Q139 / 150 · MHT CET 2026 (17 April Shift 2) · Physics
A straight line conductor of length 0.4 m is moved with a speed of $7.0 \text{ ms}^{-1}$ perpendicular to magnetic field of intensity $0.8 \text{ Wb m}^{-2}$. The induced e.m.f. across the conductor is
- Correct answerA$2 \cdot 24$ V
- B$2 \cdot 80$ V
- C$3 \cdot 20$ V
- D$5 \cdot 60$ V
Solution
Given $l = 0.4 \text{ m}$, $v = 7.0 \text{ ms}^{-1}$, and $B = 0.8 \text{ Wb m}^{-2}$.
The induced e.m.f. across a straight conductor moving perpendicular to a magnetic field is given by $e = B l v$.
Substituting the given values:
$e = 0.8 \times 0.4 \times 7.0$
$e = 2.24 \text{ V}$
Answer: $2 \cdot 24$ V
The induced e.m.f. across a straight conductor moving perpendicular to a magnetic field is given by $e = B l v$.
Substituting the given values:
$e = 0.8 \times 0.4 \times 7.0$
$e = 2.24 \text{ V}$
Answer: $2 \cdot 24$ V
Hence, the correct answer is option: 'A'.