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 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.
Show 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.
Updated On: Apr 7, 2026
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The Correct Option is A
Solution and Explanation
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
\[ \epsilon = - \frac{d\Phi}{dt} \] 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
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