Question

The current in a coil of inductance $0.2 \,H$ changes from $5A$ to $2A$ in $0.5\,sec$. The magnitude of the average induced emf in the coil is

• A. 0.6 V
• B. 1.2 V
• C. 30 V
• D. 0.3 V

Answer 1

The Correct Option is B

The induced emf in a coil is given by the formula: \[ \text{emf} = -L \frac{\Delta I}{\Delta t} \] Where:
\( L \) is the inductance of the coil,
\( \Delta I \) is the change in current,
\( \Delta t \) is the time interval during which the change occurs. Given:
 \( L = 0.2 \, \text{H} \), - \( \Delta I = 5 \, \text{A} - 2 \, \text{A} = 3 \, \text{A} \),
\( \Delta t = 0.5 \, \text{sec} \).
Substitute the values into the formula: \[ \text{emf} = -0.2 \cdot \frac{3}{0.5} = -0.2 \cdot 6 = -1.2 \, \text{V} \] The negative sign indicates the direction of the induced emf, but the magnitude is \( 1.2 \, \text{V} \).

Answer 2

The induced emf in a coil is given by Faraday's law of electromagnetic induction: \[ \varepsilon = -L \frac{\Delta I}{\Delta t} \] where:
\( \varepsilon \) is the induced emf,
\( L \) is the inductance of the coil,
\( \Delta I \) is the change in current,
\( \Delta t \) is the time interval during which the current changes. Given:
\( L = 0.2 \, \text{H} \),
Initial current \( I_{\text{initial}} = 5 \, \text{A} \),
Final current \( I_{\text{final}} = 2 \, \text{A} \),
\( \Delta t = 0.5 \, \text{sec} \).

The change in current is: \[ \Delta I = I_{\text{final}} - I_{\text{initial}} = 2 \, \text{A} - 5 \, \text{A} = -3 \, \text{A} \] Now, substituting the values into the formula for induced emf: \[ \varepsilon = -0.2 \times \frac{-3}{0.5} = 0.2 \times 6 = 1.2 \, \text{V} \] Thus, the magnitude of the average induced emf is \({1.2 \, \text{V}} \).