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: Understanding the Concept:
By Faraday's law the induced emf of an \(n\)-turn coil is \(\varepsilon=\dfrac{n\,\Delta\phi}{t}\).

Step 2: Find the emf:
\(\varepsilon=\dfrac{n(\phi_2-\phi_1)}{t}\) in magnitude.

Step 3: Find the total resistance:
The coil has resistance \(R\) and the galvanometer has \(2R\), in series, so the total is \(3R\).

Step 4: Find the current:
\(I=\dfrac{\varepsilon}{3R}=\dfrac{n(\phi_2-\phi_1)}{3Rt}\). Option C.

Step 5: Why the other options are wrong.
Options A, B and D use a total resistance of \(R\) or \(2R\) and so ignore one of the parts of the circuit.

Final Answer:
The induced current is n (phi2 - phi1)/(3Rt). \[ \boxed{\text{(C) }\dfrac{n(\phi_2-\phi_1)}{3Rt}} \]