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:
Correct Answer: 5
Approach Solution - 1
To determine the value of β in the given problem, we apply the formula for the maximum voltage (emf) induced in a rotating coil: \( E_{\text{max}} = NAB\omega \). Here:
- \( N = 200 \) (number of turns)
- \( A = 0.20 \, \text{m}^2 \) (area of the coil)
- \( B = 0.01 \, \text{T} \) (magnetic field strength)
- \( \omega \) (angular speed in radians per second needs to be calculated)
The coil is rotated at half a revolution per second. The angular speed \(\omega\) in radians per second is given by:
\(\omega = 0.5 \times 2\pi = \pi \, \text{rad/s}\)
Now, substituting these values into the formula for \( E_{\text{max}} \):
\( E_{\text{max}} = 200 \times 0.20 \times 0.01 \times \pi = 0.4\pi \, \text{V}\)
We are told \( E_{\text{max}} = \frac{2\pi}{\beta} \), so we equate and solve for β:
\( 0.4\pi = \frac{2\pi}{\beta} \)
\( \beta = \frac{2\pi}{0.4\pi} = \frac{2}{0.4} = 5 \)
Therefore, the value of β is 5.
Approach Solution -2
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.
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