Question

A magnetic field of $1$~T applied at an angle $\pi/3$ to the vertical direction is decreased to zero at a steady rate in one second. The magnitude of induced emf in a horizontally placed circular loop of radius $5$~cm is given by

• A. \( 1.25 \sqrt{3} \pi \text{ mV} \)
• B. \( 12.5 \sqrt{3} \pi \text{ V} \)
• C. \( 1.25 \pi \text{ mV} \)
• D. \( 12.5 \pi \text{ V} \)
• E. \( 25 \pi \text{ V} \)

Hint:

Always verify if the angle is with the plane or the normal. Since the loop is horizontal and the field is relative to the vertical, the vertical is the normal.

Answer 1

The Correct Option is C

Concept: According to Faraday's Law, induced emf is the rate of change of magnetic flux (\( \Phi = BA \cos \theta \)). The "vertical direction" is the normal to the "horizontal loop," so \( \theta = \pi/3 \).

Step 1: {Identify the parameters from Screenshot 2026-04-29 225540.jpg.}
Magnetic field change: \( \Delta B = 1 - 0 = 1 \text{ T} \).
Time: \( \Delta t = 1 \text{ s} \).
Radius: \( r = 5 \text{ cm} = 0.05 \text{ m} \).
Area: \( A = \pi r^2 = \pi (0.05)^2 = 0.0025\pi \text{ m}^2 \).

Step 2: {Calculate the change in flux.}
\[ \Delta \Phi = (\Delta B) A \cos(\pi/3) \] \[ \Delta \Phi = (1) (0.0025\pi) \left(\frac{1}{2}\right) = 0.00125\pi \text{ Wb} \]

Step 3: {Determine the magnitude of induced emf.}
\[ |\varepsilon| = \frac{\Delta \Phi}{\Delta t} = \frac{0.00125\pi}{1} = 0.00125\pi \text{ V} \] \[ |\varepsilon| = 1.25\pi \text{ mV} \]