A conductor of length 5 cm is moved parallel to itself with a speed of 2 m/s, perpendicular to a uniform magnetic field of 10-3 Wb/m2 . The induced e.m.f. generated is

Question

A conductor of length 5 cm is moved parallel to itself with a speed of 2 m/s, perpendicular to a uniform magnetic field of 10-3 Wb/m2 . The induced e.m.f. generated is (A)  2× 10-3V (B)  1× 10-3V  (C)  1× 10-4V  (D)  2× 10-4V

Tardigrade Admin · Accepted Answer

e=-B v l Given, l=5 cm =5 × 10-2 m, v=2 m /s B=10-3 Wb / m 2 e=10-3 × 2 × 5 × 10-2 e=10 × 10-5 V e=1 × 10-4 V

Q. A conductor of length $5 c m$ is moved parallel to itself with a speed of $2 m / s,$ perpendicular to a uniform magnetic field of $1 0^{- 3} Wb / m^{2}$ . The induced e.m.f. generated is

$2 \times 1 0^{- 3} V$

$1 \times 1 0^{- 3} V$

$1 \times 1 0^{- 4} V$

$2 \times 1 0^{- 4} V$

$e = - B v l$
Given, $l = 5 c m = 5 \times 1 0^{- 2} m, v = 2 m / s$
$B = 1 0^{- 3} Wb / m^{2}$
$e = 1 0^{- 3} \times 2 \times 5 \times 1 0^{- 2}$
$e = 10 \times 1 0^{- 5} V$
$e = 1 \times 1 0^{- 4} V$

Q. A conductor of length 5cm is moved parallel to itself with a speed of 2m/s, perpendicular to a uniform magnetic field of 10−3Wb/m2 . The induced e.m.f. generated is

2×10−3V

1×10−3V

1×10−4V

2×10−4V