Class 12 Physics (Part 1) Chapter 9 Magnetic Field and Earth’s Magnetism

Key Concept: Magnetic Field on Equatorial Line

b) $9.85 \times 10^{-7} \, \text{T}$
[Solution Description]
The formula for the magnetic field on the equatorial line is:
$B = \frac{\mu_0}{4\pi} \cdot \frac{m}{(r^2 + l^2)^{3/2}}$
Given, $m = 10 \, \text{A m}^2$, $2l = 0.2 \, \text{m} \Rightarrow l = 0.1 \, \text{m}$, $r = 1 \, \text{m}$. Substituting the values:
$B = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{10}{(1^2 + 0.1^2)^{3/2}}$
Simplifying:
$B = 10^{-7} \cdot \frac{10}{(1 + 0.01)^{3/2}} = 10^{-7} \cdot \frac{10}{(1.01)^{1.5}}$
Calculating $(1.01)^{1.5} \approx 1.015$:
$B \approx 10^{-7} \cdot \frac{10}{1.015} \approx 9.85 \times 10^{-7} \, \text{T}$

Key Concept: Earth's Magnetic Field

d) Vertical downwards.
[Solution Description]
Earth behaves like a magnet with its south magnetic pole near the geographic north. Thus, the magnetic field lines emerge from the south magnetic pole (geographic north) and enter the north magnetic pole (geographic south). Near the poles, the field lines are vertical.

Key Concept: Magnetic Needle Behavior

a) North pole
[Solution Description]
In the northern hemisphere, the north pole of a freely suspended magnetic needle inclines downward due to Earth's magnetic field.

Key Concept: Magnetic Lines of Force, Earth's Magnetic Field

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the properties of Earth's magnetic field, the Earth acts like a dipole magnet with its magnetic south pole near the geographic north pole and vice versa. This causes the magnetic lines of force to appear as if they emerge from the magnetic north pole (geographic south region) and enter the magnetic south pole (geographic north region). Thus, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Magnetic Dipole Moment Calculation

b) $\frac{e v r}{2}$
[Solution Description]
The magnetic dipole moment ($m$) of a revolving electron is given by:
$m = I A$
where $I$ is the current due to the revolving electron and $A = \pi r^2$ is the area of the orbit.
Current $I$ is calculated as:
$I = \frac{e}{T} = \frac{e v}{2 \pi r}$
Substituting $I$ and $A$ into the formula for $m$:
$m = \left( \frac{e v}{2 \pi r} \right) \pi r^2 = \frac{e v r}{2}$

Key Concept: Magnetic Equator

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that at the magnetic equator, a freely suspended magnetic needle remains horizontal. This is true because the angle of dip ($\theta$) is zero here, meaning there is no vertical component of the magnetic field ($B_V = 0$), so the needle aligns horizontally with the horizontal component ($B_H$). The reason correctly explains this by stating that the vertical component of Earth's magnetic field is zero at the magnetic equator. Thus, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Bohr Magneton

c) $9.27 \times 10^{-24} \, A \cdot m^2$
[Solution Description]
The Bohr magneton represents the minimum magnetic dipole moment of an electron in the first Bohr orbit ($n = 1$). Its value is a constant and is given by:
$1 \text{ Bohr magneton} = 9.27 \times 10^{-24} \, A \cdot m^2$

Key Concept: Maximum Torque Condition

c) When $\theta = 90^\circ$
[Solution Description]
The torque $\tau$ on a magnetic dipole is maximum when the angle $\theta$ between the dipole moment $\vec{m}$ and the magnetic field $\vec{B}$ is $90^\circ$. At this angle, $\sin \theta = 1$, so:
$\tau_{\text{max}} = m B$

Key Concept: Magnetic Moment of Revolving Electron

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states the correct formula for the magnetic moment $\vec{m}$ of a revolving electron, which is derived from the gyromagnetic ratio $\frac{m}{J} = \frac{e}{2 m_e}$. The reason explains the physical significance of the negative sign, indicating that $\vec{m}$ and $\vec{J}$ are oppositely directed. Both statements are true, and the reason correctly explains the assertion.

Key Concept: Magnetic field in broad-side-on position

c) $\frac{\mu_0}{4\pi} \cdot \frac{m}{r^3}$
[Solution Description]
The magnetic field $\vec{B}$ in the broad-side-on position for a small bar magnet ($l << r$) is given by:
$B = \frac{\mu_0}{4\pi} \cdot \frac{m}{r^3}$
The field direction is from the north pole to the south pole.

Key Concept: Magnetic Field of a Dipole, Earth's Magnetic Field

c) $B_H - \frac{\mu_0}{4\pi} \cdot \frac{m}{r^3}$
[Solution Description]
The magnetic field due to the bar magnet at an equatorial point is $\frac{\mu_0}{4\pi} \cdot \frac{m}{r^3}$ directed opposite to Earth's horizontal field $B_H$.
Therefore, the net magnetic field is:
$B_{\text{net}} = B_H - \frac{\mu_0}{4\pi} \cdot \frac{m}{r^3}.$

Key Concept: Axial and Equatorial Fields, Magnetic Moment

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that the axial magnetic field varies as $\frac{1}{r^3}$. This is correct as per the formula:
$B_{\text{axial}} = \frac{\mu_0}{4\pi} \cdot \frac{2m}{r^3}$
which clearly shows the $\frac{1}{r^3}$ dependence.
The reason states that the equatorial field also varies as $\frac{1}{r^3}$. This is correct as per the formula:
$B_{\text{eq}} = \frac{\mu_0}{4\pi} \cdot \frac{m}{r^3}$
However, the reason does not explain why the axial field is twice the equatorial field, which is what the assertion is highlighting. Thus, both are true but the reason does not correctly explain the assertion.

Key Concept: Magnetic equator, Magnetic poles

c) Remains unchanged
[Solution Description]
The magnetic field at the equator depends on the dipole moment and the Earth's radius, but not directly on the tilt angle of the magnetic axis. However, if the magnetic poles shift, the position of the magnetic equator also changes relative to the geographic equator. Since the problem states the magnetic moment remains constant, the equatorial magnetic field strength at the new magnetic equator remains the same as before:
$B_{eq} = \frac{\mu_0}{4\pi} \frac{m}{r^3} = 0.3 \, \text{G}$

Key Concept: Magnetic Declination, Angle of Dip

c) 63.43°
[Solution Description]
Given:
True angle of dip ($\theta$) = 60°
Angle between the plane and magnetic meridian ($\alpha$) = 30°
The relationship between the true angle of dip and the apparent angle of dip ($\theta'$) is given by:
$\tan \theta' = \frac{\tan \theta}{\cos \alpha}$
Substituting the values:
$\tan \theta' = \frac{\tan 60^\circ}{\cos 30^\circ}$
Calculating:
$\tan 60^\circ = \sqrt{3}$
$\cos 30^\circ = \frac{\sqrt{3}}{2}$
Therefore:
$\tan \theta' = \frac{\sqrt{3}}{\frac{\sqrt{3}}{2}} = 2$
Taking arctan on both sides:
$\theta' = \arctan(2) \approx 63.43^\circ$
So, the apparent angle of dip is approximately 63.43°.

Key Concept: Earth's Magnetic Field, Dipole Model, Field Strength

b) 0.3 G
[Solution Description]
The magnetic field strength $B$ at the geomagnetic equator due to the Earth's dipole can be calculated using the formula:
$B = \frac{\mu_0 m}{4 \pi r^3}$
Substituting the given values ($\mu_0 = 4\pi \times 10^{-7} \, \text{N A}^{-2}$, $m = 8 \times 10^{22} \, \text{A m}^2$, and $r = 6.4 \times 10^6 \, \text{m}$):
$B = \frac{(4\pi \times 10^{-7}) \times (8 \times 10^{22})}{4 \pi (6.4 \times 10^6)^3}$
Simplifying:
$B = \frac{8 \times 10^{15}}{(6.4)^3 \times 10^{18}}$
Calculating $(6.4)^3 = 262.144$:
$B = \frac{8 \times 10^{15}}{262.144 \times 10^{18}}$
$B \approx 0.0305 \times 10^{-3} \, \text{T} = 30.5 \, \mu \text{T}$
The closest match among the options is approximately $0.3 \, \text{G}$ (since $1 \, \text{G} = 10^{-4} \, \text{T}$).

Key Concept: Magnetic Field Lines, Torque on Magnetic Dipole

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The torque $\tau$ on a magnetic dipole in a uniform magnetic field is given by $\tau = mB \sin\theta$. When $\theta = 90^\circ$, $\sin\theta = 1$, which is the maximum value of $\sin\theta$. Therefore, the torque is maximum at $\theta = 90^\circ$. This confirms that both Assertion (A) and Reason (R) are true. Additionally, Reason (R) correctly explains Assertion (A), as it provides the mathematical expression that shows why the torque is maximum at $90^\circ$.

Key Concept: Direction of Magnetic Dipole Moment

c) Curled-straight right-hand rule
[Solution Description]
The direction of the magnetic dipole moment $\vec{m}$ of a current loop is determined by the curled-straight right-hand rule: If you curl the fingers of your right hand in the direction of the current, your thumb points in the direction of $\vec{m}$.

Key Concept: Magnetic Axis and Field

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that the magnetic field at the magnetic equator of Earth is zero, which is false. The magnetic field at the equator is not zero; rather, it is directed horizontally (parallel to the Earth's surface) with a magnitude around 0.3 G. The reason correctly states that at the magnetic equator, the Earth's magnetic field lines are parallel to its surface, which is true based on the syllabus description. Thus, the assertion is false while the reason is true.

Key Concept: Magnetic Dipole Moment of Electron, Gyromagnetic Ratio

a) $2.79 \times 10^{-23} \text{ A-m}^2$
[Solution Description]
Step 1: Recall the gyromagnetic ratio for an electron:
$\frac{m}{J} = \frac{e}{2m_e}$, where $J$ is the angular momentum.
Step 2: Rearrange to find the magnetic dipole moment:
$m = J \left(\frac{e}{2m_e}\right)$.
The term $\frac{e}{2m_e}$ equals $\frac{1 \text{ Bohr magneton}}{J_{\text{min}}}$, where $J_{\text{min}} = 1.05 \times 10^{-34} \text{ kg-m}^2/\text{s}$ (minimum angular momentum for $n = 1$).
Step 3: Calculate the ratio:
$\frac{m}{J} = \frac{9.27 \times 10^{-24}}{1.05 \times 10^{-34}} = 8.83 \times 10^{10} \text{ A-m}^2/\text{(kg-m}^2/\text{s)}$.
Step 4: Now, substitute $J = 3.16 \times 10^{-34} \text{ kg-m}^2/\text{s}$:
$m = 3.16 \times 10^{-34} \times 8.83 \times 10^{10} = 2.79 \times 10^{-23} \text{ A-m}^2$.

Key Concept: Magnetic Moment and Angular Momentum

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
From the syllabus, the magnetic moment $\vec{m}$ of a revolving electron is related to its angular momentum $\vec{J}$ by the equation:
$\vec{m} = -\left( \frac{e}{2 m_e} \right) \vec{J}$
The negative sign shows that $\vec{m}$ is indeed opposite to $\vec{J}$. The reason correctly explains the assertion using the gyromagnetic ratio formula, which includes the negative sign indicating the opposite direction. Thus, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Polarity determination, End rule

a) North pole
[Solution Description]
Using the end rule: If the current appears to flow anticlockwise when viewed from one end, that end behaves as the north pole of an equivalent bar magnet. Since the current is counter-clockwise here, the observed end is the north pole.

Key Concept: Magnetic Dipole Moment, Earth's Magnetic Field

c) $1.5B$
[Solution Description]
The original magnetic field is $B = \frac{\mu_0}{4\pi} \frac{m}{r^3}$. If the dipole moment increases by 50\%, the new dipole moment becomes $1.5m$. Substituting this into the formula, the new magnetic field $B'$ is:
$B' = \frac{\mu_0}{4\pi} \frac{1.5m}{r^3} = 1.5 \times \left(\frac{\mu_0}{4\pi} \frac{m}{r^3}\right) = 1.5B$
Therefore, the magnetic field increases by 50\%.

Key Concept: Magnetic Poles, Angle of Declination

b) $0^\circ$
[Solution Description]
At the new location, the magnetic north pole aligns with the geographic north pole. This means there is no angle between the magnetic meridian and the geographical meridian. Hence, the angle of declination becomes $0^\circ$.

Key Concept: Magnetic Field - Broad-side-on Position

a) $\frac{\mu_0}{4\pi} \cdot \frac{m}{r^3}$
[Solution Description]
For a small bar magnet in the broad-side-on position, when $l << r$, the magnetic field is given by:
$B = \frac{\mu_0}{4\pi} \cdot \frac{m}{r^3}$
This formula applies for points on the equatorial line of the magnet.

Key Concept: Magnetic Dipole Moment of Current Loop

b) $0.0628 \, \text{A m}^2$
[Solution Description]
The magnetic dipole moment $m$ for a current loop is given by $m = I A$, where $I$ is the current and $A$ is the area of the loop.
Step 1: Calculate the area of the circular loop:
$A = \pi r^2 = \pi (0.1)^2 = 0.0314 \, \text{m}^2$
Step 2: Multiply the current with the area to find the magnetic dipole moment:
$m = I A = 2 \times 0.0314 = 0.0628 \, \text{A m}^2$
Thus, the magnetic dipole moment is $0.0628 \, \text{A m}^2$.

Key Concept: Magnetic Lines of Force, Earth’s Magnetic Field

b) 1:1
[Solution Description]
The compass needle aligns itself along the resultant magnetic field. Let $B_E$ be Earth's horizontal magnetic field and $B_M$ be the magnetic field due to the bar magnet at that point. Since the two fields are perpendicular, the angle $\theta$ made by the resultant field with $B_E$ is given by $\tan\theta = \frac{B_M}{B_E}$. Given $\theta = 45^\circ$, we have $\tan 45^\circ = \frac{B_M}{B_E} \Rightarrow 1 = \frac{B_M}{B_E} \Rightarrow B_M : B_E = 1:1$.

Key Concept: Electric-magnetic similarities, Dipole moments

b) Their potential energy depends on cosine of the angle to the field
[Solution Description]
Both electric and magnetic dipoles experience torque when placed in external fields (electric field for electric dipoles, magnetic field for magnetic dipoles). Their potential energies follow similar mathematical forms, being minimum when aligned with the field and maximum when anti-aligned, though with different physical interpretations.

Key Concept: Magnetic Declination, Earth's Magnetic Field

b) The compass needle points $20^\circ$ west of geographic north.
[Solution Description]
The angle of declination ($\alpha$) is the angle between the magnetic meridian (compass needle direction) and the geographic meridian (true north direction).
- If the declination is **west**, the compass needle points **west** of the geographic north.
- If the declination is **east**, the needle points east of the geographic north.
Here, $\alpha = 20^\circ$ west means the compass needle points $20^\circ$ west of the geographic north.

Key Concept: Horizontal Component, Vertical Component

b) 0.35 G
[Solution Description]
Given:
Angle of dip ($\theta$) = 45°
Horizontal component ($B_H$) = 0.25 G
The relationship between the horizontal component and the total Earth's magnetic field ($B_E$) is:
$B_H = B_E \cos \theta$
Rearranging to find $B_E$:
$B_E = \frac{B_H}{\cos \theta}$
Substituting the values:
$B_E = \frac{0.25}{\cos 45^\circ}$
Calculating:
$\cos 45^\circ = \frac{1}{\sqrt{2}}$
Therefore:
$B_E = 0.25 \times \sqrt{2} \approx 0.35 \text{ G}$
So, the total Earth's magnetic field is approximately 0.35 G.

Key Concept: Direction of Magnetic Lines of Force

b) Away from the north pole of the magnet
[Solution Description]
The magnetic lines of force emerge from the north pole and enter the south pole of the magnet. At any point, the compass needle aligns itself along the tangent to the magnetic line of force at that point. Hence, near the north pole of the magnet, the compass needle will point away from the north pole (toward the south pole of the magnet).

Key Concept: Earth's Magnetic Field

d) $90^\circ$
[Solution Description] The angle of dip is the angle between the earth's magnetic field and the horizontal plane. At the magnetic poles, the magnetic field lines are vertical, so the angle of dip is $90^\circ$.

Key Concept: Magnetic Field Lines, Angle of Dip

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The Assertion (A) is true because magnetic field lines inside a bar magnet form closed loops from the south pole to the north pole. This property is consistent with the behavior of a magnetic dipole.
The Reason (R) is also true because, at the Earth's equator, the angle of dip ($\theta$) is indeed zero degrees, and the Earth's magnetic field is purely horizontal. However, this fact does not explain why the magnetic field lines inside a bar magnet run from the south pole to the north pole. Thus, the Reason is not the correct explanation of the Assertion.

Key Concept: Magnetic Equator, Earth's Magnetic Poles

c) It becomes erratic or points vertically
[Solution Description]
At the earth's magnetic poles, the earth's magnetic field is vertical (perpendicular to the surface). This means:
- The horizontal component ($B_H$) becomes zero because all field lines are vertical.
- The vertical component ($B_V$) equals the total earth's magnetic field ($B_E$).
- The angle of dip is $90^\circ$.
Under these conditions, a compass needle (which aligns with the horizontal component of the earth's magnetic field) will not have any preferred direction horizontally. Instead, it will tend to align vertically if free to move in three dimensions.
However, traditional compass needles are constrained to rotate horizontally. So, at the magnetic pole:
- The compass needle will point vertically upward or downward (depending on pole).
- It may also spin freely or become erratic due to the absence of a horizontal component.

Key Concept: Magnetic Dipole Moment, Gyromagnetic Ratio

b) $2\mu_B$
[Solution Description]
The magnetic dipole moment ($m$) of an electron revolving in an orbit is given by:
$m = \frac{e v r}{2}$
For the second Bohr orbit ($n=2$), the angular momentum is $J = 2 \cdot \frac{h}{2\pi} = \frac{h}{\pi}$. Using $J = m_e v r$, we solve for $v r$:
$v r = \frac{J}{m_e} = \frac{h}{\pi m_e}$
Substituting into the expression for $m$:
$m = \frac{e}{2} \cdot \frac{h}{\pi m_e} = \frac{e h}{2 \pi m_e}$
This matches the Bohr magneton multiplied by $2$.

Key Concept: Magnetic Dipole Moment

c) $\text{A m}^2$
[Solution Description]
The SI unit of magnetic moment $\vec{m}$ is derived from its definition as $N I A$, where $N$ is the number of turns, $I$ is the current, and $A$ is the area.
$\text{SI unit of } \vec{m} = \text{A m}^2$

Key Concept: Angle of Dip

d) $0^\circ$
[Solution Description]
The angle of dip ($\theta$) is the angle between the earth's magnetic field and the horizontal. At the equator, the earth's magnetic field is entirely horizontal (parallel to the surface), so the angle of dip is:
$\theta = 0^\circ$

Key Concept: Uniform Magnetic Field

b) Parallel and equidistant
[Solution Description]
A uniform magnetic field has constant strength and direction at every point. According to the properties of magnetic lines of force, in such a field, the lines are parallel and equally spaced because the field strength is the same everywhere. This is observed, for example, in Earth's magnetic field at a place.

Key Concept: Field Dependence on Angle

c) $90^\circ$
[Solution Description]
The general field due to a dipole is $E = \frac{p}{4\pi \epsilon_0 r^3} \sqrt{3\cos^2\theta + 1}$. Setting $E = E_{\text{eq}} = \frac{p}{4\pi \epsilon_0 r^3}$, solve $\sqrt{3\cos^2\theta + 1} = 1$, giving $\theta = \cos^{-1}(0)$.

Key Concept: Magnetic Equator, Dip Needle Behavior

c) Increases from 0° to 90°
[Solution Description]
At the magnetic equator, the angle of dip is 0° (needle is horizontal). As the needle moves towards the magnetic poles, the angle increases. At the magnetic north pole, the angle becomes 90° (needle is vertical).

Key Concept: Earth's Magnetism

b) Movement of molten iron and nickel in Earth's outer core
[Solution Description]
Earth's magnetism is primarily caused by the movement of molten iron and nickel in its outer core, which generates electric currents and thus a magnetic field.

Key Concept: Magnetic Axis of Earth

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
According to the syllabus, the magnetic axis of the earth indeed makes an angle of $11.5^\circ$ with the earth’s rotational axis, so the Assertion is true. The Reason states that the positions of the earth’s magnetic poles change gradually, which is also mentioned in the syllabus as an experimental observation. However, the gradual change in the positions of the magnetic poles does not explain why the magnetic axis makes a specific angle of $11.5^\circ$ with the rotational axis. Therefore, both statements are true, but the Reason does not correctly explain the Assertion.

Key Concept: Magnetic field equivalence, Far axial field

b) Doubles
[Solution Description]
The far axial magnetic field of a solenoid is given by $B = \frac{\mu_0}{4\pi} \frac{2m}{r^3}$. If the magnetic moment $m$ is doubled, the field also doubles because $B \propto m$.

Key Concept: Magnetic Moment, Torque

a) 1.06 N m
[Solution Description]
The magnetic moment of the coil is calculated as $m = NIA = 50 \times 3 \times 0.02 = 3.0 \text{A m}^2$. The torque is given by $\tau = mB\sin\theta = 3.0 \times 0.5 \times \sin(45^\circ) = 1.5 \times 0.7071 \approx 1.06 \text{N m}$.

Key Concept: Angle of Dip and Field Components

c) $40\,\mu T$
[Solution Description]
The Earth's magnetic field can be resolved into horizontal ($B_H$) and vertical ($B_V$) components using the relations:
$B_H = B_E \cos \theta$
Given $B_H = 20\,\mu T$ and $\theta = 60^\circ$, we first find $B_E$ as:
$B_E = \frac{B_H}{\cos \theta} = \frac{20\,\mu T}{\cos 60^\circ} = \frac{20\,\mu T}{0.5} = 40\,\mu T$

Key Concept: Historical perspective, Oersted experiment

c) That a compass needle deflected when placed near a current-carrying wire
[Solution Description]
Oersted's experiment demonstrated that an electric current could deflect a nearby compass needle, showing for the first time that electric currents produce magnetic effects. This proved that electricity and magnetism were interconnected phenomena, not independent as previously thought.

Key Concept: Axial and Equatorial Fields, Inverse Cube Law

d) 8
[Solution Description]
Step 1: Recall that the axial magnetic field intensity is given by:
$$ B_{\text{axial}} = \frac{\mu_0}{4\pi} \cdot \frac{2m}{r^3} $$
Step 2: The field at $r_1 = 2\, \text{cm}$ is:
$$ B_1 \propto \frac{1}{(0.02)^3} $$
Step 3: The field at $r_2 = 4\, \text{cm}$ is:
$$ B_2 \propto \frac{1}{(0.04)^3} $$
Step 4: Calculate the ratio $\frac{B_1}{B_2}$:
$$ \frac{B_1}{B_2} = \left( \frac{0.04}{0.02} \right)^3 = 8 $$
Thus, the ratio is $8$.

Key Concept: Horizontal Component of Earth's Magnetic Field

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the horizontal component of the earth's magnetic field is $B_H = B_E \cos \theta$, which is correct as per the definition. The reason states that the angle of dip $\theta$ is the angle between the direction of the earth's magnetic field and the horizontal, which is also correct. Moreover, the reason correctly explains why the horizontal component is calculated using cosine of the angle of dip. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Vertical Component, Horizontal Component of Earth's Magnetic Field

d) $0.8$ G
[Solution Description]
Given: Horizontal component $B_H = 0.4$ G, angle of dip $\theta = 60^\circ$. The horizontal component is related to the total field by:
$B_H = B_E \cos \theta$
Solve for $B_E$:
$B_E = \frac{B_H}{\cos \theta} = \frac{0.4}{\cos 60^\circ} = \frac{0.4}{0.5} = 0.8 \, \text{G}$

Key Concept: Horizontal and Vertical Components

c) $25 \sqrt{3}$ $\mu$T
[Solution Description]
Given the relationship $B_H = B_E \cos \theta$, substitute $B_E = 50 \, \mu T$ and $\theta = 30^\circ$. So,
$B_H = 50 \cos 30^\circ = 50 \times \frac{\sqrt{3}}{2} = 25 \sqrt{3} \, \mu T.$

Key Concept: Angle of Dip

d) $0^\circ$
[Solution Description]
At the Earth's magnetic equator, the magnetic field is horizontal (parallel to the surface). Hence, the angle of dip, which is the angle between the direction of Earth's magnetic field and the horizontal, is $0^\circ$.

Key Concept: Earth's Magnetic Field

a) $0^\circ$
[Solution Description]
The angle of dip ($\theta$) is the angle between the direction of the earth's magnetic field and the horizontal in the magnetic meridian. At the magnetic equator, the needle becomes horizontal, so the angle of dip is $0^\circ$.

Key Concept: Field Strength

c) Axial field is double of equatorial field
[Solution Description]
The magnetic field at the axial point (\$B\_a\$) is stronger than the field at the equatorial point (\$B\_e\$). The formula for axial field is \$B\_a = \frac{\mu\_0}{4\pi}\frac{2m}{r^3}\$, while for the equatorial field it is \$B\_e = \frac{\mu\_0}{4\pi}\frac{m}{r^3}\$. Thus, \$B\_a = 2B\_e\$.

Key Concept: Maximum Torque

b) $3 \times 10^{-3}$ A m$^2$
[Solution Description]
The maximum torque $\tau_{\text{max}}$ acting on a bar magnet is given by:
$\tau_{\text{max}} = m B$
Here, $B = 0.5$ T and $\tau_{\text{max}} = 1.5 \times 10^{-3}$ Nm. Rearranging the formula for magnetic moment $m$, we get:
$m = \frac{\tau_{\text{max}}}{B} = \frac{1.5 \times 10^{-3}}{0.5} = 3 \times 10^{-3} \text{ A m}^2$
Therefore, the magnetic moment is $3 \times 10^{-3}$ A m$^2$.

Key Concept: Magnetic Declination

c) Assertion is true, but Reason is false.
[Solution Description]
The angle of declination is defined as the acute angle between the magnetic meridian and geographical meridian at a place. The assertion is true because declination does vary depending on the location since Earth's magnetic field is not uniform. The reason is false because magnetic and geographical meridians do not coincide everywhere - they only coincide at places with zero declination. Thus, the correct relationship is that the assertion is true but the reason is false.

Key Concept: Equilibrium states, Torque

c) At $\theta = 90^\circ$, the potential energy is zero
[Solution Description]
Analyze equilibrium conditions for a magnetic dipole:
1. **Stable Equilibrium**: Occurs when $\theta = 0^\circ$. Here, torque $\tau = mB\sin(0^\circ) = 0$ and potential energy $U = -mB = -0.06 \, \text{J}$.
2. **Unstable Equilibrium**: Occurs when $\theta = 180^\circ$. Here, torque $\tau = mB\sin(180^\circ) = 0$ and potential energy $U = +mB = +0.06 \, \text{J}$.
3. At $\theta = 90^\circ$, torque is maximum ($\tau_{\text{max}} = mB = 0.06 \, \text{J}$) and potential energy $U = 0$.
The correct statement must align with these properties.

Key Concept: Horizontal Component

b) $0.433 \, \text{G}$
[Solution Description]
The horizontal component of the earth's magnetic field is given by $B_H = B_E \cos\theta$. Substituting the given values:
$B_H = 0.5 \times \cos30^\circ = 0.5 \times \frac{\sqrt{3}}{2} \approx 0.433 \, \text{G}.$

Key Concept: Apparent Angle of Dip

b) $67.8^\circ$
[Solution Description]
The apparent angle of dip ($\theta'$) is related to the true angle of dip ($\theta$) and the angle ($\alpha$) between the dip circle and magnetic meridian by:
$\tan \theta' = \frac{\tan \theta}{\cos \alpha}.$
Given $\theta = 60^\circ$ and $\alpha = 45^\circ$, we have:
$\tan \theta' = \frac{\tan 60^\circ}{\cos 45^\circ} = \frac{\sqrt{3}}{\frac{1}{\sqrt{2}}} = \sqrt{6}.$
Thus, $\theta' = \tan^{-1}(\sqrt{6}) \approx 67.8^\circ$.

Key Concept: Angle of Declination, Angle of Dip

c) $49.1^\circ$
[Solution Description]
The apparent angle of dip ($\theta'$) is given by the formula:
$\tan \theta' = \frac{\tan \theta}{\cos \alpha}$
Here, $\theta = 45^\circ$ (true angle of dip), and $\alpha = 30^\circ$ (angle between the vertical plane of the dip circle and the magnetic meridian).
First, calculate $\tan 45^\circ = 1$.
Then, $\cos 30^\circ = \frac{\sqrt{3}}{2} = 0.866$.
Now, substitute these values into the formula:
$\tan \theta' = \frac{1}{0.866} \approx 1.1547$
Taking the inverse tangent:
$\theta' = \tan^{-1}(1.1547) \approx 49.1^\circ$
So, the apparent angle of dip is approximately $49.1^\circ$.

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The angle of dip ($\theta$) is defined as the angle between the earth's magnetic field ($B_E$) and the horizontal in the magnetic meridian. At the magnetic equator, $\theta = 0^\circ$, which means the magnetic field is entirely horizontal, i.e., $B_V = B_E \sin \theta = 0$ and $B_H = B_E \cos \theta = B_E$. Therefore, the Assertion is true. The Reason correctly explains why the angle of dip is $0^\circ$ at the magnetic equator because $B_V$ vanishes there, leaving only $B_H$.

Key Concept: Angle of Dip, Horizontal Component

d) 60 $\mu$T
[Solution Description]
The total Earth's magnetic field $B_E$ can be found using the relation $B_H = B_E \cos \theta$, where $\theta$ is the angle of dip. Given $B_H = 30 \, \mu T$ and $\theta = 60^\circ$, we can solve for $B_E$:
$B_E = \frac{B_H}{\cos \theta} = \frac{30 \, \mu T}{\cos 60^\circ} = \frac{30}{0.5} = 60 \, \mu T$
So, the total Earth's magnetic field is 60 $\mu$T.

Key Concept: Oersted's Experiment

c) Compass needle deflected perpendicular to the wire
[Solution Description]
Oersted's experiment showed that an electric current produces a magnetic field. When current flows through a wire, it deflects the nearby compass needle perpendicular to the wire.
Reasoning:
1. Before the current flows, the compass needle aligns with Earth's magnetic field.
2. When current flows, a circular magnetic field is generated around the wire.
3. This field interacts with the compass needle, causing deflection.
4. The direction of deflection depends on the current direction (right-hand rule).

Key Concept: Neutral Points, Earth's Magnetic Field

b) 10 cm
[Solution Description]
At a neutral point, the magnetic field due to the two magnets cancels the Earth's horizontal field. For a point at distance $x$ from one magnet and $(0.2 - x)$ from the other:
$\frac{\mu_0}{4\pi} \left( \frac{2M}{x^3} \right) = \frac{\mu_0}{4\pi} \left( \frac{2M}{(0.2 - x)^3} \right) + B_H$
Since the magnets are identical and oppositely oriented, their fields add up. Solving for equilibrium:
$\frac{2 \times 5}{x^3} - \frac{2 \times 5}{(0.2 - x)^3} = 0.35 \times 10^{-4} \, (\text{converting G to T})$
Solving numerically, we find $x \approx 0.1 \, \text{m}$.

Key Concept: Geomagnetic poles, Magnetic axis

c) They are offset from the geographical poles by $11.5^\circ$.
[Solution Description]
The geomagnetic poles are the points where the Earth's magnetic field lines are perpendicular to the surface. Since the magnetic axis is inclined at $11.5^\circ$ to the rotational axis, the geomagnetic poles are not aligned with the geographical poles but are offset by approximately $11.5^\circ$.

Key Concept: Gyromagnetic Ratio

b) \$\frac{e}{2 m_e}\$
[Solution Description]
The gyromagnetic ratio is defined as the ratio of the magnetic dipole moment (\$m\$) to the orbital angular momentum (\$J\$). For an electron, this is given by:
$\frac{m}{J} = \frac{e}{2 m_e}$
where \$e\$ is the charge of the electron and \$m_e\$ is its mass.

Key Concept: Angle of Dip

a) $0^\circ$
[Solution Description]
The angle of dip ($\theta$) is the angle between the axis of a freely suspended dip needle and the horizontal direction. At the magnetic equator, the earth's magnetic field is purely horizontal, meaning the needle aligns horizontally. Thus, the angle of dip at the equator is $0^\circ$.

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The horizontal component of Earth's magnetic field is given by $B_H = B_E \cos \theta$, where $\theta$ is the angle of dip. At the magnetic poles, the angle of dip is $90^\circ$ because the magnetic field lines are vertical. Substituting $\theta = 90^\circ$ into the equation, we get $B_H = B_E \cos 90^\circ = 0$. Thus, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Elements of Earth's Magnetic Field

b) 0.25 G
[Solution Description]
The horizontal component of the earth's magnetic field is given by:
$B_H = B_E \cos \theta$.
Substituting the values:
$B_H = 0.5 \times \cos 60^\circ = 0.5 \times 0.5 = 0.25 \, \text{G}$.

Key Concept: Magnetic Equator - Dip Angle

d) $0^\circ$
[Solution Description]
The angle of dip ($\theta$) is the angle between the earth's magnetic field and the horizontal. At the magnetic equator, the earth's magnetic field is purely horizontal. Therefore:
$B_V = B_E \sin \theta$
Since $B_V = 0$ at the magnetic equator,
$0 = B_E \sin \theta \implies \theta = 0^\circ$
Hence, the angle of dip is zero at the magnetic equator.

Key Concept: Relationship between Axial and Equatorial Fields

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Step 1: The assertion states that the axial field is twice the equatorial field for the same distance and magnetic moment. This is correct because, as per the syllabus, the axial field formula is $B_{\text{axial}} = \frac{\mu_0}{4\pi} \cdot \frac{2m}{r^3}$, and the equatorial field formula is $B_{\text{eq}} = \frac{\mu_0}{4\pi} \cdot \frac{m}{r^3}$. Clearly, $B_{\text{axial}} = 2B_{\text{eq}}$ when the distance $r$ and magnetic moment $m$ are the same.
Step 2: The reason provides the exact formulas for both fields, which confirms the relationship mentioned in the assertion. Since the reason is not just true but also correctly explains the assertion, the correct option is (a).

Key Concept: Oersted's Experiment

b) An electric current produces a magnetic field
[Solution Description]
Oersted's experiment in 1820 showed that an electric current produces a magnetic field, thereby establishing a connection between electricity and magnetism.

Key Concept: Field Components

a) $30^\circ$
[Solution Description]
The net field at an angle $\theta$ is:
$E = \frac{kp}{r^3} \sqrt{3\cos^2\theta + 1}$
The axial field is $E_a = \frac{2kp}{r^3}$. Setting $E = \frac{E_a}{2}$, we solve for $\theta$.

Key Concept: Geomagnetic Poles

b) The magnetic south pole is located in the northern hemisphere
[Solution Description]
The Earth's magnetic south pole is located in the northern hemisphere near the geographic north pole, and the magnetic north pole is in the southern hemisphere near the geographic south pole. This is because opposite magnetic poles attract.

Key Concept: Magnetic Lines of Force, Properties of Magnetic Lines of Force

d) No magnetic lines of force will pass through the neutral point.
[Solution Description]
A neutral point is a location where the net magnetic field is zero. According to the properties of magnetic lines of force, no lines of force will pass through a neutral point because the field strength is zero there. Therefore, the only correct statement is that no magnetic lines of force will pass through the neutral point.

Key Concept: Field Produced by 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, which is true because the field lines inside an ideal solenoid are parallel and evenly spaced. The reason provided is also true as it correctly explains why the field is uniform—the straight and equidistant field lines indicate a constant magnetic field strength and direction. Therefore, the reason is the correct explanation of the assertion.

Key Concept: Earth's Magnetic Field

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a freely suspended magnetic needle aligns itself in the north-south direction, which is true because of Earth's magnetic field. The reason correctly explains this behavior by stating that Earth acts like a giant magnet with its south pole near the geographic north pole and vice versa. This alignment happens because the north pole of the needle is attracted to Earth's magnetic south pole (geographic north). Hence, both the Assertion and Reason are true, and the Reason is the correct explanation of the Assertion.

Key Concept: Apparent Angle of Dip, True Angle of Dip

c) 49.1\textdegree
[Solution Description]
The apparent angle of dip ($\theta'$) is related to the true angle of dip ($\theta$) and the angle ($\alpha$) between the plane of the dip circle and the magnetic meridian by the formula:
$\tan \theta' = \frac{\tan \theta}{\cos \alpha}$
Given $\theta = 45^\circ$ and $\alpha = 30^\circ$,
$\tan \theta' = \frac{\tan 45^\circ}{\cos 30^\circ} = \frac{1}{0.866} \approx 1.155$
Therefore, $\theta' = \tan^{-1}(1.155) \approx 49.1^\circ$.

Key Concept: Horizontal Component

b) $0.35 \, \text{gauss}$
[Solution Description]
The horizontal component of the Earth's magnetic field is given by $B_H = B_E \cos \theta$. Substituting the values:
$B_H = 0.4 \cdot \cos(30^\circ) = 0.4 \cdot \frac{\sqrt{3}}{2} = 0.4 \cdot 0.866 = 0.3464 \, \text{gauss}.$
Therefore, the horizontal component is approximately $0.35 \, \text{gauss}$.

Key Concept: Magnetic Axis of Earth

b) 11.5°
[Solution Description]
The magnetic axis of the earth makes an angle of 11.5° with the earth’s rotational axis. This is a fundamental property of the earth's geomagnetic field.

Key Concept: Direction of Torque

c) $90^\circ$
[Solution Description]
The torque $\tau$ on a magnetic dipole in a magnetic field is given by:
$\tau = m B \sin \theta$
The torque is maximum when $\sin \theta = 1$, which occurs at $\theta = 90^\circ$.

Key Concept: Angle of Dip

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the angle of dip is $90^\circ$ at the earth's magnetic poles, which is true because at the poles, the magnetic field lines are perpendicular to the horizontal surface, making the angle of dip $90^\circ$. The reason given is that at the poles, the earth's magnetic field is vertical, which is also true and correctly explains why the angle of dip is $90^\circ$ there. Hence, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Earth's magnetic field, Angle of dip, Horizontal component

b) $0.25 \, \text{G}$
[Solution Description]
The horizontal component ($B_H$) of the Earth's magnetic field ($B_E$) is given by:
$B_H = B_E \cos \theta$
Here, $B_E = 0.5 \, \text{G}$ and $\theta = 60^\circ$. Substituting the values:
$B_H = 0.5 \times \cos 60^\circ = 0.5 \times 0.5 = 0.25 \, \text{G}$

Key Concept: Torque on Magnetic Dipole

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description] The torque on a magnetic dipole in a uniform magnetic field is given by $\tau = mB\sin\theta$, where $m$ is the magnetic moment, $B$ is the magnetic field strength, and $\theta$ is the angle between the magnetic moment vector and the field direction. The sine function reaches its maximum value of 1 when $\theta = 90^\circ$. Therefore, both the Assertion and Reason are correct, and the Reason properly explains why the Assertion is true.

Key Concept: Definition of Magnetic Declination

a) Angle between magnetic meridian and geographical meridian
[Solution Description]
The angle of declination ($\alpha$) is the acute angle between the magnetic meridian and the geographical meridian at a given location. It represents the deviation of the compass needle from the true geographic north.

Key Concept: Relationship Between the Components

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the apparent angle of dip ($\theta'$) is greater than the true angle of dip ($\theta$) when the magnetic needle is not aligned with the magnetic meridian. This is correct because, as per the given syllabus, $tan\theta' = \frac{tan\theta}{\cos\alpha}$, where $\alpha$ is the angle between the plane and the magnetic meridian. Since $\cos\alpha tan\theta$ or $\theta' > \theta$.
The reason explains that the horizontal component ($B_H$) decreases when the needle is not aligned with the magnetic meridian, changing to $B_H \cos\alpha$. This is also correct as per the syllabus. Additionally, the reason correctly justifies why the apparent angle increases since $tan\theta' = \frac{B_V}{B_H \cos\alpha}$.
Thus, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Relationship Between Axial and Equatorial Fields

a) $\frac{B_a}{2}$
[Solution Description]
For a short magnet, the relationship between the axial ($B_a$) and equatorial ($B_e$) fields is given by:
$\vec{B}_a = -2 \vec{B}_e$
This means the magnitude of the axial field is twice the equatorial field, but they are in opposite directions. Thus:
$B_e = \frac{B_a}{2}$

Key Concept: Magnetic Dipole Moment

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The magnetic dipole moment of a current-carrying loop is defined as the product of the current $I$ and the area $A$ of the loop, which matches the assertion. The reason is also correct because the direction of the magnetic dipole moment is determined by curling the fingers of the right hand in the direction of the current; the thumb then points in the direction of $\vec{m}$. Thus, the reason correctly explains the assertion.

Key Concept: Magnetic Dipole Field

c) Near the poles of the magnet
[Solution Description]
The magnetic field intensity is strongest near the poles of the magnet where the field lines are most densely packed. According to formula $B = \frac{\mu_0}{4\pi} \cdot \frac{2m}{r^3}$ for axial position, as r decreases, B increases significantly.

Key Concept: Magnetic Moment of Revolving Electron

c) $\frac{e v r}{2}$
[Solution Description]
The magnetic dipole moment $\vec{m}$ of an electron revolving in a circular orbit is given by:
$m = I \times A$
where $I$ is the current and $A$ is the area of the loop. The current due to the electron's motion is $I = \frac{e v}{2 \pi r}$, and the area of the orbit is $A = \pi r^2$.
Substituting these values, we get:
$m = \left( \frac{e v}{2 \pi r} \right) \times \pi r^2 = \frac{e v r}{2}$
Therefore, the correct expression for the magnetic dipole moment is $\frac{e v r}{2}$.

Key Concept: Magnetic Equator, Angle of Dip

c) Horizontal: $\frac{B}{\sqrt{2}}$, Vertical: $\frac{B}{\sqrt{2}}$
[Solution Description]
The angle of dip $\theta$ is the angle between the Earth's magnetic field and the horizontal component. The horizontal component ($B_H$) and vertical component ($B_V$) are given by:
$B_H = B \cos \theta$
$B_V = B \sin \theta$
Substituting $\theta = 45°$:
$B_H = B \cos 45° = \frac{B}{\sqrt{2}}$
$B_V = B \sin 45° = \frac{B}{\sqrt{2}}$
Hence, both horizontal and vertical components are equal to $\frac{B}{\sqrt{2}}$.

Key Concept: Gyromagnetic Ratio

b) $\frac{e}{2 m_e}$
[Solution Description]
The gyromagnetic ratio is defined as the ratio of the magnetic dipole moment ($m$) to the angular momentum ($J$). For an electron, it is given as $\frac{m}{J} = \frac{e}{2 m_e}$, where $e$ is the charge of the electron and $m_e$ is its mass.

Key Concept: Field Due to Magnetic Dipole, Axial Line

c) $8 \times 10^{-6} \text{ T}$
[Solution Description]
Step 1: The magnetic dipole moment ($m$) is given by:
$m = \text{pole strength} \times \text{magnetic length} = 50 \times 0.1 = 5 \text{ A-m}^2$.
Step 2: Since $l << r$ ($0.1 < 0.5$), use the simplified axial field formula:
$B = \frac{\mu_0}{4\pi} \frac{2m}{r^3}$.
Step 3: Substitute the values:
$B = 10^{-7} \times \frac{2 \times 5}{(0.5)^3} = 10^{-7} \times \frac{10}{0.125} = 10^{-7} \times 80 = 8 \times 10^{-6} \text{ T}$.

Key Concept: Torque, Magnetic dipole moment

b) $\pi$ N·m
[Solution Description]
To find the torque:
Step 1: Calculate area ($A$) of loop = $\pi r^2$ = $\pi (2)^2$ = $4\pi$ m$^2$
Step 2: Magnetic dipole moment ($m$) = $I \times A$ = $5 \times 4\pi$ = $20\pi$ A·m$^2$
Step 3: Torque ($\tau$) = $mB\sin\theta$ = $(20\pi)(0.1)\sin(30°)$ = $(20\pi)(0.1)(0.5)$ = $\pi$ N·m

Key Concept: Direction of magnetic moment, Angular momentum

c) $180^\circ$
[Solution Description]
From the relation $\vec{m} = -\left( \frac{e}{2m_e} \right) \vec{J}$, the negative sign indicates they are antiparallel. Therefore, the angle between them is $180^\circ$.
Step-by-step:
1) The equation shows $\vec{m}$ is proportional to $-\vec{J}$
2) This means they point in exactly opposite directions
3) Hence angle between them is $180^\circ$

Key Concept: Magnetic Lines of Force, Properties

b) The field is uniform in both magnitude and direction.
[Solution Description]
Parallel and equidistant magnetic lines of force indicate a uniform magnetic field. This is because the density of the lines (which represents the field strength) is constant everywhere. Additionally, the direction of the field is the same at all points since the lines are parallel.

Key Concept: Angle of Dip and Apparent Angle of Dip

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The apparent angle of dip ($\theta'$) is related to the true angle of dip ($\theta$) by the formula:
$\tan \theta' = \frac{\tan \theta}{\cos \alpha}$
where $\alpha$ is the angle between the dip circle and the magnetic meridian. Since $\cos \alpha \tan \theta$, which implies $\theta' > \theta$. Thus, the Assertion is true. The Reason states that the horizontal component decreases when the dip circle is rotated, which is correct because $B_H' = B_H \cos \alpha$. This decrease causes $\tan \theta'$ to increase, making the Reason also true and the correct explanation of the Assertion.

Key Concept: Direction of Magnetic Field

c) From north pole to south pole
[Solution Description]
For a bar magnet in the broad-side-on position, the magnetic field $\vec{B}$ at any point on the equatorial line is parallel to the magnetic axis and directed from the north pole (N) to the south pole (S) of the magnet. This is unlike the end-on position where $\vec{B}$ points from S to N.

Key Concept: Gyromagnetic Ratio

b) $8.79 \times 10^{10} \, \text{C/kg}$
[Solution Description]
The gyromagnetic ratio is given by $\frac{m}{J} = \frac{e}{2 m_e}$.
Step 1: Since the gyromagnetic ratio is independent of the speed and radius of the orbit, we can directly use the formula:
$\frac{m}{J} = \frac{1.6 \times 10^{-19}}{2 \times 9.1 \times 10^{-31}} = \frac{1.6 \times 10^{-19}}{1.82 \times 10^{-30}} = 8.79 \times 10^{10} \, \text{C/kg}$
Thus, the ratio is $8.79 \times 10^{10} \, \text{C/kg}$.

Key Concept: Bohr Magneton, Quantization of Angular Momentum

c) Doubles
[Solution Description]
The Bohr magneton $\mu_B$ is defined as:
$\mu_B = \frac{e h}{4 \pi m_e}$
If $h$ is doubled ($h' = 2h$), the new Bohr magneton becomes:
$\mu_B' = \frac{e (2h)}{4 \pi m_e} = 2 \left(\frac{e h}{4 \pi m_e}\right) = 2\mu_B$
Therefore, the Bohr magneton doubles.

Key Concept: Magnetic Moment of Solenoid

b) $m = (n \cdot 2l) I \cdot \pi a^2$
[Solution Description]
The magnetic moment of a solenoid is calculated using the formula:
$m = (n \cdot 2l) I \cdot \pi a^2$
where $n$ is the number of turns per unit length, $2l$ is the length of the solenoid, $I$ is the current, and $a$ is the radius of the solenoid.

Key Concept: Polarity of a Solenoid

b) South pole
[Solution Description]
Using the end rule, if the current flows clockwise when viewed from an end, that end acts as the **south pole** of the solenoid. Therefore, the polarity of the given end is south.

Key Concept: Magnetic Lines of Force

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that magnetic lines of force never intersect each other, which is a known property of magnetic lines of force as per the syllabus. The reason explains that if they intersect, there would be two directions of the magnetic field at the point of intersection, which is impossible. This correctly explains why the assertion is true. Thus, both assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Far Axial Field Calculation

b) 0.0157 A·m\textsuperscript{2}
[Solution Description]
The magnetic moment $m$ of a solenoid is given by:
$m = (n \cdot 2l) \cdot I \cdot \pi a^2$
Substituting the given values:
$m = (1000 \times 0.2) \times 0.5 \times \pi (0.01)^2 = 100 \times 0.5 \times \pi \times 10^{-4} = 0.0157 \, \text{A·m}^2$

Key Concept: Emergence and Termination, Non-Intersecting Nature

d) They never intersect
[Solution Description]
Magnetic lines of force never intersect each other. If they did intersect, it would imply two directions of the magnetic field at the same point, which is impossible. This property arises from the fundamental nature of magnetic fields where a single unique direction exists at any given point.

Key Concept: Torque on a Current Loop

b) $\pi \, \text{Nm}$
[Solution Description]
First, calculate the area $A$ of the loop:
$A = \pi r^2 = \pi (2)^2 = 4\pi \, \text{m}^2$
Next, find the magnetic dipole moment $m$:
$m = I A = 5 \times 4\pi = 20\pi \, \text{A m}^2$
The angle $\theta$ between the dipole moment $\vec{m}$ and $\vec{B}$ is $90^\circ - 60^\circ = 30^\circ$ because the plane of the loop makes $60^\circ$ with $\vec{B}$, but $\vec{m}$ is perpendicular to the plane.
Now, the torque $\tau$ is:
$\tau = m B \sin \theta = 20\pi \times 0.1 \times \sin 30^\circ = 2\pi \times 0.5 = \pi \, \text{Nm}$

Key Concept: Magnetic Dipole Moment - Torque Calculation

b) $0.005\pi \, \text{N m}$
[Solution Description]
Given:
$r = 0.1 \, \text{m}, \quad I = 2 \, \text{A}, \quad B = 0.5 \, \text{T}, \quad \theta = 30^\circ$
The area of the loop is:
$A = \pi r^2 = \pi (0.1)^2 = 0.01\pi \, \text{m}^2$
The magnetic dipole moment is:
$m = I A = 2 \times 0.01\pi = 0.02\pi \, \text{A m}^2$
The torque is given by:
$\tau = m B \sin \theta = 0.02\pi \times 0.5 \times \sin 30^\circ$
Since $\sin 30^\circ = 0.5$,
$\tau = 0.02\pi \times 0.5 \times 0.5 = 0.005\pi \, \text{N m}$

Key Concept: Magnetic Poles, Magnetic Needle Behavior

d) Near the geomagnetic south pole
[Solution Description]
A magnetic needle aligns itself with the Earth's magnetic field. At the geomagnetic poles:
- The north pole of the needle points vertically downward in the southern hemisphere (near the geomagnetic south pole).
- The south pole of the needle points vertically downward in the northern hemisphere (near the geomagnetic north pole).
Thus, the north pole points vertically downward at the geomagnetic south pole, which is located in the northern hemisphere.

Key Concept: Gyromagnetic ratio, Magnetic moment

b) It doubles
[Solution Description]
The gyromagnetic ratio $\frac{m}{J} = \frac{e}{2m_e}$ is constant for an electron. Since $\vec{m} = -\left( \frac{e}{2m_e} \right)\vec{J}$, when angular momentum $J$ is doubled, the magnetic moment $m$ will also double.
Step-by-step:
1) Initial magnetic moment: $m_1 = \frac{e v r}{2}$
2) Initial angular momentum: $J_1 = m_e v r$
3) Final angular momentum: $J_2 = 2J_1 = 2m_e v r$
4) Using gyromagnetic ratio: $\frac{m_2}{J_2} = \frac{e}{2m_e}$
5) Thus: $m_2 = \left( \frac{e}{2m_e} \right) J_2 = \left( \frac{e}{2m_e} \right) (2m_e v r) = e v r$
6) Comparing $m_2 = 2 \times \frac{e v r}{2} = 2m_1$

Key Concept: Magnetic Declination, Magnetic Meridian

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that the angle of declination is not defined at the magnetic poles, which is true because the concept of magnetic declination relies on the existence of a non-zero angle between the magnetic and geographical meridians. At the magnetic poles, the magnetic field lines are vertical, and hence, the magnetic meridian is undefined, making the angle of declination also undefined. The reason claims that the magnetic meridian coincides with the geographical meridian at the magnetic poles, which is false because the magnetic meridian itself is undefined there due to the vertical alignment of the magnetic field. Thus, the assertion is true, but the reason is false.

Key Concept: Earth's Magnetic Field

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the horizontal component of Earth's magnetic field ($B_H$) is zero at the magnetic poles. This is true because, at the poles, the magnetic field lines are nearly vertical, resulting in an angle of dip ($\theta$) of 90$^\circ$. Using the formula $B_H = B_E \cos \theta$, substituting $\theta = 90^\circ$ gives $B_H = B_E \cos 90^\circ = 0$.
The reason correctly explains why $B_H$ is zero at the poles by relating it to the angle of dip being 90$^\circ$. Thus, both the assertion and reason are true, and the reason logically justifies the assertion.

Key Concept: Angle of dip determination

c) $60^\circ$
[Solution Description]
The relationship between $B_H$ and $B_E$ is:
$B_H = B_E \cos \theta$
Rearranging to find $\theta$:
$\cos \theta = \frac{B_H}{B_E} = \frac{0.3}{0.6} = 0.5$
Thus,
$\theta = \cos^{-1}(0.5) = 60^\circ$

Key Concept: Torque on a Magnetic Dipole, Equivalent Solenoid

a) $7.85 \times 10^{-3} \, \text{N m}$
[Solution Description]
The torque on a magnetic dipole in a uniform magnetic field is:
$\tau = mB \sin\theta$
For a current loop, $m = I \times A = I \pi r^2$.
Here, $\theta = 90^\circ$ (since the plane is perpendicular to $\vec{B}$), so $\sin\theta = 1$.
Substituting values:
$m = 2 \times \pi \times (0.05)^2 = 0.0157 \, \text{A m}^2$
Now, torque:
$\tau = 0.0157 \times 0.5 \times 1 = 7.85 \times 10^{-3} \, \text{N m}$

Key Concept: Magnetic Axis and Equator

c) Perpendicular to the magnetic axis and where the magnetic needle is horizontal
[Solution Description]
The magnetic equator is the circle where the magnetic needle is parallel to the Earth's surface, formed by the intersection of the Earth's surface with a plane perpendicular to the magnetic axis.

Key Concept: Comparison with end-on position, direction analysis

b) $3.61 \times 10^{-5} \, \text{T}$
[Solution Description]
From the given ratio $\vec{B}_{\text{end-on}} = -2 \vec{B}_{\text{broad-side-on}}$, we know the fields are antiparallel. In broad-side-on position, the magnet's field ($B_m = 2 \times 10^{-5} \, \text{T}$) is perpendicular to Earth's field ($B_E = 3 \times 10^{-5} \, \text{T}$). The resultant field is the vector sum:
$B_{\text{resultant}} = \sqrt{B_m^2 + B_E^2} = \sqrt{(2 \times 10^{-5})^2 + (3 \times 10^{-5})^2} = \sqrt{4 \times 10^{-10} + 9 \times 10^{-10}} = \sqrt{13 \times 10^{-10}} = \sqrt{13} \times 10^{-5} \approx 3.606 \times 10^{-5} \, \text{T}$

Key Concept: Earth's Magnetic Field

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
At the geomagnetic equator, the earth's magnetic field is parallel to the surface of the earth, meaning the vertical component $B_V = B_E \sin \theta$ becomes zero as $\theta = 0^\circ$. Therefore, the magnetic needle aligns itself horizontally, confirming the Assertion is true. The Reason correctly explains the Assertion because the zero angle of dip ($\theta = 0^\circ$) implies no vertical component of the earth's magnetic field, making the needle horizontal. Hence, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Magnetic Field of Solenoid

b) $B = \frac{\mu_0}{4\pi} \cdot \frac{2m}{r^3}$
[Solution Description]
The magnetic field at a far axial point due to a solenoid is given by:
$B = \frac{\mu_0}{4\pi} \cdot \frac{2m}{r^3}$
This formula shows the equivalence between the solenoid and a bar-magnet.

Key Concept: Magnetic moment

b) $0.36 \, \text{A m}^2$
[Solution Description]
The magnetic moment $\vec{m}$ for a current-loop is given by:
$\vec{m} = N I \vec{A}$
Here, $N = 50$, $I = 3$ A, and $\vec{A} = 0.04 \times 0.06 = 0.0024 \, \text{m}^2$. Thus:
$m = 50 \times 3 \times 0.0024 = 0.36 \, \text{A m}^2$

Key Concept: Magnetic Lines of Force, Neutral Point

c) Increases by a factor of $2^{1/3}$
[Solution Description]
At the neutral point, the magnetic field due to both magnets cancels Earth's magnetic field. For two identical magnets with strength $M$, the field at a distance $d$ from each magnet is $\frac{\mu_0}{4\pi}\frac{2M}{d^3}$ along the perpendicular bisector. Doubling the magnet strength means the field becomes $\frac{\mu_0}{4\pi}\frac{4M}{d'^3}$. For the neutral point condition to hold, $\frac{2M}{d^3} = \frac{4M}{d'^3} \Rightarrow d'^3 = 2d^3 \Rightarrow d' = 2^{1/3}d$. Thus, the distance increases by a factor of $2^{1/3}$.

Key Concept: Relationship between Axial and Equatorial Fields

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the axial magnetic field ($B_{axial}$) of a small magnet is twice the equatorial magnetic field ($B_{equatorial}$) for the same distance and magnetic moment. This is true because, from the given formulas, $B_{axial} = \frac{\mu_0}{4\pi} \cdot \frac{2m}{r^3}$ and $B_{equatorial} = \frac{\mu_0}{4\pi} \cdot \frac{m}{r^3}$. Dividing the two equations gives $\frac{B_{axial}}{B_{equatorial}} = 2$. Thus, $B_{axial} = 2 \cdot B_{equatorial}$, proving the assertion. The reason provides the correct formulas used to derive this relationship, making it the correct explanation of the assertion.

Key Concept: Horizontal Component of Earth’s Magnetic Field

b) 25 $\mu T$
[Solution Description]
The horizontal component of the earth's magnetic field is given by:
$B_H = B_E \cos \theta$
Substituting $B_E = 50 \, \mu T$ and $\theta = 60^\circ$:
$B_H = 50 \times \cos 60^\circ = 50 \times 0.5 = 25 \, \mu T$
Thus, the horizontal component is 25 $\mu T$.

Key Concept: Earth's magnetism, Historical context

b) Permanent magnets existed in Earth's iron-rich crust
[Solution Description]
Prior to 1820, Earth's magnetism was often attributed to permanent magnetization of materials within the Earth or to some inherent property of the Earth's core, completely separate from electrical phenomena. There was no concept of electromagnetism as we know it today before Oersted's discovery.

Key Concept: Magnetic Lines of Force

b) From north pole to south pole
[Solution Description]
Magnetic lines of force emerge from the north pole and terminate at the south pole of a bar magnet. Outside the magnet, the direction is from the north pole to the south pole.

Key Concept: Magnetic Equator and Dip Angle

c) $0^\circ$
[Solution Description]
The angle of dip (\theta) is defined as the angle between the Earth's magnetic field and the horizontal component. At the magnetic equator, the Earth's magnetic field is parallel to the surface, meaning the vertical component $B_V$ is zero. Therefore, from the equation $\theta = \tan^{-1} \left( \frac{B_V}{B_H} \right)$, when $B_V = 0$, $\theta = \tan^{-1}(0) = 0^\circ$.

Key Concept: Torque on Magnetic Dipole, Stable Equilibrium

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The torque acting on a magnetic dipole in a uniform magnetic field is given by $\tau = m B \sin \theta$, where $m$ is the magnetic moment, $B$ is the magnetic field strength, and $\theta$ is the angle between the magnetic moment vector $\vec{m}$ and the magnetic field vector $\vec{B}$. When the magnet is parallel to the field ($\theta = 0^\circ$ or $180^\circ$), $\sin \theta = 0$, so $\tau = 0$.
For stable equilibrium, the potential energy of the dipole $U = -m B \cos \theta$ must be minimized. This occurs when $\theta = 0^\circ$ ($\cos \theta = 1$) because $U$ reaches its most negative value ($-m B$). Thus, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Earth's Magnetic Moment Calculation

b) $8 \times 10^{22} \, \text{J T}^{-1}$
[Solution Description]
Rearrange the given formula to solve for $m$:
$B = \frac{\mu_0}{4\pi} \frac{m}{r^3}$
Substitute the values: $B = 0.3 \times 10^{-4} \, \text{T}$, $r = 6.4 \times 10^6 \, \text{m}$, and $\mu_0/4\pi = 10^{-7} \, \text{T m A}^{-1}$:
$m = B \cdot r^3 / (\mu_0/4\pi) = (0.3 \times 10^{-4}) \times (6.4 \times 10^6)^3 / 10^{-7}$
Calculate $r^3 = 2.62144 \times 10^{20} \, \text{m}^3$. Then:
$m = 0.3 \times 10^{-4} \times 2.62144 \times 10^{20} / 10^{-7} = 7.86432 \times 10^{22} \, \text{J T}^{-1}$
This approximates to $8 \times 10^{22} \, \text{J T}^{-1}$.

Key Concept: True Angle of Dip, Apparent Angle of Dip

b) 26.56°
[Solution Description]
Given:
Apparent angles of dip ($\theta_1$ and $\theta_2$) = 30° and 45°
The relationship between the true angle of dip ($\theta$) and the apparent angles is:
$\cot^2 \theta = \cot^2 \theta_1 + \cot^2 \theta_2$
Substituting the values:
$\cot^2 \theta = \cot^2 30^\circ + \cot^2 45^\circ$
Calculating:
$\cot 30^\circ = \sqrt{3}$
$\cot 45^\circ = 1$
Therefore:
$\cot^2 \theta = (\sqrt{3})^2 + (1)^2 = 3 + 1 = 4$
Taking square root:
$\cot \theta = 2$
Therefore:
$\theta = \arctan\left(\frac{1}{2}\right) \approx 26.56^\circ$
So, the true angle of dip is approximately 26.56°.

Key Concept: Magnetic Dipole Moment Formula

b) \$m = \frac{e v r}{2}\$
[Solution Description]
The magnetic dipole moment of an electron revolving in a circular orbit is given by:
$m = I A$
where \$I\$ is the current due to the revolving electron and \$A\$ is the area of the orbit.
Current \$I\$ is calculated as:
$I = \frac{e v}{2 \pi r}$
Area \$A\$ of the orbit is:
$A = \pi r^2$
Substituting these values, we get:
$m = \left(\frac{e v}{2 \pi r}\right) \times \pi r^2 = \frac{e v r}{2}$
Hence, the correct formula is \$m = \frac{e v r}{2}\$.

Key Concept: Magnetic Moment Calculation

b) $0.6 \, \text{A m}^2$
[Solution Description]
The maximum torque occurs when $\theta = 90^\circ$ ($\sin 90^\circ = 1$). Using:
$\tau_{\text{max}} = m B$
Given $\tau_{\text{max}} = 0.09 \, \text{Nm}$ and $B = 0.15 \, \text{T}$, we get:
$m = \frac{\tau_{\text{max}}}{B} = \frac{0.09}{0.15} = 0.6 \, \text{A m}^2$

Key Concept: Assertion and Reason

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The magnetic dipole moment of an electron in orbit is given by $m = \frac{evr}{2}$. For the ground state of a hydrogen atom, substituting the values of $e$, $v$, and $r$ yields the minimum value of magnetic moment as the Bohr magneton ($9.27 \times 10^{-24} A m^2$). However, the gyromagnetic ratio $\frac{m}{J} = \frac{e}{2m_e}$ is a general relationship and does not necessarily imply that every electron has a magnetic moment exactly equal to the Bohr magneton. Thus, Assertion is true only for specific cases (like the ground state of hydrogen), but Reason is universally true.
Hence, Assertion is conditionally true, but Reason is universally true and correctly explains the concept behind the assertion under specific conditions.

Key Concept: Torque on a Magnetic Dipole in a Uniform Magnetic Field

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The torque acting on a magnetic dipole (bar magnet) in a uniform magnetic field is given by $\tau = m B \sin \theta$. When the axis of the magnet is perpendicular to the field, $\theta = 90^\circ$, and thus $\sin \theta = 1$. Therefore, the torque is maximum at this position, which validates the Assertion (A). The Reason (R) correctly states the formula for torque on a magnetic dipole, which explains why the torque is maximum when $\theta = 90^\circ$. Hence, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Magnetic Equator, Uniform Magnetic Field

c) Over a small area like a laboratory
[Solution Description]
The Earth's magnetic field lines are parallel and equidistant over a small area like that of a laboratory. This uniformity occurs because the magnetic field does not vary significantly in magnitude or direction over such a small region.

Key Concept: Vertical Component Calculation

b) 0.4 G
[Solution Description]
Given: \$B_H = 0.4\$ G, \$\theta = 45^\circ\$.
Using the relation \$B_V = B_H \tan \theta\$, we get:
$B_V = B_H \tan \theta = 0.4 \times \tan 45^\circ = 0.4 \times 1 = 0.4 \text{ G}$

Key Concept: Lines of Force Near Magnetic Poles

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the properties of magnetic lines of force near poles, two lines of force never intersect each other because if they did, there would be two directions of the magnetic field at the point of intersection, which is not possible. Hence, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Magnetic Dipole Moment of Current Loop

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the torque on a current-carrying loop in a magnetic field is maximum when the plane of the loop is perpendicular to the field, which is true because the torque is given by $\tau = m B \sin \theta$. The torque is maximum when $\theta = 90^\circ$, i.e., when the plane of the loop is perpendicular to the field. The reason correctly explains the assertion by stating the formula for torque and the condition for its maximum value.

Key Concept: Behavior of solenoids

b) They will repel each other
[Solution Description]
Current-carrying solenoids behave like bar magnets, exhibiting attraction between unlike poles and repulsion between like poles. Since like poles (N-N or S-S) face each other here, they will repel.

Key Concept: Polarity determination, Solenoid end rule

b) It is the south pole.
[Solution Description]
According to the end rule:
- If the current appears anticlockwise when looking at an end, it is the north pole.
- If the current appears clockwise, it is the south pole.
Here, the current appears clockwise at the given end, so it is the south pole.

Key Concept: Magnetic field calculation, broad-side-on position

a) $1.41 \times 10^{-6} \, \text{T}$
[Solution Description]
First, we need to use the formula for magnetic field in broad-side-on position: $B = \frac{\mu_0}{4\pi} \frac{m}{(r^2 + l^2)^{3/2}}$. Given:
- $m = 4.8 \times 10^{-2} \, \text{J T}^{-1}$
- $2l = 2 \, \text{cm} \Rightarrow l = 1 \, \text{cm} = 0.01 \, \text{m}$
- $r = 15 \, \text{cm} = 0.15 \, \text{m}$
Now calculate denominator: $(r^2 + l^2)^{3/2} = (0.15^2 + 0.01^2)^{3/2} = (0.0225 + 0.0001)^{3/2} = (0.0226)^{3/2} = (0.0226)^{1} * \sqrt{0.0226} = 0.0226 * 0.1503 \approx 0.003398$
Now plug values into formula: $B = 10^{-7} * \frac{4.8 \times 10^{-2}}{0.003398} \approx 10^{-7} * 14.126 \approx 1.4126 \times 10^{-6} \, \text{T}$

Key Concept: Magnetic Dipole Moment - Direction

d) Negative $z$-axis
[Solution Description]
Using the right-hand rule for the magnetic dipole moment, if we grasp the loop with fingers curling in the direction of the current (clockwise here), the thumb points in the direction of $\overrightarrow{m}$. For a clockwise current in the $xy$-plane, the thumb points along the negative $z$-axis.

Key Concept: Maximum torque, Current loop configuration

c) 10$\sqrt{2}$ N·m
[Solution Description]
Solution approach:
Step 1: When two dipoles are perpendicular, their vector sum is $m_{total}$ = $\sqrt{m_1^2 + m_2^2}$
Step 2: Here $m_{total}$ = $\sqrt{5^2 + 5^2}$ = $\sqrt{50}$ = $5\sqrt{2}$ A·m$^2$
Step 3: Maximum torque occurs when $\theta=90°$ so $\sinθ=1$
Step 4: $\tau_{max}$ = $m_{total}B$ = $(5\sqrt{2})(2)$ = $10\sqrt{2}$ N·m

Key Concept: Solenoid field calculation, Non-axial point

b) $\frac{\mu_0nI}{2}$
[Solution Description]
For points very close to solenoid end but slightly off-axis:
1) Near-end axial field ≈ $\frac{\mu_0nI}{2}$
2) Small perpendicular component appears due to off-axis position
3) Total field B ≈ $\sqrt{B_{axial}^2 + B_{perp}^2}$
4) Since δ << R, perpendicular component is small compared to axial
The field will be slightly less than but very close to $\frac{\mu_0nI}{2}$

Key Concept: Magnetic Needle Behavior

c) At Earth's magnetic poles
[Solution Description]
A magnetic needle aligns vertically when the angle of dip is $90^\circ$. From the syllabus, this occurs at Earth's magnetic poles where the magnetic field is completely vertical.

Key Concept: Earth’s Magnetic Field and Dipole Model

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states the formula for the magnetic field on Earth's equatorial line, which is derived from the dipole model of Earth's magnetic field. The reason explains why Earth can be approximated as a dipole, with its magnetic south pole near the geographic north pole. Both statements are true, and the reason correctly explains why the assertion holds because the dipole model is the basis for the formula provided in the assertion.

Key Concept: Special Cases of Magnetic Declination

c) At the magnetic poles
[Solution Description]
The angle of declination is not defined at the Earth's magnetic poles because the magnetic field lines are vertical there, and thus, there is no meaningful horizontal component to define the magnetic meridian.