Question

The magnetic flux \( \phi \) through a stationary loop of wire having a resistance \( R \) varies with time as \( \phi = 4t^2 + 3t \). The average emf and total charge flowing in the loop in the time interval \( t = 0 \) to \( t = \tau \) respectively are

• A. \( \tau + 3, \; \frac{4\tau^2 + 3\tau}{R} \)
• B. \( 3\tau + 4, \; \frac{4\tau^2 + 3\tau}{R} \)
• C. \( 4\tau + 3, \; \frac{4\tau^2 + 3\tau}{R} \)
• D. \( 4\tau + 3, \; \frac{4\tau^2 + 3\tau}{R} \)

Hint:

Average emf over time interval can be found using integration, while total charge is obtained by integrating current \(I = \frac{\mathcal{E}}{R}\).

Answer 1

The Correct Option is D

Step 1: Use Faraday’s law of electromagnetic induction.
\[ \mathcal{E} = -\frac{d\phi}{dt} \]

Step 2: Differentiate the flux.
\[ \phi = 4t^2 + 3t \] \[ \frac{d\phi}{dt} = 8t + 3 \]

Step 3: Expression for instantaneous emf.
\[ \mathcal{E} = -(8t + 3) \]

Step 4: Find average emf from \(0\) to \(\tau\).
\[ \mathcal{E}_{avg} = \frac{1}{\tau} \int_{0}^{\tau} (8t + 3)\, dt \]
\[ \mathcal{E}_{avg} = \frac{1}{\tau} \left[4t^2 + 3t\right]_0^{\tau} \]
\[ \mathcal{E}_{avg} = \frac{4\tau^2 + 3\tau}{\tau} = 4\tau + 3 \]

Step 5: Find total charge flown.
Using \( I = \frac{\mathcal{E}}{R} \), total charge is:
\[ Q = \int I\, dt = \frac{1}{R} \int \mathcal{E}\, dt \]
\[ Q = \frac{1}{R} \int_{0}^{\tau} (8t + 3)\, dt \]

Step 6: Evaluate the integral.
\[ Q = \frac{1}{R} \left[4t^2 + 3t\right]_0^{\tau} \]
\[ Q = \frac{4\tau^2 + 3\tau}{R} \]

Step 7: Final conclusion.
\[ \boxed{4\tau + 3, \; \frac{4\tau^2 + 3\tau}{R}} \] Hence, correct answer is option (D).