Question

A rectangular wire loop of sides 8 cm and 3 cm with a small cut, is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the plane of the loop. The emf developed across the cut, if the velocity of the loop is 2 cm s⁻¹, in a direction normal to the shorter side of the loop, will be: ____.

• A. 4.8 $\times$ 10⁻⁴ volt
• B. 1.3 $\times$ 10⁻⁴ volt
• C. 1.2 $\times$ 10⁻⁴ volt
• D. 1.8 $\times$ 10⁻⁴ volt

Hint:

If velocity is normal to the short side, it means the loop is moving along the direction of the short side, so the long side acts as the "cutting" length $L$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When a conducting loop moves through a magnetic field, a motional electromotive force (emf) is induced across the segments of the loop that cut the magnetic field lines.

Step 2: Key Formula or Approach:
\[ e = BvL \] Where $L$ is the length of the side that is perpendicular to the velocity and cutting the field lines.

Step 3: Detailed Explanation:
1. Given: $B = 0.3$ T, $v = 2 \text{ cm/s} = 0.02 \text{ m/s}$. 2. The velocity is normal to the shorter side (3 cm). This means the longer side (8 cm) is the one cutting the magnetic field lines as it exits. 3. Therefore, $L = 8 \text{ cm} = 0.08 \text{ m}$. 4. Calculate emf: \[ e = 0.3 \times 0.02 \times 0.08 \] \[ e = 0.3 \times 0.0016 = 0.00048 \text{ V} \] \[ e = 4.8 \times 10^{-4} \text{ V} \]

Step 4: Final Answer:
The emf developed is 4.8 $\times$ 10⁻⁴ volt.