Question

A coil having effective area $A$, is held with its plane normal to magnetic field of induction $B$. The magnetic induction is quickly reduced by $25%$ of its initial value in $2$ second. Then the e.m.f. induced across the coil will be

• A. \frac{AB}{8}
• B. \frac{AB}{2}
• C. \frac{3AB}{4}
• D. \frac{3AB}{8}

Hint:

Induced EMF is directly proportional to the rate of change of magnetic flux. Use $e = \frac{\Delta \phi}{\Delta t}$ to find the average induced EMF.

Answer 1

The Correct Option is A

Step 1: The magnetic flux $\phi$ through a coil of area $A$ in a magnetic field $B$ is given by: \[ \phi = BA \cos(\theta) \] Since the plane of the coil is normal to the magnetic field, $\theta = 0^\circ$, so the initial flux is $\phi_1 = BA$.
Step 2: The magnetic field is reduced by $25%$ of its initial value, so the final magnetic field $B_2$ is: \[ B_2 = B - \frac{25}{100}B = B - \frac{1}{4}B = \frac{3}{4}B \]
Step 3: The final magnetic flux $\phi_2$ through the coil is: \[ \phi_2 = B_2 A = \frac{3}{4}BA \]
Step 4: According to Faraday's law of electromagnetic induction, the magnitude of the induced e.m.f. $e$ is the rate of change of magnetic flux: \[ e = \frac{\Delta \phi}{\Delta t} = \frac{\phi_1 - \phi_2}{\Delta t} \]
Step 5: Substituting the values and given time $\Delta t = 2$ seconds: \[ e = \frac{BA - \frac{3}{4}BA}{2} = \frac{\frac{1}{4}BA}{2} = \frac{AB}{8} \]