308. A solenoid with 200 turns is placed in a uniform magnetic field of 0.5 T. If the magnetic flux through the solenoid changes uniformly from 0.02 Wb to 0.08 Wb in 2 seconds, what is the magnitude and direction of the induced current if the resistance of the solenoid is 4 $\Omega$?
Key Concept: Faraday's law, Lenz's law
d) 1.5 A, clockwise
[Solution Description]
First, calculate the change in magnetic flux: $\Delta \phi = 0.08\, \text{Wb} - 0.02\, \text{Wb} = 0.06\, \text{Wb}$.
Next, calculate the rate of change of flux: $\frac{d\phi}{dt} = \frac{0.06\, \text{Wb}}{2\, \text{s}} = 0.03\, \text{Wb/s}$.
Using Faraday's law, the induced emf is $e = -N \frac{d\phi}{dt} = -200 \times 0.03 = -6\, \text{V}$. The negative sign indicates the direction according to Lenz's law.
The magnitude of the current is $I = \frac{e}{R} = \frac{6}{4} = 1.5\, \text{A}$. Since the flux is increasing, the induced current will oppose the increase, creating a magnetic field opposite to the applied field.
Your Answer is correct.
d) 1.5 A, clockwise
[Solution Description]
First, calculate the change in magnetic flux: $\Delta \phi = 0.08\, \text{Wb} - 0.02\, \text{Wb} = 0.06\, \text{Wb}$.
Next, calculate the rate of change of flux: $\frac{d\phi}{dt} = \frac{0.06\, \text{Wb}}{2\, \text{s}} = 0.03\, \text{Wb/s}$.
Using Faraday's law, the induced emf is $e = -N \frac{d\phi}{dt} = -200 \times 0.03 = -6\, \text{V}$. The negative sign indicates the direction according to Lenz's law.
The magnitude of the current is $I = \frac{e}{R} = \frac{6}{4} = 1.5\, \text{A}$. Since the flux is increasing, the induced current will oppose the increase, creating a magnetic field opposite to the applied field.