Question:
A coil having \(n\) turns and resistance \(R\Omega\) is connected with a galvanometer of resistance \(4R\Omega\). This combination is moved in time \(t\) second from a magnetic field \(w_1\) weber to \(w_2\) weber. The induced current in the circuit is
A coil having \(n\) turns and resistance \(R\Omega\) is connected with a galvanometer of resistance \(4R\Omega\). This combination is moved in time \(t\) second from a magnetic field \(w_1\) weber to \(w_2\) weber. The induced current in the circuit is
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Remember to include total resistance (coil + galvanometer) in the circuit.
Updated On: Apr 20, 2026
- \(\frac{-(w_2 - w_1)}{5Rt}\)
- \(\frac{-n(w_2 - w_1)}{5Rt}\)
- \(\frac{-(w_2 - w_1)}{Rnt}\)
- \(\frac{-n(w_2 - w_1)}{Rt}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Induced emf: \(E = -n\frac{d\phi}{dt}\), current: \(I = \frac{E}{R_{\text{total}}}\).
Step 2: Detailed Explanation:
Change in flux: \(E = -n\frac{w_2 - w_1}{t}\). Total resistance: \(R_{\text{total}} = R + 4R = 5R\). Current: \[ I = \frac{E}{5R} = \frac{-n(w_2 - w_1)}{5Rt} \]
Step 3: Final Answer:
\[ \boxed{\frac{-n(w_2 - w_1)}{5Rt}} \]
Induced emf: \(E = -n\frac{d\phi}{dt}\), current: \(I = \frac{E}{R_{\text{total}}}\).
Step 2: Detailed Explanation:
Change in flux: \(E = -n\frac{w_2 - w_1}{t}\). Total resistance: \(R_{\text{total}} = R + 4R = 5R\). Current: \[ I = \frac{E}{5R} = \frac{-n(w_2 - w_1)}{5Rt} \]
Step 3: Final Answer:
\[ \boxed{\frac{-n(w_2 - w_1)}{5Rt}} \]
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