Q137 / 180 · NEET 2026 (Cancelled) · Physics
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$1.2 \times 10^{-4}$ volt
- C$1.3 \times 10^{-4}$ volt
- Correct answerD$1.8 \times 10^{-4}$ volt
Solution
Given:
Magnetic field, $B = 0.3$ T
Velocity of the loop, $v = 2$ cm s$^{-1} = 2 \times 10^{-2}$ m s$^{-1}$
Length of the shorter side, $l = 3$ cm $= 3 \times 10^{-2}$ m
Since the loop is moving in a direction normal to the shorter side, the velocity vector is perpendicular to the shorter side. The motional emf is induced across the side that is perpendicular to the direction of motion.
The induced emf $e$ is given by the formula:
$e = B l v$
Substituting the given values:
$e = 0.3 \times (3 \times 10^{-2}) \times (2 \times 10^{-2})$
$e = 1.8 \times 10^{-4}$ V
Answer: $1.8 \times 10^{-4}$ volt
Magnetic field, $B = 0.3$ T
Velocity of the loop, $v = 2$ cm s$^{-1} = 2 \times 10^{-2}$ m s$^{-1}$
Length of the shorter side, $l = 3$ cm $= 3 \times 10^{-2}$ m
Since the loop is moving in a direction normal to the shorter side, the velocity vector is perpendicular to the shorter side. The motional emf is induced across the side that is perpendicular to the direction of motion.
The induced emf $e$ is given by the formula:
$e = B l v$
Substituting the given values:
$e = 0.3 \times (3 \times 10^{-2}) \times (2 \times 10^{-2})$
$e = 1.8 \times 10^{-4}$ V
Answer: $1.8 \times 10^{-4}$ volt
Hence, the correct answer is option: 'D'.