Question

A coil of n turns and resistance $R \Omega$ is connected in series with resistance $R/4$. The combination is moved for time t second through magnetic flux $\phi_1$ to $\phi_2$. The induced current in the circuit is

• A. $\frac{2n(\phi_1 - \phi_2)}{5Rt}$
• B. $\frac{4n(\phi_1 - \phi_2)}{5Rt}$
• C. $\frac{3n(\phi_1 - \phi_2)}{4Rt}$
• D. $\frac{5n(\phi_1 - \phi_2)}{3Rt}$

Hint:

Induced current depends on the total resistance of the closed loop, including external resistors.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
A change in magnetic flux through a coil induces an electromotive force (EMF) based on Faraday's law. This EMF causes a current to flow through the total resistance of the circuit.
Step 2: Key Formula or Approach:
1. Induced EMF for $n$ turns: $e = -n \frac{\Delta \phi}{\Delta t} = n \frac{\phi_1 - \phi_2}{t}$ (magnitude)
2. Total resistance $R_{total} = R + \frac{R}{4}$
3. Induced current $I = \frac{e}{R_{total}}$
Step 3: Detailed Explanation:
The total resistance of the circuit is:
\[ R_{total} = R + \frac{R}{4} = \frac{5R}{4} \]
The magnitude of induced EMF in the coil is:
\[ e = n \frac{|\phi_2 - \phi_1|}{t} = n \frac{(\phi_1 - \phi_2)}{t} \]
Induced current is:
\[ I = \frac{n (\phi_1 - \phi_2) / t}{5R/4} \]
\[ I = \frac{4n (\phi_1 - \phi_2)}{5Rt} \]
Step 4: Final Answer:
The induced current is $\frac{4n(\phi_1 - \phi_2)}{5Rt}$.