Question

A square loop of side 2 cm makes angle 60° with magnetic field \( \mathbf{B} = 0.4 \sin(300t) \, T \). Find maximum induced EMF (in mV) in loop.

• A. 24
• B. 36
• C. 48
• D. 16

Hint:

To find the maximum induced EMF, differentiate the magnetic flux with respect to time and use the fact that the maximum value of the cosine function is 1.

Answer 1

The Correct Option is A

The induced EMF in a loop is given by Faraday's Law of Induction:
\[ \epsilon = - \frac{d\Phi}{dt} \] Where \( \Phi \) is the magnetic flux given by:
\[ \Phi = B A \cos \theta \] Here: - \( B = 0.4 \sin(300t) \, T \) (time-dependent magnetic field) - \( A = (2 \, \text{cm})^2 = (0.02 \, \text{m})^2 = 4 \times 10^{-4} \, \text{m}^2 \) (area of the square loop) - \( \theta = 60^\circ \) (angle between the magnetic field and the plane of the loop) Step 1: Calculate the magnetic flux.
\[ \Phi = B A \cos \theta = (0.4 \sin(300t)) \times (4 \times 10^{-4}) \times \cos(60^\circ) \] Since \( \cos(60^\circ) = \frac{1}{2} \), we get:
\[ \Phi = 0.4 \times 4 \times 10^{-4} \times \sin(300t) \times \frac{1}{2} \] \[ \Phi = 8 \times 10^{-5} \sin(300t) \, \text{Wb} \]
Step 2: Differentiate the flux to find the induced EMF.
\[ \epsilon = - \frac{d\Phi}{dt} \] \[ \epsilon = - \frac{d}{dt} \left( 8 \times 10^{-5} \sin(300t) \right) \] \[ \epsilon = - 8 \times 10^{-5} \times 300 \cos(300t) \] \[ \epsilon = - 2.4 \times 10^{-2} \cos(300t) \, \text{V} \]
Step 3: Find the maximum EMF.
The maximum value of \( \cos(300t) \) is 1, so the maximum EMF is:
\[ \epsilon_{\text{max}} = 2.4 \times 10^{-2} \, \text{V} = 24 \, \text{mV} \] Thus, the maximum induced EMF is 24 mV.
Final Answer: 24 mV