Question

A coil of 200 turns and area $0.20 \, \text{m}^2$ is rotated at half a revolution per second and is placed in a uniform magnetic field of $0.01 \, \text{T}$ perpendicular to the axis of rotation of the coil. The maximum voltage generated in the coil is \[\frac{2\pi}{\beta} \, \text{volt}.\]The value of $\beta$ is:

Answer 1

Correct Answer: 5

Given: - Number of turns: \( N = 200 \) - Area of the coil: \( A = 0.20 \, \text{m}^2 \) - Magnetic field: \( B = 0.01 \, \text{T} \) - Frequency of rotation: \( f = 0.5 \, \text{Hz} \)

Step 1: Calculating the Angular Velocity

The angular velocity \( \omega \) is given by:

\[ \omega = 2\pi f \]

Substituting the given frequency:

\[ \omega = 2\pi \times 0.5 = \pi \, \text{rad/s} \]

Step 2: Calculating the Maximum EMF

The maximum induced EMF (voltage) in the rotating coil is given by Faraday's law of electromagnetic induction:

\[ \text{EMF}_{\text{max}} = NAB\omega \]

Substituting the given values:

\[ \text{EMF}_{\text{max}} = 200 \times 0.20 \, \text{m}^2 \times 0.01 \, \text{T} \times \pi \, \text{rad/s} \] \[ \text{EMF}_{\text{max}} = 200 \times 0.002 \times \pi \] \[ \text{EMF}_{\text{max}} = 0.4\pi \, \text{volt} \]

Step 3: Relating the Maximum EMF to the Given Expression

The given expression for the maximum voltage is:

\[ \text{EMF}_{\text{max}} = \frac{2\pi}{\beta} \, \text{volt} \]

Equating the two expressions:

\[ 0.4\pi = \frac{2\pi}{\beta} \]

Cancelling \( \pi \) from both sides:

\[ 0.4 = \frac{2}{\beta} \]

Rearranging to find \( \beta \):

\[ \beta = \frac{2}{0.4} \] \[ \beta = 5 \]

Conclusion:

The value of \( \beta \) is 5.