 \times 10^{-19})^2 (1.2)^2 (0.4)^2}{2 \times 3.34 \times 10^{-27}}$
Calculate each component step-by-step:
- $(1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38}$
- $(1.2)^2 = 1.44$
- $(0.4)^2 = 0.16$
- $2 \times 3.34 \times 10^{-27} = 6.68 \times 10^{-27}$
Now multiply the numerators:
$2.56 \times 10^{-38} \times 1.44 \times 0.16 = 5.89824 \times 10^{-39}$
Finally, divide by the denominator:
$K = \frac{5.89824 \times 10^{-39}}{6.68 \times 10^{-27}} = 8.83 \times 10^{-13} \, \text{J}$
Convert to MeV (given 1 MeV = $1.602 \times 10^{-13}$ J):
$K = \frac{8.83 \times 10^{-13}}{1.602 \times 10^{-13}} = 5.51 \, \text{MeV}$

Key Concept: Limitations of Cyclotron

b) Neutrons
[Solution Description]
Due to their very small mass and rapid acquisition of high speeds, electrons quickly become relativistic, causing their mass to increase and making them go out of sync with the alternating potential difference. Hence, cyclotrons are unsuitable for accelerating electrons.

Key Concept: Lorentz Force and Helical Motion

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the Lorentz force causes a charged particle to move in a helical path when it enters a uniform magnetic field at an angle other than $0^\circ$ or $90^\circ$. This is correct because the velocity $\vec{v}$ can be resolved into components $v_{\parallel} = v \cos \theta$ (parallel) and $v_{\perp} = v \sin \theta$ (perpendicular). The parallel component causes linear motion, while the perpendicular component results in circular motion due to the Lorentz force, combining to form helical motion. The reason correctly explains how these components lead to helical motion.

Key Concept: Direction of Force Between Parallel Currents

b) Repulsive
[Solution Description]
According to the rule for forces between parallel currents:
- If currents are in the **same** direction, they **attract**.
- If currents are in **opposite** directions, they **repel**.
Here, since $I_1$ and $I_2$ are in opposite directions, the force is repulsive.

Key Concept: Magnetic Force on a Moving Charge, Direction of Force

d) Assertion is false, but Reason is true.
[Solution Description]
The Assertion states that the direction of the magnetic force on a positively charged particle moving perpendicular to a uniform magnetic field is tangential to its circular path. However, this is incorrect because the magnetic force is always perpendicular to both $\overrightarrow{v}$ and $\overrightarrow{B}$, which means it acts radially inward (centripetal force), not tangentially. The Reason correctly states that the magnetic force provides the centripetal force but does not explain the Assertion, which is false.
Therefore, the Assertion is false, and the Reason is true.

Key Concept: Lorentz Force

b) $Bqv \sin \theta$
[Solution Description]
The magnetic force acting on a moving charge is given by $F = Bqv \sin \theta$. This formula represents the magnitude of the Lorentz force when a charge moves through a magnetic field.
Step 1: Identify the given parameters - charge ($q$), velocity ($v$), magnetic field ($B$), and the angle ($\theta$) between velocity and magnetic field.
Step 2: Directly apply the formula for magnetic force on a moving charge: $F = Bqv \sin \theta$.

Key Concept: Motion of Charged Particles

d) Circular
[Solution Description]
When a charged particle moves perpendicular to a uniform magnetic field ($\theta = 90^\circ$), it describes a circular path because the force is always perpendicular to its velocity. The centripetal force required for circular motion is provided by the magnetic force $F = Bqv$, and the radius is given by $r = \frac{mv}{qB}$.

Key Concept: Magnetic Field at Centre for N-turn Coil, Resultant Field for Combined Segments

c) $12.1875\mu_0\,\text{T}$
[Solution Description]
First, calculate the magnetic field due to the circular coil:
$B_1 = \frac{\mu_0 N I}{2a} = \frac{\mu_0 \times 10 \times 5}{2 \times 2} = \frac{50\mu_0}{4} = 12.5\mu_0\,\text{T}$
Next, find the field due to the semi-circular loop:
$B_2 = \frac{\mu_0 I}{4a} = \frac{\mu_0 \times 5}{4 \times 4} = \frac{5\mu_0}{16} = 0.3125\mu_0\,\text{T}$
Since the currents are in opposite directions, subtract $B_2$ from $B_1$:
$B_{\text{net}} = B_1 - B_2 = 12.5\mu_0 - 0.3125\mu_0 = 12.1875\mu_0\,\text{T}$

Key Concept: Faraday's Law of Induction

b) $0.3 \, \text{V}$
[Solution Description]
According to Faraday's law of induction, the induced emf $\epsilon$ is given by:
$\epsilon = -\frac{\Delta\phi}{\Delta t}$
Here, $\Delta\phi = 0.05 - 0.02 = 0.03 \, \text{Wb}$ and $\Delta t = 0.1 \, \text{s}$. So,
$\epsilon = \frac{0.03}{0.1} = 0.3 \, \text{V}$

Key Concept: Right-hand Thumb Rule, Maxwell’s Right-hand Screw Rule

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
Let us evaluate the Assertion and Reason step-by-step:
1. **Assertion Analysis**:
- Maxwell’s right-hand screw rule states that if a screw is rotated in the direction of the current, the direction in which the screw advances gives the direction of the magnetic field.
- If the current-carrying conductor is rotated **clockwise**, it implies the conventional current flows downward (as per the rule). For a point **below** the conductor, the magnetic field direction would indeed be **into the plane of the paper**.
- Thus, the Assertion is **true**.
2. **Reason Analysis**:
- The right-hand thumb rule correctly describes that for a straight current-carrying conductor, the curled fingers give the direction of the circular magnetic field lines when the thumb points in the direction of current.
- Hence, the Reason is also **true**.
3. **Explanation Check**:
- While both statements are true, the Reason does **not** explain why the Assertion is true. The Assertion relies on Maxwell’s screw rule, whereas the Reason describes the thumb rule for a straight conductor. These are two distinct rules.
- Therefore, the Reason is **NOT** the correct explanation of the Assertion.
Final Answer: Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

Key Concept: Maxwell’s Right-hand Screw Rule, Right-hand Palm Rule No. 1

b) Downward
[Solution Description]
Using Maxwell’s Right-hand Screw Rule:
1. Rotate the screw in the direction of the current (clockwise as viewed from below).
2. The direction of the advancing screw gives the magnetic field direction (downward).
Alternatively, using Right-hand Palm Rule No. 1:
1. Align your right-hand fingers along the current (clockwise).
2. The perpendicular from the palm points downward at the center.
Thus, the field direction is "Downward."

Key Concept: Force Between Parallel Current-Carrying Conductors

b) $1.5 \times 10^{-4}$ N/m
[Solution Description]
The force per unit length between two parallel current-carrying conductors is given by:
$\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2 \pi r}$
Here, $I_1 = 3$ A, $I_2 = 5$ A, $r = 0.02$ m. Substituting the values:
$\frac{F}{l} = \frac{(4\pi \times 10^{-7})(3)(5)}{2 \pi (0.02)} = 1.5 \times 10^{-4} \text{ N/m}$

Key Concept: Definition of Magnetic Field

a) Weber/metre$^2$ ($\text{Wb m}^{-2}$)
[Solution Description]
From the given equation $B = \frac{F}{qv \sin \theta}$, when $F = 1$ N, $q = 1$ C, $v = 1$ m/s, and $\theta = 90^\circ$, the SI unit of $B$ is derived as newton per ampere-metre ($\text{N A}^{-1} \text{m}^{-1}$), known as tesla (T).

Key Concept: Oersted’s Experiment

b) It produces a magnetic field
[Solution Description]
Oersted's experiment showed that a current-carrying conductor produces a magnetic field around it, causing a nearby magnetic needle to deflect. Reversing the current reverses the deflection.

Key Concept: Magnetic Field due to a Long Straight Wire

c) $\frac{\mu_0 I}{2 \pi r}$
[Solution Description]
Using Ampere's circuital law for a long straight wire, we have $\oint \vec{B} \cdot d\vec{l} = \mu_0 I$. For a circular path of radius $r$ around the wire, $\vec{B}$ is constant and tangential, so $B(2 \pi r) = \mu_0 I$. Solving for $B$, we get $B = \frac{\mu_0 I}{2 \pi r}$.

Key Concept: Torque due to magnetic field, Solenoid-coil interaction

b) $3.96 \times 10^{-5} \, \text{N m}$
[Solution Description]
First, find the magnetic field inside the solenoid:
$B = \mu_0 n I_0 = (4\pi \times 10^{-7}) \times 2500 \times 5 = 1.57 \times 10^{-2} \, \text{T}.$
The torque on the coil is given by:
$\tau = N I A B \sin\theta,$
where $A = \pi r^2 = \pi (0.02)^2 = 1.26 \times 10^{-3} \, \text{m}^2$ and $\theta = 30^\circ$.
Substituting:
$\tau = 50 \times 0.8 \times 1.26 \times 10^{-3} \times 1.57 \times 10^{-2} \times \sin 30^\circ.$
Simplifying:
$\tau = 50 \times 0.8 \times 1.26 \times 1.57 \times 0.5 \times 10^{-5} = 3.96 \times 10^{-5} \, \text{N m}.$

Key Concept: Magnetic Field due to a Straight Current-Carrying Conductor

c) Circular paths centered on the conductor
[Solution Description]
According to the right-hand thumb rule, if you grasp the conductor with your right hand such that the thumb points in the direction of the current, the curled fingers indicate the direction of the magnetic field lines around the conductor.

Key Concept: Cyclotron Resonance Condition

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In a cyclotron, the resonance condition requires that the frequency $n_0$ of the applied potential difference matches the cyclotron frequency $\nu = \frac{qB}{2\pi m}$. This ensures the particle is accelerated each time it crosses the gap between the dees. The reason states that the time $t = \frac{\pi m}{qB}$ to complete a semi-circle is independent of speed and radius, which is true as derived from the cyclotron motion equations. The Reason correctly explains why the frequency matching is necessary for continuous acceleration, making both Assertion and Reason true with the Reason being the correct explanation.

Key Concept: Biot-Savart Law

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Biot-Savart law states that the magnetic field $dB$ due to a current element $I \vec{dl}$ at a point is given by:
$dB = \frac{\mu_0}{4\pi} \cdot \frac{I dl \sin \theta}{r^2}$
Here, $dB$ is indeed inversely proportional to $r^2$, as stated in Assertion (A). The Reason (R) correctly explains why this dependence exists: the field strength diminishes with distance following an inverse-square relationship, which is a fundamental characteristic of fields produced by point-like sources. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Assertion-Reason

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The Biot-Savart law states that the magnetic field $dB$ due to a small current element $I \, dl$ at a distance $r$ is given by $dB = \frac{\mu_0}{4\pi} \cdot \frac{I \, dl \sin \theta}{r^2}$. For a straight conductor, the net field at a perpendicular distance $r$ can be derived by integrating this expression, which results in $B \propto \frac{I}{r}$. Thus, the Assertion is true. The Reason provides the formula for the magnetic field due to an infinitesimal current element, but it does not directly explain why the field is inversely proportional to the perpendicular distance for a finite conductor. Hence, both are true, but the Reason does not correctly explain the Assertion.

Key Concept: Magnetic Field Definition, Direction

a) Upwards.
[Solution Description]
Using the right-hand rule for the cross product $\overrightarrow{F} = q (\overrightarrow{v} \times \overrightarrow{B})$:
1. Point fingers in the direction of $\overrightarrow{v}$ (to the right).
2. Curl fingers into the direction of $\overrightarrow{B}$ (into the plane).
3. Thumb points in the direction of $\overrightarrow{F}$ (upwards for positive charge).
Thus, the force acts upwards.

Key Concept: Basic definition of magnetic force on moving charge

c) Charge, velocity, and magnetic field
[Solution Description]
From the formula $\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})$, the magnetic force depends on:
1) Charge $q$ of the particle
2) Velocity $\overrightarrow{v}$ of the particle
3) Magnetic field $\overrightarrow{B}$
However, it does not depend on the mass of the particle.

Key Concept: Magnetic Field for Semi-Circular Loop, Direction of Field

c) $90^\circ$
[Solution Description]
The field due to the semi-circular loop is:
$B_{\text{loop}} = \frac{\mu_0 I}{4a} = \frac{\mu_0 \times 10}{4 \times 5} = 0.5\mu_0\,\text{T}$
The field due to the straight wire (using Biot-Savart law for an infinite wire) is:
$B_{\text{wire}} = \frac{\mu_0 I}{2\pi r} = \frac{\mu_0 \times 10}{2\pi \times 0} \rightarrow \infty\,\text{(invalid)}$
However, the question implies the wire is at the centre but finite. For simplicity, assume the wire contributes a field perpendicular to the loop's plane. Thus, the angle between $\vec{B}_{\text{loop}}$ (along the axis) and $\vec{B}_{\text{wire}}$ (perpendicular to the axis) is $90^\circ$.

Key Concept: Definition

c) Weber
[Solution Description]
The SI unit of magnetic flux is the Weber (Wb). It measures the total magnetic field passing through a given area.

Key Concept: Ampere’s Circuital Law

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion claims that the line integral of $\vec{B}$ around any closed path is always zero, which is incorrect. According to Ampere’s circuital law, the line integral $\oint \vec{B} \cdot d\vec{l}$ equals $\mu_0 I$, where $I$ is the net current passing through the loop. This means the line integral is non-zero if there is a current enclosed by the path. Therefore, the assertion is false. The reason correctly states Ampere’s circuital law, making it true. Hence, the correct answer is that the Assertion is false, but the Reason is true.

Key Concept: Force between two parallel current-carrying conductors

c) $\frac{\mu_0 I_1 I_2}{2 \pi d}$
[Solution Description]
The force per unit length between two parallel current-carrying conductors is given by:
$\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2 \pi d}$
Since the currents are in the same direction, the force is attractive. Substituting the values, we get:
$\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2 \pi d}$

Key Concept: Right-hand Thumb Rule

c) Into the plane of the paper (\$\times\$)
[Solution Description]
Using the Right-hand Thumb Rule, grasp the conductor with the right hand such that the thumb points upwards (direction of current). The curled fingers represent the direction of $\overrightarrow{B}$. At a point east of the conductor, the field direction is into the plane of the paper.

Key Concept: Magnetic field inside a solenoid

b) $6\pi \times 10^{-4} \, \text{T}$
[Solution Description]
The magnetic field inside a long solenoid is given by:
$B = \mu_0 n I$
Here, $n = 500 \, \text{turns/m}$ and $I = 3 \, \text{A}$. Substituting these values:
$B = (4\pi \times 10^{-7}) \times 500 \times 3 = 6\pi \times 10^{-4} \, \text{T}$

Key Concept: Dimensions of Magnetic Field

b) $\text{M T}^{-2} \text{A}^{-1}$
[Solution Description]
The dimensions of magnetic field $\overrightarrow{B}$ can be derived from the equation $[B] = \frac{[F]}{[q][v]}$. Given that $[F] = \text{M L T}^{-2}$, $[q] = \text{A T}$, and $[v] = \text{L T}^{-1}$, we have $[B] = \text{M T}^{-2} \text{A}^{-1}$.

Key Concept: Magnetic Field Inside a Solenoid

d) 7.54 \times 10^{-4} T
[Solution Description]
The magnetic field inside a long solenoid is given by:
$B = \mu_0 n I$
Given:
Number of turns per unit length, $n = 200$ turns/m
Current, $I = 3$ A
Permeability of free space, $\mu_0 = 4\pi \times 10^{-7}$ Tm/A
Substituting the values:
$B = (4\pi \times 10^{-7}) \times 200 \times 3 = 2400 \pi \times 10^{-7} = 2.4 \pi \times 10^{-4}$ T ≈ $7.54 \times 10^{-4}$ T (since $\pi \approx 3.14$)

Key Concept: Biot-Savart Law - Total field integration

b) $4\pi \times 10^{-6}$ T
[Solution Description]
For a circular loop, each current element $I dl$ contributes to the field at the center. The angle $\theta$ between $dl$ and $r$ is always $90^\circ$. Integrating over the entire loop:
$B = \frac{\mu_0 I}{2R} = \frac{4\pi \times 10^{-7} \times 2}{2 \times 0.1} = 4\pi \times 10^{-6} \text{ T}$

Key Concept: Force on Current-Carrying Conductor, Torque Calculation

c) $4.3 \times 10^{-3}$ Nm
[Solution Description]
Given:
- Dimensions of loop: 0.05 m $\times$ 0.10 m (area $A = 0.05 \times 0.10 = 0.005$ m$^2$)
- Current $I = 2$ A
- Magnetic field $B = 0.5$ T
- Angle between plane and field $\theta = 30^\circ$. Note: Actual angle between normal to the plane and field is $60^\circ$ ($90^\circ - 30^\circ$).
The torque $\tau$ on a current loop is given by:
$\tau = NIAB \sin \phi$
where $\phi$ is the angle between the normal to the plane and $\vec{B}$. Here, $\phi = 60^\circ$, $N = 1$.
Substituting values:
$\tau = 1 \times 2 \times 0.005 \times 0.5 \times \sin 60^\circ$
Calculate each step:
$2 \times 0.005 = 0.01 \\ \\ 0.01 \times 0.5 = 0.005 \\ \\ \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \\ \\ 0.005 \times 0.866 = 0.00433 \text{ Nm}$
Rounding to two significant figures:
$\tau \approx 4.3 \times 10^{-3} \text{ Nm}$

Key Concept: Concept: Direction of field

c) The magnetic field is perpendicular to both the current element and the line joining it to the observation point.
[Solution Description]
The electric field (\$\vec{E}\$) due to a point charge is always along the line joining the charge and the observation point, as per Coulomb’s Law. However, the magnetic field (\$\vec{B}\$) due to a current element, according to Biot-Savart’s Law, is perpendicular to both the current element \$I \vec{dl}\$ and the line joining it to the observation point (given by the right-hand rule).

Key Concept: Ampere's Circuital Law, Solenoid Magnetic Field

b) $2.51 \times 10^{-3} \text{ T}$
[Solution Description]
The magnetic field inside a long solenoid is given by:
$B = \mu_0 n I,$
where $n$ is the number of turns per unit length. Substituting $\mu_0 = 4\pi \times 10^{-7} \text{ T·m/A}$, $n = 1000 \text{ turns/m}$, and $I = 2 \text{ A}$:
$B = (4\pi \times 10^{-7}) \times 1000 \times 2 = 8\pi \times 10^{-4} \text{ T}.$
Numerically, this evaluates to approximately $2.51 \times 10^{-3} \text{ T}$.

Key Concept: Direction of Magnetic Field

c) Perpendicular to both $\vec{dl}$ and $\vec{r}$
[Solution Description]
The direction of the magnetic field $\vec{dB}$ is given by the cross product of the current element vector $I \vec{dl}$ and the position vector $\vec{r}$. Hence, $\vec{dB}$ is perpendicular to both $\vec{dl}$ and $\vec{r}$.

Key Concept: Definition of Ampere

b) 1 ampere
[Solution Description]
By definition, when two long, straight, and parallel conductors carry equal currents and are placed 1 metre apart, producing a force per unit length of $2 \times 10^{-7}$ N/m, the current in each conductor is defined as 1 ampere.

Key Concept: Magnetic Field of Solenoid, Directionality

b) $4\pi^2 \times 10^{-6} \, \text{Wb}$
[Solution Description]
First, calculate the magnetic field inside the solenoid: $B = \mu_0 n I$, where $n = \frac{1000}{0.5} = 2000 \text{ turns/m}$ and $I = 2 \text{ A}$:
$B = 4\pi \times 10^{-7} \times 2000 \times 2 = 16\pi \times 10^{-4} \text{ T}.$
The area of the loop is $A = \pi r^2 = \pi (0.05)^2 = 0.0025\pi \text{ m}^2$. The flux is:
$\Phi = B \cdot A = 16\pi \times 10^{-4} \times 0.0025\pi = 4\pi^2 \times 10^{-6} \text{ Wb}.$

Key Concept: Magnetic Field Due to a Current Loop

b) $\frac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}}$
[Solution Description]
The magnetic field at distance $x$ along the axis of a circular loop of radius $a$ carrying current $I$ is derived by integrating contributions from all current elements.
From Biot-Savart law and symmetry considerations, the field is:
$B = \frac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}}.$
Here, $\mu_0$ is the permeability of free space, and $(a^2 + x^2)^{1/2}$ is the distance from any point on the loop to the point on the axis.

Key Concept: Time Period and Frequency

b) $5.59 \times 10^{10} \, Hz$
[Solution Description]
The frequency of revolution is given by:
$\nu = \frac{qB}{2\pi m}$
Substituting the values (ignoring the sign of charge as frequency is always positive):
$\nu = \frac{(1.6 \times 10^{-19} \, C)(2 \, T)}{2\pi (9.11 \times 10^{-31} \, kg)}$
Simplifying step-by-step:
$\nu = \frac{3.2 \times 10^{-19}}{5.72 \times 10^{-30}} = 5.59 \times 10^{10} \, Hz$

Key Concept: Motion in Magnetic Field

c) Straight line
[Solution Description]
When a charged particle moves parallel ($\theta = 0$) or antiparallel ($\theta = 180^\circ$) to the magnetic field, the angle between $\overrightarrow{v}$ and $\overrightarrow{B}$ makes $\sin \theta = 0$, which means the force $F = Bqv \sin \theta = 0$. Hence, the electron moves undeviated along a straight line.

Key Concept: Circular motion in magnetic field, Time period

b) $m_p : m_e$
[Solution Description]
The time period of revolution in a magnetic field is given by:
$T = \frac{2 \pi m}{q B}$
Given both particles have the same kinetic energy:
$\frac{1}{2} m_e v_e^2 = \frac{1}{2} m_p v_p^2$
But since $m_p \approx 1836 m_e$, their velocities are different.
However, the time period depends only on mass and charge, not velocity.
For electron: $T_e = \frac{2 \pi m_e}{e B}$
For proton: $T_p = \frac{2 \pi m_p}{e B}$
Thus, the ratio is:
$\frac{T_p}{T_e} = \frac{m_p}{m_e} \approx 1836$

Key Concept: Magnetic Force on a Current-Carrying Conductor

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The force on a current-carrying conductor in a magnetic field is given by $F = BIl \sin \theta$. When the conductor is perpendicular to the magnetic field ($\theta = 90^\circ$), $\sin \theta = 1$, resulting in the maximum force $F = BIl$. Thus, both Assertion (A) and Reason (R) are true, and Reason correctly explains why the force is maximum when the conductor is perpendicular to the field.

Key Concept: Magnetic Field Due to Current-Carrying Wires

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The formula for the magnetic field at the center of a semi-circular current loop is correctly stated as $B = \frac{\mu_0 I}{4} \left( \frac{1}{a_1} + \frac{1}{a_2} \right)$, which means the Assertion (A) is true. The Reason (R) correctly states that the direction of the magnetic field is perpendicular to the plane of the paper and directed downwards, which is also true. Moreover, the Reason provides the correct explanation for the Assertion because the direction is indeed determined by the right-hand rule for the current loop. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Relativistic Effects, Limitation of Cyclotron

c) 9.15 MHz
[Solution Description]
The relativistic mass $m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$. Given $m = 2m_0$, we get $\sqrt{1 - \frac{v^2}{c^2}} = 0.5$. The original frequency was set for non-relativistic mass: $n_0 = \frac{qB}{2\pi m_0} = \frac{(1.6 \times 10^{-19})(1.2)}{2\pi (1.67 \times 10^{-27})} \approx 18.3$ MHz (matches given). For relativistic mass $m = 2m_0$, the new frequency should be $n_{new} = \frac{qB}{2\pi m} = \frac{(1.6 \times 10^{-19})(1.2)}{2\pi (2 \times 1.67 \times 10^{-27})} \approx 9.15$ MHz to maintain resonance.

Key Concept: Biot-Savart Law, Finite Straight Wire

b) $\frac{\mu_0 I L}{4\pi d \sqrt{(L/2)^2 + d^2}}$
[Solution Description]
For a finite straight wire, the magnetic field at a perpendicular distance $d$ from its midpoint is calculated using Biot-Savart law. The field is perpendicular to the plane containing the wire and the point P. The magnitude of the field is:
$B = \frac{\mu_0 I L}{4\pi d \sqrt{(L/2)^2 + d^2}}.$
This is obtained by integrating the contributions of infinitesimal current elements along the wire, taking into account the angle $\theta$ and distance $r$ for each element.

Key Concept: Ampere's Circuital Law - Basic Definition

c) $\mu_0 I$
[Solution Description]
According to Ampere's circuital law, the line integral of the magnetic field $\vec{B}$ around a closed loop is equal to $\mu_0$ times the net current passing through the area enclosed by the loop.
Step 1: The mathematical expression for Ampere's circuital law is:
$\oint \vec{B} \cdot d\vec{l} = \mu_0 I$
where $\mu_0$ is the permeability of free space and $I$ is the net current enclosed by the loop.
Therefore, the correct answer corresponds to this relation.

Key Concept: Magnetic field inside a solenoid

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The magnetic field inside a long solenoid is indeed uniform because the field lines are equidistant and parallel, indicating a constant magnitude and direction. The reason correctly explains why the assertion is true.

Key Concept: Motion of Charged Particles in Uniform Magnetic Field

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force acts as a centripetal force. The force magnitude is given by $F = qvB$ since $\sin(90^\circ) = 1$. This force is always perpendicular to the velocity, so it changes the direction of motion but not the speed, resulting in a circular path. Thus, both Assertion (A) and Reason (R) are true, and Reason correctly explains Assertion.

Key Concept: Earth's Magnetic Field

c) The Earth's magnetic field is from geographic south to geographic north.
[Solution Description] The north pole of a freely suspended magnetic needle points towards the Earth's geographic north, indicating that Earth's magnetic field acts from its geographic south to geographic north. This is because opposite magnetic poles attract each other.

Key Concept: Cyclotron Working Principle

c) Charged particles gain energy in steps by passing through alternating electric fields.
[Solution Description]
The cyclotron works on the principle that a charged particle, when subjected to an electric potential difference (\$V\$), gains kinetic energy equal to \$qV\$. A magnetic field is used to bring the particle back to the gap between the dees for repeated acceleration.

Key Concept: Application of Biot-Savart Law

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
According to Biot-Savart law, the magnetic field $dB$ at a point due to a small current element is given by:
$dB = \frac{\mu_0}{4\pi} \cdot \frac{I \, dl \sin \theta}{r^2}$
where $\theta$ is the angle between the current element ($d\vec{l}$) and the position vector ($\vec{r}$). The field is maximized when $\sin \theta = 1$, i.e., when $\theta = 90^\circ$. This aligns with Assertion (A). Additionally, Reason (R) correctly describes the relationship stated in the Biot-Savart law. However, Reason (R) does not explain why the field is maximum at $\theta = 90^\circ$, as it only restates the proportionality condition from the law. Thus, both are true, but Reason (R) is not the correct explanation for Assertion (A).

Key Concept: Magnetic Field of a Long Straight Solenoid

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the magnetic field inside an infinitely long solenoid is uniform, which is true because the field lines inside the solenoid are straight and parallel, resulting in a constant magnetic field. The reason given is also true as it correctly explains why the field is uniform: the magnetic field lines inside the solenoid are indeed parallel and uniformly spaced, leading to a uniform magnetic field strength throughout. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Ampere's Circuital Law, Toroid

b) $1.2 \times 10^{-3} \, \text{T}$
[Solution Description]
The magnetic field inside a toroidal solenoid is given by:
$B = \frac{\mu_0 N I}{2 \pi r}$
where $N$ is the total number of turns, $I$ is the current, and $r$ is the mean radius. Given $N = 1000$, $I = 3 \, \text{A}$, $r = 0.5 \, \text{m}$, and $\mu_0 = 4 \pi \times 10^{-7} \, \text{N A}^{-2}$:
$B = \frac{4 \pi \times 10^{-7} \times 1000 \times 3}{2 \pi \times 0.5} = \frac{12 \times 10^{-4}}{1} = 1.2 \times 10^{-3} \, \text{T}$

Key Concept: Effect of Reversing Current in Oersted's Experiment

a) Deflection direction reverses
[Solution Description]
Reversing the current direction changes the direction of the magnetic field produced around the conductor (as per the right-hand rule). Consequently, the deflection direction of the magnetic needle also reverses. For example, if the needle initially deflected east, it will now deflect west under the same conditions.

Key Concept: Magnetic Field at Axis

a) $4.44 \times 10^{-6}$ T
[Solution Description]
Given formula: $B = \frac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}}$
Here, $a = 2$ cm = 0.02 m, $I = 5$ A, $x = 4$ cm = 0.04 m
Substitute values:
$B = \frac{(4\pi \times 10^{-7}) \times 5 \times (0.02)^2}{2[(0.02)^2 + (0.04)^2]^{3/2}}$
Calculate numerator: $(4\pi \times 10^{-7}) \times 5 \times 0.0004 = 8\pi \times 10^{-10}$
Calculate denominator: $2[0.0004 + 0.0016]^{3/2} = 2(0.002)^{3/2} = 2 \times 2.828 \times 10^{-4} = 5.656 \times 10^{-4}$
Final calculation: $B = \frac{8\pi \times 10^{-10}}{5.656 \times 10^{-4}} \approx 4.44 \times 10^{-6}$ T

Key Concept: Magnetic Field Inside a Solenoid

a) $7.54 \times 10^{-4}$ T
[Solution Description]
The magnetic field inside a long solenoid is given by:
$B = \mu_0 n I$
Given: $n = 200 \, \text{turns/m}$, $I = 3 \, \text{A}$, and $\mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}$. Substitute these values:
$B = (4\pi \times 10^{-7})(200)(3) = 24\pi \times 10^{-5} \, \text{T}$
Convert to Tesla:
$24\pi \times 10^{-5} = 24 \times 3.14 \times 10^{-5} \approx 7.54 \times 10^{-4} \, \text{T}$

Key Concept: Magnetic Field Strength, Force on Moving Charge, Magnetic Field Due to Current-Carrying Wire

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For two parallel wires carrying equal currents $I$ in opposite directions, the magnetic field at a point equidistant ($d$) from both wires can be calculated using $B = \frac{\mu_0 I}{2 \pi d}$. Since the currents are in opposite directions, the right-hand rule indicates their magnetic fields will also be opposite in direction. At the midpoint, the fields are equal in magnitude but opposite in direction, resulting in cancellation. Thus, the net magnetic field is zero. Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Biot-Savart Law

d) Biot-Savart law.
[Solution Description] The Biot-Savart law gives the magnetic field $dB$ produced by a small current element $Idl$ at a distance $r$. It is given by:
$dB = \frac{\mu_0}{4\pi} \cdot \frac{I \, dl \, \sin \theta}{r^2}$

Key Concept: Lorentz Force, Cyclotron Frequency

a) $\pi m v$
[Solution Description]
First, recall the formula for the Lorentz force:
$F = q v B \sin\theta$, where $\theta = 30^\circ$.
The cyclotron frequency is given by $f = \frac{q B}{2 \pi m}$.
The ratio $\frac{F}{f}$ is computed as:
$\frac{F}{f} = \frac{q v B \sin 30^\circ}{\frac{q B}{2 \pi m}} = 2 \pi m v \sin 30^\circ$.
Since $\sin 30^\circ = \frac{1}{2}$, we have $\frac{F}{f} = \pi m v$.

Key Concept: Magnetic Field and Force on Moving Charge

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that a charged particle moving perpendicular ($\theta = 90^\circ$) to a uniform magnetic field experiences maximum force. This is true because the force equation $F = Bqv \sin \theta$ reaches its maximum value when $\theta = 90^\circ$, since $\sin 90^\circ = 1$.
The reason provides the formula for the magnetic force acting on a moving charged particle, which is correct. However, while the formula justifies the assertion's truth, it does not directly explain why the maximum force occurs at $\theta = 90^\circ$—it merely states the relationship. Thus, both are true, but the reason does not fully explain the assertion.

Key Concept: Magnetic Field inside a Solenoid

c) $\mu_0 n I_0$
[Solution Description]
The magnetic field inside a long solenoid can be derived using Ampere's law. For an ideal solenoid, the field outside is negligible, and inside it is uniform and parallel to the axis.
Step 1: The formula for the magnetic field inside the solenoid is:
$B = \mu_0 n I_0$
where $\mu_0$ is the permeability of free space, $n$ is the number of turns per unit length, and $I_0$ is the current.
Therefore, the correct answer corresponds to this relation.

Key Concept: Biot-Savart Law, Right-hand Rules

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that the magnetic field at point P is perpendicular to the plane formed by the conductor and the point P. This is true as the direction of the magnetic field $\vec{B}$ is determined by the right-hand rules, which state that the field is perpendicular to the plane containing the current element and the position vector $\vec{r}$.
The Reason explains that this perpendicularity arises from the cross product $d\vec{l} \times \vec{r}$ in the Biot-Savart Law, which indeed produces a vector perpendicular to both $d\vec{l}$ and $\vec{r}$. Thus, the Reason correctly explains the Assertion.
Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Lorentz Force

c) Either along $+\hat{i}$ or $-\hat{i}$ depending on $E$ and $vB$
[Solution Description]
The Lorentz force is given by $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$. Here, $\vec{v} = v \hat{k}$, $\vec{E} = E \hat{i}$, and $\vec{B} = B \hat{j}$.
Cross product: $\vec{v} \times \vec{B} = v \hat{k} \times B \hat{j} = -vB \hat{i}$.
Thus, $\vec{F} = q(E \hat{i} - vB \hat{i}) = q(E - vB)\hat{i}$.
Direction depends on $E$ and $vB$: If $E > vB$, force is along $+\hat{i}$; if $E < vB$, force is along $-\hat{i}$.

Key Concept: Force Direction

b) Repulsive
[Solution Description]
When currents flow in opposite directions in two parallel conductors, the magnetic fields around them interact to produce a repulsive force.

Key Concept: Magnetic Field Inside a Solenoid

b) $7.54 \times 10^{-3}$ T
[Solution Description]
For a long solenoid, the magnetic field $B$ inside is given by:
$B = \mu_0 n I$
Here, $n = 3000 \, \text{turns/m}$, $I = 2 \, \text{A}$. Substituting:
$B = 4\pi \times 10^{-7} \times 3000 \times 2 = 24\pi \times 10^{-4} \, \text{T} \approx 7.54 \times 10^{-3} \, \text{T}$

Key Concept: Magnetic Field at the Centre of a Circular Loop

c) $\frac{\mu_0 I}{2a}$
[Solution Description]
The magnetic field at the center of a circular loop is given by the formula:
$B = \frac{\mu_0 I}{2a}$
Here, $\mu_0$ is the permeability of free space, $I$ is the current, and $a$ is the radius of the loop.

Key Concept: Circular Motion of Charged Particles

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the angular frequency $\omega$ of a charged particle moving in a uniform magnetic field is independent of its velocity. This is true because $\omega$ depends only on the charge $q$, magnetic field $B$, and mass $m$ of the particle, as given by $\omega = \frac{qB}{m}$. The reason correctly provides the formula for $\omega$, which explicitly shows no dependence on velocity $v$. Therefore, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Force on a Current-Carrying Conductor in a Uniform Magnetic Field

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that a current-carrying conductor in a magnetic field experiences a force perpendicular to both the current and the magnetic field, which is true as per the syllabus. The Reason provides the correct formula for this force, $\vec{F} = I \vec{l} \times \vec{B}$, which is also true. Moreover, the Reason correctly explains the Assertion because the formula directly describes the force acting on the conductor due to the magnetic field. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Direction of Force

c) No force acts on the wire
[Solution Description]
According to the formula:
$F = I B l \sin\theta$
If the wire is parallel to the field ($\theta = 0^\circ$ or $180^\circ$), then $\sin\theta = 0$, and thus $F = 0$.

Key Concept: Ampere's Circuital Law, Long Straight Wire

b) $1 \times 10^{-5} \, \text{T}$
[Solution Description]
To find the magnetic field at a distance $r$ from a long straight wire carrying current $I$, we use Ampere's law. The formula is:
$B = \frac{\mu_0 I}{2 \pi r}$
Given $I = 5 \, \text{A}$, $r = 0.1 \, \text{m}$, and $\mu_0 = 4 \pi \times 10^{-7} \, \text{N A}^{-2}$. Substituting the values:
$B = \frac{4 \pi \times 10^{-7} \times 5}{2 \pi \times 0.1} = \frac{20 \times 10^{-7}}{0.2} = 1 \times 10^{-5} \, \text{T}$

Key Concept: Magnetic Force on Moving Charge, Helical Motion

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
When a charged particle moves in a magnetic field, the force acting on it is given by the Lorentz force $\vec{F} = q (\vec{v} \times \vec{B})$. For a velocity $\vec{v}$ at an angle $\theta$ to $\vec{B}$, we can resolve $\vec{v}$ into two components: $v_{\parallel} = v \cos \theta$ (parallel to $\vec{B}$) and $v_{\perp} = v \sin \theta$ (perpendicular to $\vec{B}$).
The magnetic force acts only on the perpendicular component $v_{\perp}$ because $\sin 0^\circ = 0$ for the parallel component. Thus, the particle undergoes circular motion due to $v_{\perp}$ (since $\vec{F}$ provides the centripetal force), while the parallel component $v_{\parallel}$ remains constant, leading to a helical path.
Therefore, Assertion (A) is true. Reason (R) correctly explains why the path is helical, so it is also true and is the correct explanation.

Key Concept: Ampere's Law, Solenoid

a) $\mu_0 n I$
[Solution Description]
Using Ampere's law for a solenoid:
$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$
For a solenoid, the enclosed current is $I_{\text{enc}} = n I L$, where $L$ is the length of the Amperian loop. The field inside is uniform and parallel to the axis, so:
$B L = \mu_0 n I L$
Simplifying:
$B = \mu_0 n I$

Key Concept: Lorentz Force

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Lorentz force $\overrightarrow{F} = q (\overrightarrow{v} \times \overrightarrow{B})$ is always perpendicular to the velocity $\overrightarrow{v}$. Since work done $W = \overrightarrow{F} \cdot \overrightarrow{d}$ and displacement $\overrightarrow{d}$ is along $\overrightarrow{v}$, the dot product becomes zero when $\overrightarrow{F}$ is perpendicular to $\overrightarrow{v}$. Hence, no work is done by the Lorentz force. Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Application of Ampere's Law

b) 4 \times 10^{-4} T
[Solution Description]
For a toroidal solenoid, the magnetic field is given by:
$B = \frac{\mu_0 N I}{2 \pi r}$
Given:
Number of turns, $N = 500$
Current, $I = 2$ A
Mean radius, $r = 0.5$ m
Permeability of free space, $\mu_0 = 4\pi \times 10^{-7}$ Tm/A
Substituting the values:
$B = \frac{(4\pi \times 10^{-7}) \times 500 \times 2}{2 \pi \times 0.5} = \frac{4000 \pi \times 10^{-7}}{\pi} = 4 \times 10^{-4}$ T

Key Concept: Motion in combined electric and magnetic fields, Helical motion

a) $\frac{E}{B v_y}$
[Solution Description]
For helical motion with axis parallel to the x-axis, the proton must have no net force along the y-axis. The forces acting are:
1. Electric force: $\vec{F}_E = q \vec{E} = q E \hat{j}$
2. Magnetic force: $\vec{F}_B = q (\vec{v} \times \vec{B}) = q (v_x \hat{i} + v_y \hat{j}) \times B \hat{k} = q B (-v_x \hat{j} + v_y \hat{i})$
For no net force along y-axis: $F_E + F_{B_y} = 0$
$q E - q B v_x = 0$
Hence, $v_x = \frac{E}{B}$
Since the motion is helical, $v_y$ must be non-zero. Thus, the required ratio is $\frac{v_x}{v_y} = \frac{E}{B v_y}$. However, since $v_y$ is arbitrary (only condition is $v_y \neq 0$), the simplest form is $\frac{v_x}{v_y} = \frac{E}{B v_y}$. But given the options, we consider standard cases where $v_y$ is related to other parameters.
After solving, the correct ratio is $\frac{v_x}{v_y} = \frac{E}{B v_y}$ but in standard cases, it simplifies to $\frac{E}{B v_y} = 1$ if $v_y = \frac{E}{B}$.
Therefore, the correct answer is when $\frac{v_x}{v_y} = \frac{E}{B v_y}$ or simplified forms.

Key Concept: Biot-Savart Law Proportionality

c) $\frac{I dl \sin \theta}{r^2}$
[Solution Description]
The Biot-Savart law states that the magnetic field $dB$ at a point due to a current element $I dl$ depends on four factors: it is directly proportional to the current $I$, the length of the element $dl$, the sine of the angle $\theta$ between the element and the line joining the element to the point, and inversely proportional to the square of the distance $r$ between the element and the point. Combining these, we get $dB \propto \frac{I dl \sin \theta}{r^2}$.

Key Concept: Magnetic Field at the Centre of a Circular Current-Carrying Loop

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion (A) correctly states the formula for the magnetic field at the center of a circular current-carrying loop, which is derived from integrating the contributions from all infinitesimal segments of the loop using Biot-Savart's law. The reason (R) correctly explains the direction of the magnetic field using the right-hand palm rule, which states that if the fingers curl in the direction of the current, the thumb points in the direction of the magnetic field at the center. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that the magnetic field inside a long solenoid is uniform and independent of its diameter. This is true because, for an ideal solenoid (infinitely long), the field inside is uniform and axial. The Reason correctly cites Ampere's circuital law, which gives the formula $B = \mu_0 n I$ for the magnetic field inside the solenoid. This formula shows that the field depends only on the current $I$ and the number of turns per unit length $n$, not on the diameter. Hence, the Reason is the correct explanation for the Assertion.

Key Concept: Force on a Moving Charge in a Uniform Magnetic Field

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a positively charged particle moving perpendicular to a uniform magnetic field experiences a force, which is true because the magnetic force formula is $F = Bqv \sin \theta$. When $\theta = 90^\circ$, $\sin \theta = 1$, so the force is maximum and non-zero. The reason correctly explains why the assertion is true by referring to the formula and the angle condition. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Force Between Parallel Current-Carrying Conductors, Direction of Force

b) $2 \times 10^{-4} \, \text{N/m}$
[Solution Description]
Given:
Current in wire 1, $I_1 = 5 \, \text{A}$,
Current in wire 2, $I_2 = 10 \, \text{A}$,
Separation, $r = 0.1 \, \text{m}$.
The force per unit length is given by:
$$ \frac{F}{l} = \frac{\mu_0 I_1 I_2}{2 \pi r} $$
Substituting the values ($\mu_0 = 4\pi \times 10^{-7} \, \text{Tm/A}$):
$$ \frac{F}{l} = \frac{(4\pi \times 10^{-7}) \times 5 \times 10}{2\pi \times 0.1} $$
Simplifying:
$$ \frac{F}{l} = \frac{2 \times 10^{-5}}{0.1} = 2 \times 10^{-4} \, \text{N/m} $$
Since the currents are in opposite directions, the force is repulsive.

Key Concept: Right-hand Thumb Rule, Magnetic Field Direction

b) Into the plane of the paper
[Solution Description]
Using the Right-hand Thumb Rule:
1. Point your right thumb upward (direction of the current).
2. Curl your fingers around the conductor.
3. The fingers curl in the direction of the magnetic field lines.
4. For a point to the east of the conductor, the field direction will be into the plane of the paper (\times) because the fingers curl clockwise when viewed from above.
Therefore, the correct answer is "Into the plane of the paper."

Key Concept: Magnetic Force, Work Done

c) The work done is always zero.
[Solution Description]
The magnetic force $\overrightarrow{F} = q (\overrightarrow{v} \times \overrightarrow{B})$ is always perpendicular to the velocity $\overrightarrow{v}$ of the particle. Since work done is given by $W = \overrightarrow{F} \cdot \overrightarrow{d}$ and $\overrightarrow{F}$ is perpendicular to displacement $\overrightarrow{d}$, the work done by the magnetic force is zero. This holds true for any angle between $\overrightarrow{v}$ and $\overrightarrow{B}$.

Key Concept: Force on a Moving Charge in Magnetic Field, Direction of Force

b) $3.0 \times 10^{-5}$ N
[Solution Description]
Given:
- Charge $q = 3 \times 10^{-6}$ C
- Velocity $\vec{v} = (4\mathbf{\hat{i}} + 3\mathbf{\hat{j}}) \times 10^{5}$ m/s
- Magnetic field $\vec{B} = 2 \mathbf{\hat{k}}$ T
The magnetic force is given by $\vec{F} = q(\vec{v} \times \vec{B})$.
First, compute the cross product $\vec{v} \times \vec{B} = (4\mathbf{\hat{i}} + 3\mathbf{\hat{j}}) \times 2\mathbf{\hat{k}} = 4 \times 2 (\mathbf{\hat{i}} \times \mathbf{\hat{k}}) + 3 \times 2 (\mathbf{\hat{j}} \times \mathbf{\hat{k}})$.
Using the right-hand rule:
\begin{itemize}

Key Concept: Vector Form of Biot-Savart Law

c) $\vec{dB} = \frac{\mu_0}{4\pi} \cdot \frac{I (\vec{dl} \times \vec{r})}{r^3}$
[Solution Description]
The vector form of the Biot-Savart law is derived using the cross product of the current element $I \vec{dl}$ and the unit vector $\hat{r}$ (or $\vec{r}$ scaled by $r^3$).
The formula is:
$\vec{dB} = \frac{\mu_0}{4\pi} \cdot \frac{I (\vec{dl} \times \vec{r})}{r^3}.$
This includes the permeability of free space $\mu_0$ and accounts for the direction of $\vec{dB}$ being perpendicular to both $\vec{dl}$ and $\vec{r}$.

Key Concept: Force Between Current-Carrying Conductors

d) $8 \times 10^{-4}$ N/m
[Solution Description]
The force per unit length between two parallel current-carrying conductors is given by:
$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi r}$
Substituting the values:
$\frac{F}{L} = \frac{(4\pi \times 10^{-7}\,\text{Tm/A})(10\,\text{A})(20\,\text{A})}{2 \pi (0.05\,\text{m})} = 8 \times 10^{-4}\,\text{N/m}$

Key Concept: Radius of Circular Path, Time Period of Charged Particle

c) It doubles
[Solution Description]
The radius of the circular path is $r = \frac{m v}{q B}$.
If the radius doubles ($r' = 2r$), then $2r = \frac{m v}{q B'}$ implies $B' = \frac{B}{2}$.
The time period is $T = \frac{2 \pi m}{q B}$.
For the new magnetic field $B' = \frac{B}{2}$, the time period becomes:
$T' = \frac{2 \pi m}{q \cdot \frac{B}{2}} = 2 \left( \frac{2 \pi m}{q B} \right) = 2 T$.
Thus, the time period also doubles.

Key Concept: Biot-Savart Law, Direction of Magnetic Field

d) Assertion is false, but Reason is true.
[Solution Description]
The Assertion states that the magnetic field at the center of a circular loop is maximum when the point lies in the plane of the loop. However, this is incorrect because the field at the center of a circular loop is always maximum regardless of the plane, as all current elements contribute constructively.
The Reason correctly states that the magnitude of the magnetic field depends on $\sin\theta$ according to Biot-Savart law, but this does not explain the Assertion since the field at the center of a loop is maximized due to symmetric contributions from all elements, not just $\sin\theta$.
Therefore, Assertion is false, but Reason is true.

Key Concept: Magnetic Field at the End of a Solenoid

b) $9.42 \times 10^{-4} \, \text{T}$
[Solution Description]
The magnetic field at the end of a long solenoid is half of that at its center:
$B_{\text{end}} = \frac{1}{2} \mu_0 n I_0$
Given: $n = 1000 \, \text{m}^{-1}$, $I_0 = 1.5 \, \text{A}$, $\mu_0 = 4\pi \times 10^{-7} \, \text{N A}^{-2}$. Substituting the values:
$B_{\text{end}} = \frac{1}{2} \times (4\pi \times 10^{-7}) \times 1000 \times 1.5$
Simplifying step-by-step:
$B_{\text{end}} = 2 \times 3.1416 \times 10^{-7} \times 1500$
$B_{\text{end}} = 9424.77 \times 10^{-7} \, \text{N A}^{-1} \, \text{m}^{-1} \approx 9.42 \times 10^{-4} \, \text{T}$
The magnetic field at the end of the solenoid is approximately $9.42 \times 10^{-4}$ Tesla.

Key Concept: Magnetic Force on a Moving Charge

a) $F = Bqv$
[Solution Description]
The force experienced by a moving charge in a magnetic field is given by $F = Bqv \sin \theta$. Since the velocity is perpendicular to the magnetic field ($\theta = 90^\circ$), $\sin \theta = 1$. Thus, $F = Bqv$.

Key Concept: Force Between Two Parallel Current-Carrying Conductors

a) Attractive
[Solution Description]
According to Ampere's law, two parallel current-carrying conductors attract each other when the currents are in the same direction and repel when they are in opposite directions. The force per unit length is given by $\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2 \pi d}$.

Key Concept: Magnetic Field due to Current-Carrying Conductor, Right-Hand Thumb Rule

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The magnetic field around a current-carrying wire indeed forms concentric circles, as demonstrated by Oersted's experiment. The direction of this field is given by the right-hand thumb rule, where if the thumb points in the direction of the current, the fingers curl in the direction of the magnetic field lines. Thus, Assertion (A) is true.
The Reason (R) states that the magnetic field is inversely proportional to the square of the distance from the current element according to Biot-Savart’s law. This is also correct because Biot-Savart’s law is expressed as:
$dB \propto \frac{I \, dl \, \sin \theta}{r^2}$
Here, $dB$ depends on $\frac{1}{r^2}$. However, the reason does not explain why the field forms concentric circles or the direction follows the right-hand thumb rule—it only describes the magnitude dependence on distance. Hence, both (A) and (R) are true, but (R) is NOT the correct explanation of (A).

Key Concept: Magnetic Field due to Current

c) The deflection increases
[Solution Description]
In Oersted's experiment, increasing the current in the wire increases the strength of the magnetic field produced around it. As a result, the deflection of the magnetic needle increases because the force acting on it becomes stronger.

Key Concept: Magnetic Field Direction, Biot-Savart Law

a) Perpendicular to the plane of the semicircle and outward
[Solution Description]
According to Biot-Savart law, for each small current element $Idl$ along the semicircular arc, the magnetic field at $O$ points perpendicular to both $dl$ and the position vector from $dl$ to $O$. Since all current elements are in the plane and symmetrical about the vertical axis, their horizontal components cancel out, leaving only the vertical component pointing out of the plane.

Key Concept: Biot-Savart Law, Solenoid

a) $\mu_0 n I$
[Solution Description]
For an ideal solenoid, the magnetic field inside is uniform and given by:
$B = \mu_0 n I$
This is derived using Ampere's law or integrating Biot-Savart contributions from all turns.

Key Concept: Direction of Magnetic Field

c) Parallel or antiparallel to the field
[Solution Description]
According to the definition, if $\theta = 0^\circ$, then $F = qvB \sin 0^\circ = 0$. This means the particle must move parallel or antiparallel to the magnetic field.

Key Concept: Magnetic Force on Moving Charge, Direction of Magnetic Field

b) $8 \times 10^{-14} \, \text{N}$
[Solution Description]
Given:
Charge of proton, $q = 1.6 \times 10^{-19} \, \text{C}$,
Velocity, $v = 2 \times 10^6 \, \text{m/s}$,
Magnetic field, $B = 0.5 \, \text{T}$,
Angle, $\theta = 30^\circ$.
The magnetic force is given by:
$$ F = qvB \sin \theta $$
Substituting the values:
$$ F = (1.6 \times 10^{-19}) \times (2 \times 10^6) \times 0.5 \times \sin 30^\circ $$
$$ \sin 30^\circ = 0.5 $$
$$ F = 1.6 \times 10^{-19} \times 2 \times 10^6 \times 0.5 \times 0.5 $$
$$ F = 8 \times 10^{-14} \, \text{N} $$

Key Concept: Direction of Magnetic Field

c) Parallel or antiparallel to the direction of motion
[Solution Description]
From the formula for magnetic force, $F = qvB \sin \theta$.
If there is no deflecting force, $F = 0$, which implies $\sin \theta = 0$.
This occurs when $\theta = 0^\circ$ or $\theta = 180^\circ$, meaning the particle's velocity is either parallel or antiparallel to the magnetic field.

Key Concept: Helical Motion, Lorentz Force

c) $\frac{\sqrt{2} \pi m v}{q B}$
[Solution Description]
The pitch $p$ of the helical path is given by $p = v \cos \theta \times T$, where $T = \frac{2 \pi m}{q B}$. Here, $\theta = 45^\circ$.
Step 1: Compute the time period $T = \frac{2 \pi m}{q B}$.
Step 2: Resolve the velocity into components. The parallel component is $v_{\parallel} = v \cos 45^\circ = \frac{v}{\sqrt{2}}$.
Step 3: Calculate pitch $p = v_{\parallel} \times T = \frac{v}{\sqrt{2}} \times \frac{2 \pi m}{q B} = \frac{\sqrt{2} \pi m v}{q B}$.

Key Concept: Biot-Savart Law, Magnetic Field Calculation

c) Four times the original field
[Solution Description]
The magnetic field at the center of a circular loop is given by:
$B = \frac{\mu_0 I}{2R}$
If the current is doubled ($I' = 2I$) and the radius is halved ($R' = R/2$), the new magnetic field $B'$ is:
$B' = \frac{\mu_0 (2I)}{2(R/2)} = \frac{\mu_0 I}{R/2} = \frac{2\mu_0 I}{R}$
Comparing with the original field $B = \frac{\mu_0 I}{2R}$, the new field is 4 times the original field.

Key Concept: Magnetic Force on a Moving Charge

b) $0.03$ N
[Solution Description]
The force acting on the moving charge in a magnetic field is given by:
$F = Bqv \sin \theta$
Given:
- $q = 2 \times 10^{-6}$ C,
- $v = 3 \times 10^4$ m/s,
- $B = 0.5$ T,
- $\theta = 90^\circ$ (since $\overrightarrow{v}$ is perpendicular to $\overrightarrow{B}$).
Substituting these values:
$F = (0.5)(2 \times 10^{-6})(3 \times 10^4) \sin 90^\circ = 0.5 \times 2 \times 3 \times 10^{-2} \times 1 = 3 \times 10^{-2} N = 0.03 N$

Key Concept: Direction of Magnetic Force

b) Toward the left
[Solution Description]
Using the right-hand rule for negative charges like electrons, point the thumb opposite to the direction of motion (downward), fingers in the direction of the magnetic field (into the plane). The palm pushes in the direction of the force, which is toward the left.

Key Concept: Ampere's Circuital Law, Toroid Magnetic Field

b) $6 \times 10^{-3} \text{ T}$
[Solution Description]
For a toroid, the magnetic field inside is given by:
$B = \frac{\mu_0 N I}{2 \pi r},$
where $N$ is the total number of turns and $r$ is the average radius. Substituting $\mu_0 = 4\pi \times 10^{-7} \text{ T·m/A}$, $N = 500$, $I = 3 \text{ A}$, and $r = 0.05 \text{ m}$:
$B = \frac{(4\pi \times 10^{-7}) \times 500 \times 3}{2 \pi \times 0.05} = \frac{6000 \pi \times 10^{-7}}{0.1 \pi} = 6 \times 10^{-3} \text{ T}.$

Key Concept: Magnetic Field due to a Long Straight Wire

a) $5 \times 10^{-7} \, \text{T}$
[Solution Description]
The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $I$ is given by:
$B = \frac{\mu_0 I}{2 \pi r}$
Here, $\mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1}$, $I = 5 \, \text{A}$, and $r = 2 \, \text{m}$.
Substituting the values:
$B = \frac{(4\pi \times 10^{-7}) \times 5}{2 \pi \times 2} = \frac{20\pi \times 10^{-7}}{4\pi} = 5 \times 10^{-7} \, \text{T}$

Key Concept: Helical Motion

b) Perpendicular component ($v_{\perp}$)
[Solution Description]
In helical motion, the velocity $\vec{v}$ has two components: parallel ($v_{\parallel} = v \cos \theta$) and perpendicular ($v_{\perp} = v \sin \theta$) to the magnetic field $\vec{B}$. The perpendicular component $v_{\perp}$ causes circular motion, and the radius $r$ is given by $r = \frac{m v_{\perp}}{q B}$. Thus, the perpendicular component determines the radius.

Key Concept: Force Between Parallel Wires

b) $5 \times 10^{-4}\,\text{N/m}$
[Solution Description]
The force per unit length between two parallel wires is given by:
$\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2 \pi d}$
Substituting the values:
$I_1 = 5\,\text{A}$,
$I_2 = 10\,\text{A}$,
$d = 0.02\,\text{m}$,
$\mu_0 = 4\pi \times 10^{-7}\,\text{T}\cdot\text{m/A}$.
So,
\begin{equation*}
\frac{F}{l} = \frac{(4\pi \times 10^{-7}) \times 5 \times 10}{2 \pi \times 0.02} = 5 \times 10^{-4}\,\text{N/m}.
\end{equation*}

Key Concept: Force between parallel conductors, Definition of Ampere

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the force per unit length between two infinitely long parallel conductors carrying 1 A current each and placed 1 m apart in vacuum is $2 \times 10^{-7}$ N/m. This is correct as per the definition of the ampere. The reason states that this experimental observation led to the definition of the ampere, which is also true. Furthermore, the reason correctly explains why the assertion holds because the ampere is defined based on this specific experimental setup and observed force value. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Circular Motion of Charged Particles in a Magnetic Field

a) Becomes half of the original radius
[Solution Description]
The radius $r$ of the circular path of a charged particle moving perpendicular to a magnetic field is given by:
$r = \frac{mv}{qB}$
Here, if $B$ is doubled ($B' = 2B$), the new radius $r'$ becomes:
$r' = \frac{mv}{q(2B)} = \frac{r}{2}$
Thus, the radius is halved.

Key Concept: Motion of Charged Particles in a Uniform Magnetic Field

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The magnetic force $\vec{F} = q (\vec{v} \times \vec{B})$ is always perpendicular to the velocity $\vec{v}$ of the charged particle, as per the cross product definition. Since work done by a force is given by $W = \vec{F} \cdot \vec{d}$, and $\vec{F}$ is perpendicular to $\vec{v}$, no work is done on the particle. Therefore, the kinetic energy of the particle remains unchanged. Thus, both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Key Concept: Magnetic field at axis, Direction of field

b) 4:1
[Solution Description]
The magnetic field at a distance $x$ from the center is given by:
$B = \frac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}}$
For $x = 2$ m:
$B_1 = \frac{\mu_0 \times 5 \times 2^2}{2(2^2 + 2^2)^{3/2}} = \frac{20\mu_0}{2(8)^{3/2}} = \frac{10\mu_0}{22.627} = 0.442\mu_0$
For $x = 4$ m:
$B_2 = \frac{\mu_0 \times 5 \times 2^2}{2(2^2 + 4^2)^{3/2}} = \frac{20\mu_0}{2(20)^{3/2}} = \frac{10\mu_0}{89.443} = 0.112\mu_0$
Ratio $B_1:B_2 = 0.442:0.112 = 3.95:1 \approx 4:1$

Key Concept: Pitch of Helical Path

d) $\frac{2 \pi m v \cos\theta}{qB}$
[Solution Description]
The pitch of the helical path is the distance travelled along the direction of the magnetic field in one complete revolution. It is given by $p = v \cos\theta \times T$, where $T = \frac{2 \pi m}{qB}$ is the time period. Thus, $p = \frac{2 \pi m v \cos\theta}{qB}$.

Key Concept: Concept of Magnetic Field

c) In the direction from south to north pole of the magnet
[Solution Description]
The concept states that a freely-suspended magnetic needle aligns itself in the direction of the magnetic field of the bar magnet, pointing from the south pole to the north pole of the magnet.

Key Concept: Motion of charged particles in a magnetic field

c) $\frac{mv}{eB}$
[Solution Description]
The radius of the circular path of a charged particle moving perpendicular to a magnetic field is given by:
$r = \frac{mv}{qB}$
For an electron, $q = e$. Substituting the values:
$r = \frac{mv}{eB}$

Key Concept: Comparison of Coulomb’s Law and Biot-Savart’s Law

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the magnetic field due to a current element is maximum when the angle $\theta = 90^\circ$. This is true because, from Biot-Savart’s law, the magnitude of the magnetic field is given by $dB = \frac{\mu_0}{4\pi} \cdot \frac{I dl \sin \theta}{r^2}$. The term $\sin \theta$ attains its maximum value of 1 when $\theta = 90^\circ$, making the magnetic field maximum at this angle. The reason correctly explains the assertion by referring to the dependence of the magnetic field on $\sin \theta$ as per Biot-Savart’s law. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Force on Charged Particle

c) $qvB \sin\theta$
[Solution Description]
The force on a charged particle moving in a magnetic field is given by $F = qvB \sin\theta$. When $\theta = 90^\circ$, $\sin\theta = 1$, resulting in maximum force. For any other angle, the force depends on $\sin\theta$.

Key Concept: Ampere’s Circuital Law

d) Assertion is false, but Reason is true.
[Solution Description]
The Assertion claims that the line integral of the magnetic field around any closed path is always zero, which is false. According to Ampere’s circuital law, the line integral of the magnetic field $\oint \vec{B} \cdot d\vec{l}$ equals $\mu_0$ times the net current threading through the area enclosed by the path. Therefore, it is not always zero but depends on the enclosed current. The Reason correctly states Ampere’s circuital law formula and its condition.

Key Concept: Magnetic Force on Moving Charge

c) $7.2 \times 10^{-13}$ N
[Solution Description]
The magnetic force on a moving charge is given by $F = Bqv\sin\theta$. Here, $\theta = 90^\circ$, so $\sin\theta = 1$. Substituting the values:
$F = (1.5\,\text{T})(1.6 \times 10^{-19}\,\text{C})(3 \times 10^6\,\text{m/s}) = 7.2 \times 10^{-13}\,\text{N}$

Key Concept: Right-hand Palm Rule No. 1

c) Direction of magnetic field
[Solution Description]
The right-hand palm rule No. 1 states that if the thumb points in the direction of the current ($I$) and the fingers point towards the point where the magnetic field ($\vec{B}$) is to be determined, then the perpendicular drawn from the palm gives the direction of $\vec{B}$.

Key Concept: Ampere’s Circuital Law Definition

c) $\mu_0$ times the net current threading the loop
[Solution Description]
According to Ampere's circuital law, the line integral of the magnetic field $\vec{B}$ around any closed path is given by $\oint \vec{B} \cdot d\vec{l} = \mu_0 I$, where $\mu_0$ is the permeability of free space and $I$ is the current enclosed by the loop.

Key Concept: Magnetic Field for a Coil with N Turns

d) $\frac{\mu_0 N I}{2a}$
[Solution Description]
The magnetic field at the center of a coil with $N$ turns is increased by a factor of $N$ compared to a single loop. The formula is:
$B = \frac{\mu_0 N I}{2a}$
Thus, the field is directly proportional to the number of turns $N$.

Key Concept: Magnetic Force on Moving Charge

c) $1.6 \times 10^{-13} \, \text{N}$
[Solution Description]
The magnetic force on a moving charge is given by $F = qvB \sin\theta$. Here, $q = 1.6 \times 10^{-19} \, \text{C}$, $v = 2.0 \times 10^6 \, \text{m/s}$, $B = 0.5 \, \text{T}$, and $\theta = 90^\circ$ (since the velocity is perpendicular to the field). Substituting these values:
$F = (1.6 \times 10^{-19})(2.0 \times 10^6)(0.5)\sin 90^\circ = 1.6 \times 10^{-13} \, \text{N}$

Key Concept: Ampere's Law, Solenoid

b) 1.51 T
[Solution Description]
The magnetic field inside a solenoid with core material is given by:
$B = \mu_0 \mu_r n I$
Where $\mu_r$ is relative permeability, $n$ is turns per unit length, and $I$ is current.
Given: $\mu_0 = 4\pi \times 10^{-7}\,\text{T m/A}$, $\mu_r = 1200$, $n = 500\,\text{turns/m}$, $I = 2\,\text{A}$.
Substituting values:
$B = (4\pi \times 10^{-7})(1200)(500)(2) = 4.8\pi \times 10^{-1}\,\text{T} = 1.508\,\text{T}$

Key Concept: Magnetic Force Calculation

c) 0.16 N
[Solution Description]
The formula for the magnetic force on a current-carrying conductor is:
$F = I B l \sin\theta$
Here, $\theta = 90^\circ$, so $\sin\theta = 1$.
Given:
$I = 8\,\text{A}$,
$B = 0.4\,\text{T}$,
$l = 0.05\,\text{m}$.
Substituting the values:
$F = 8 \times 0.4 \times 0.05 \times 1 = 0.16\,\text{N}$.

Key Concept: Definition of Ampere

b) Current producing $2 \times 10^{-7} \, \text{N/m}$ force at 1 m separation
[Solution Description]
One ampere is defined as the current which, when flowing through two infinitely long, parallel conductors placed 1 metre apart in vacuum, produces a force per unit length of $2 \times 10^{-7} \, \text{N/m}$ between them.

Key Concept: Concept: Inverse-square nature

c) Both are inverse-square laws but for different types of fields.
[Solution Description]
Both Coulomb’s Law and Biot-Savart’s Law are inverse-square laws, meaning their magnitudes vary with the inverse square of the distance (\$r\$). However, they describe different fields: Coulomb's Law describes the electric field due to charges, while Biot-Savart’s Law describes the magnetic field due to current elements.

Key Concept: Magnetic Field of a Long Straight Wire

c) $4 \times 10^{-5}$ T
[Solution Description]
Using Ampere's circuital law for a long straight wire, the magnetic field $B$ at a distance $r$ from the wire is given by:
$B = \frac{\mu_0 I}{2\pi r}$
Substituting the given values:
$B = \frac{4\pi \times 10^{-7} \times 8}{2\pi \times 0.04} = \frac{32\pi \times 10^{-7}}{0.08\pi} = 4 \times 10^{-5} \, \text{T}$

Key Concept: Force Calculation

c) $6 \times 10^{-6} \text{N/m}$
[Solution Description]
Using the formula for force per unit length:
$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi r}$
Substitute $I_1 = 3 \text{A}$, $I_2 = 5 \text{A}$, $r = 0.5 \text{m}$, and $\mu_0 = 4\pi \times 10^{-7} \text{N/A}^2$:
$\frac{F}{L} = \frac{(4\pi \times 10^{-7})(3)(5)}{2 \pi (0.5)} = \frac{60\pi \times 10^{-7}}{\pi} = 6 \times 10^{-6} \text{N/m}$

Key Concept: Straight Conductor, Current Element

b) $3.2 \times 10^{-5}\,\text{T}$
[Solution Description]
The magnetic field due to each wire at midpoint is calculated using:
$B = \frac{\mu_0 I}{2\pi r}$
For first wire (3 A): distance = 5 cm = 0.05 m
$B_1 = \frac{(4\pi \times 10^{-7})(3)}{2\pi (0.05)} = 1.2 \times 10^{-5}\,\text{T}$
For second wire (5 A):
$B_2 = \frac{(4\pi \times 10^{-7})(5)}{2\pi (0.05)} = 2.0 \times 10^{-5}\,\text{T}$
Since currents are in opposite directions, fields add up:
$B_{total} = B_1 + B_2 = 3.2 \times 10^{-5}\,\text{T}$

Key Concept: Magnetic Flux Calculation

b) $0.0157 \, \text{Wb}$
[Solution Description]
The magnetic flux $\phi$ through a loop is given by $\phi = BA\cos\theta$, where $B$ is the magnetic field, $A$ is the area, and $\theta$ is the angle between $B$ and the normal to the plane of the loop.
Here, $B = 0.5$ T, $\theta = 0^\circ$ (perpendicular), and the area $A = \pi r^2 = \pi (0.1)^2 = 0.0314 \, \text{m}^2$.
So, $\phi = 0.5 \times 0.0314 \times \cos 0^\circ = 0.0157 \, \text{Wb}$

Key Concept: Resonance Condition

b) The frequency of the applied potential difference must equal the particle's revolution frequency.
[Solution Description]
For efficient operation, the frequency of the applied alternating potential difference (\$n_0\$) must match the cyclotron frequency (\$\nu\$) of the charged particle, given by \$\nu = \frac{qB}{2\pi m}\$, where \$B\$ is the magnetic field and \$m\$ is the mass of the particle.

Key Concept: Units and Dimensions

c) Newton/(ampere-metre)
[Solution Description]
The SI unit of magnetic field is tesla (T), which is equivalent to newton per ampere-meter ($\text{N/(A}\cdot\text{m)}$) or weber per square meter ($\text{Wb/m}^2$).

Key Concept: Field Due to a Current in a Solenoid

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the magnetic field inside a long solenoid is uniform and parallel to its axis when current flows through it, which is true as per the theory of solenoids. The reason gives the correct formula for the magnetic field inside a long solenoid, which is also true. Moreover, the reason correctly explains why the field is uniform and axial because the formula shows that the field depends only on $\mu_0$, $n$, and $I_0$, making it constant and uniformly distributed along the axis. Thus, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Definition of Ampere

c) The current producing a force of $2 \times 10^{-7} \text{N/m}$ at 1 m distance
[Solution Description]
The ampere is defined as the constant current that, if maintained in two straight parallel conductors of infinite length, placed 1 meter apart in a vacuum, would produce a force per unit length of $2 \times 10^{-7} \text{N/m}$ between them.

Key Concept: Magnetic Field Calculation, Superposition

b) $1 \times 10^{-5} \, T$ out of the plane
[Solution Description]
Using the formula for the magnetic field due to an infinite straight conductor: $B = \frac{\mu_0 I}{2 \pi r}$. For wire 1 ($I_1 = 5A$), $B_1 = \frac{4 \pi \times 10^{-7} \times 5}{2 \pi \times 0.10} = 1 \times 10^{-5} \, T$ (direction into the plane). For wire 2 ($I_2 = 10A$), $B_2 = \frac{4 \pi \times 10^{-7} \times 10}{2 \pi \times 0.10} = 2 \times 10^{-5} \, T$ (direction out of the plane). Since the fields are in opposite directions, the net field is $B_{\text{net}} = B_2 - B_1 = 1 \times 10^{-5} \, T$ out of the plane.

Key Concept: Ampere's Circuital Law - Magnetic Field of a Long Straight Wire

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion (A) states the correct formula for the magnetic field around a long straight wire, which is derived using Ampere’s circuital law due to the symmetry of the problem.
The reason (R) correctly explains why Ampere’s law can be applied here—because the symmetry allows the simplification of the line integral $\oint \overrightarrow{B} \cdot \overrightarrow{dl}$ to $B \cdot 2\pi r$.
Therefore, both (A) and (R) are true, and (R) correctly explains (A).

Key Concept: Field Due to a Current in a Circular Loop, Direction of Magnetic Field

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that doubling the number of loops doubles the magnetic field at the center if currents flow in the same direction. This is correct because for $N$ loops, the magnetic field at the center is $B = \frac{\mu_0 N I}{2a}$. Thus, for $N = 2$, the field becomes twice that of a single loop ($B_{\text{total}} = 2 \times \frac{\mu_0 I}{2a} = \frac{\mu_0 I}{a}$).
The reason correctly provides the formula for the magnetic field at the center of a single loop and mentions that fields add vectorially when multiple loops are present. Since both statements are true and the reason explains the assertion, the correct choice is:

Key Concept: Magnetic field inside a solenoid

b) $2.51 \, \text{T}$
[Solution Description]
The magnetic field inside a solenoid is given by $B = \mu_0 \mu_r n I_0$, where $\mu_0$ is the permeability of free space, $\mu_r$ is the relative permeability, $n$ is the number of turns per unit length, and $I_0$ is the current. Given: $n = 800 \, \text{m}^{-1}$, $I_0 = 5 \, \text{A}$, $\mu_r = 500$. Substituting these values:
$B = (4 \pi \times 10^{-7} \, \text{T m A}^{-1}) \times 500 \times 800 \, \text{m}^{-1} \times 5 \, \text{A} = 2.51 \, \text{T}.$

Key Concept: Right-hand Thumb Rule

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states the correct application of the right-hand thumb rule, where the thumb indicates the current direction and the curled fingers show the magnetic field direction. The reason correctly identifies the purpose of the right-hand thumb rule, which is to find the direction of the magnetic field due to a current-carrying conductor. Moreover, the reason directly explains why the assertion is true, as the rule connects the current direction to the magnetic field direction.