A rod of length 1.0 m is rotated in a plane perpendicular to a uniform magnetic field of induction 0.25 T with a frequency of 12 rev / s. The induced emf across the ends of the rod is

Question

A rod of length 1.0 m is rotated in a plane perpendicular to a uniform magnetic field of induction 0.25 T with a frequency of 12 rev / s. The induced emf across the ends of the rod is (A) 18.89 V (B) 3 V (C) 15 V (D) 9.42 V

Tardigrade Admin · Accepted Answer

Induced emf e=π l2 B n=(l2 B ω/2) =((1.0)2 × 0.25 × 2 × 3.14 × 12/2)=9.42 V

Q. A rod of length $1.0 m$ is rotated in a plane perpendicular to a uniform magnetic field of induction $0.25 T$ with a frequency of $12 re v / s$ . The induced emf across the ends of the rod is

18.89 V

3 V

15 V

9.42 V

Induced emf $e = π l^{2} B n = \frac{l ^{2} B ω}{2}$
$= \frac{( 1.0 ) ^{2} \times 0.25 \times 2 \times 3.14 \times 12}{2} = 9.42 V$

Q. A rod of length 1.0m is rotated in a plane perpendicular to a uniform magnetic field of induction 0.25T with a frequency of 12rev/s. The induced emf across the ends of the rod is

18.89 V

3 V

15 V

9.42 V

Induced emf e=πl2Bn=2l2Bω​ =2(1.0)2×0.25×2×3.14×12​=9.42V

Q. A rod of length $1.0 m$ is rotated in a plane perpendicular to a uniform magnetic field of induction $0.25 T$ with a frequency of $12 re v / s$ . The induced emf across the ends of the rod is

Induced emf $e = π l^{2} B n = \frac{l ^{2} B ω}{2}$
$= \frac{( 1.0 ) ^{2} \times 0.25 \times 2 \times 3.14 \times 12}{2} = 9.42 V$