Question

A conducting circular loop is placed in a uniform magnetic field $ B = 0.125 \, \text{T} $ with its plane perpendicular to the loop. If the radius of the loop is made to shrink at a constant rate of 2 mm/s, then the induced emf when the radius is 4 cm is:

• A. \( 0.52 \, \mu V \)
• B. \( 20 \, \mu V \)
• C. \( \frac{3}{5} \, \mu V \)
• D. \( \frac{3}{2} \, \mu V \)

Hint:

In problems involving changing magnetic flux, apply Faraday's Law to compute the induced emf. Pay attention to the rate of change of area and the magnetic field.

Answer 1

The Correct Option is B

The induced emf in a conducting loop moving in a magnetic field is given by Faraday's Law of Induction: \[ \text{emf} = - \frac{d\Phi}{dt} \] Where: - \( \Phi \) is the magnetic flux, - \( \frac{d\Phi}{dt} \) is the rate of change of flux. The magnetic flux is given by: \[ \Phi = B \cdot A = B \cdot \pi r^2 \] Here: - \( B = 0.125 \, \text{T} \) is the magnetic field, - \( r = 4 \, \text{cm} = 0.04 \, \text{m} \) is the radius of the loop. Since the radius is shrinking at a rate of \( \frac{dr}{dt} = -2 \, \text{mm/s} = -2 \times 10^{-3} \, \text{m/s} \), we can differentiate the flux: \[ \frac{d\Phi}{dt} = B \cdot \frac{d}{dt} \left( \pi r^2 \right) = B \cdot 2\pi r \cdot \frac{dr}{dt} \] Substitute the values: \[ \frac{d\Phi}{dt} = 0.125 \cdot 2\pi \cdot 0.04 \cdot (-2 \times 10^{-3}) \] Simplifying this: \[ \frac{d\Phi}{dt} = -0.125 \cdot 2\pi \cdot 0.04 \cdot 2 \times 10^{-3} = - 20 \times 10^{-6} \, \text{V} = - 20 \, \mu V \]
Thus, the induced emf is \( 20 \, \mu V \).