Question

A square loop of side 50 cm is placed in a uniform magnetic field of 3.0 T acting perpendicular to the plane of the loop. If the loop is rotated through an angle of \(90^\circ\) in 0.3 s, the value of emf induced in the loop would be :

• A. 0.25 V
• B. 0.50 V
• C. 0.75 V
• D. 1.0 V

Hint:

Always remember that magnetic flux depends on the orientation of the loop: \(\Phi = BA\cos\theta\). When a loop is rotated by \(90^\circ\) from a perpendicular field position, the flux drops completely to zero, making the change in flux equal to the initial flux.

Answer 1

The Correct Option is C

Concept: Faraday's law: the average induced emf is the rate of change of magnetic flux, \(e = \left|\dfrac{\Delta\Phi}{\Delta t}\right|\), where \(\Phi = BA\cos\theta\) and \(\theta\) is the angle between \(\vec{B}\) and the area vector of the loop.

Step 1: Area of the loop.
Side \(= 50\text{ cm} = 0.5\text{ m}\), so \(A = (0.5)^2 = 0.25\text{ m}^2\).

Step 2: Initial and final flux.
The field is perpendicular to the plane of the loop, so initially \(\theta = 0^\circ\):
\[ \Phi_1 = BA\cos 0^\circ = 3.0 \times 0.25 = 0.75\text{ Wb} \]
After rotation through \(90^\circ\), \(\theta = 90^\circ\):
\[ \Phi_2 = BA\cos 90^\circ = 0 \]
Change in flux: \(|\Delta\Phi| = 0.75\text{ Wb}\).

Step 3: Induced emf.
\[ e = \frac{|\Delta\Phi|}{\Delta t} = \frac{0.75}{0.3} = 2.5\text{ V} \]

Step 4: Final Answer
With the data exactly as printed (side 50 cm, \(B = 3.0\text{ T}\), \(\Delta t = 0.3\text{ s}\)) the average induced emf is \(2.5\text{ V}\). This value does not appear among the printed options (A) 0.25 V, (B) 0.50 V, (C) 0.75 V, (D) 1.0 V, so the question as printed has no exactly matching option.