Question

A coil having 'n' turns and resistance 'R' is connected with a galvanometer of resistance '2R'. This combination is moved in time 't' from flux \(φ_1\) to \(φ_2\) Wb. The induced current in the circuit is

• A. \(\frac{n(φ_1-φ_2)}{Rt}\)
• B. \(\frac{n(φ_2-φ_1)}{Rt}\)
• C. \(\frac{n(φ_2-φ_1)}{3Rt}\)
• D. \(\frac{n(φ_2-φ_1)}{2Rt}\)

Hint:

emf = n (change in flux)/t and the total resistance is R + 2R.

Answer 1

The Correct Option is C

Step 1: Use charge flow idea:
Total charge through the circuit is $q=\dfrac{n\Delta\phi}{R_{tot}}$, and average current is $q/t$.

Step 2: Compute:
$R_{tot}=R+2R=3R$, so $I=\dfrac{q}{t}=\dfrac{n(\phi_2-\phi_1)}{3Rt}$.

Step 3: Pick:
Option C.

Final Answer:
The emf is n times the flux change over t and the resistance is 3R. \[ \boxed{\text{(C) }\dfrac{n(\phi_2-\phi_1)}{3Rt}} \]