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Physics Question on Electromagnetic induction

A coil of 200 turns and area 0.20m20.20 \, \text{m}^2 is rotated at half a revolution per second and is placed in a uniform magnetic field of 0.01T0.01 \, \text{T} perpendicular to the axis of rotation of the coil. The maximum voltage generated in the coil is 2πβvolt.\frac{2\pi}{\beta} \, \text{volt}.The value of β\beta is:

Answer

Given: - Number of turns: N=200N = 200 - Area of the coil: A=0.20m2A = 0.20 \, \text{m}^2 - Magnetic field: B=0.01TB = 0.01 \, \text{T} - Frequency of rotation: f=0.5Hzf = 0.5 \, \text{Hz}

Step 1: Calculating the Angular Velocity

The angular velocity ω\omega is given by:

ω=2πf\omega = 2\pi f

Substituting the given frequency:

ω=2π×0.5=πrad/s\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:

EMFmax=NABω\text{EMF}_{\text{max}} = NAB\omega

Substituting the given values:

EMFmax=200×0.20m2×0.01T×πrad/s\text{EMF}_{\text{max}} = 200 \times 0.20 \, \text{m}^2 \times 0.01 \, \text{T} \times \pi \, \text{rad/s} EMFmax=200×0.002×π\text{EMF}_{\text{max}} = 200 \times 0.002 \times \pi EMFmax=0.4πvolt\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:

EMFmax=2πβvolt\text{EMF}_{\text{max}} = \frac{2\pi}{\beta} \, \text{volt}

Equating the two expressions:

0.4π=2πβ0.4\pi = \frac{2\pi}{\beta}

Cancelling π\pi from both sides:

0.4=2β0.4 = \frac{2}{\beta}

Rearranging to find β\beta:

β=20.4\beta = \frac{2}{0.4} β=5\beta = 5

Conclusion:

The value of β\beta is 5.