Question

A coil of 'n' turns and resistance \(R\,\Omega\) is connected in series with a resistance \(R/4\). The combination is moved for time 't' second through flux \(\Phi _1\) to \(\Phi _2\). The induced current in the circuit is

• A. \(\frac{4n(\Phi _1-\Phi _2)}{5RT}\)
• B. \(\frac{5n^2(\Phi _1-\Phi _2)}{4RT}\)
• C. \(\frac{2n(\Phi _1-\Phi _2)}{3RT}\)
• D. \(\frac{3n(\Phi _1-\Phi _2)}{4RT}\)

Hint:

EMF is n times flux change over t; resistance is R + R/4.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
By Faraday's law, the induced emf in a coil of \(n\) turns is \(\varepsilon=n\dfrac{\Delta\Phi}{t}\). By Ohm's law, the current is \(I=\dfrac\varepsilon{R_{total}}\).

Step 2: EMF:
The flux changes from \(\Phi_1\) to \(\Phi_2\) in time \(t\), so the magnitude is
\[ \varepsilon=\frac{n(\Phi_1-\Phi_2)}{t} \]

Step 3: Total resistance:
The coil resistance \(R\) is in series with \(\dfrac R4\), so \(R_{total}=\dfrac{5R}4\).

Step 4: Current:
\[ I=\frac{n(\Phi_1-\Phi_2)/t}{5R/4}=\frac{4n(\Phi_1-\Phi_2)}{5Rt} \]

Step 5: Choose:
Option (A). Option (B) has \(n^2\), which would be wrong since the emf has only one factor of \(n\).

Final Answer:
The induced current is 4n(flux change)/(5Rt). \[ \boxed{I=\frac{4n(\Phi_1-\Phi_2)}{5Rt}} \]