A circuit ABCD is held perpendicular to the uniform magnetic field of B=5× 10- 2T extending over the region PQRS and directed into the plane of the paper. The circuit is moving out of the field at a uniform speed of 0.2ms- 1 for 1.5s . During this time, the current in the 5Ω resistor is

Question

A circuit ABCD is held perpendicular to the uniform magnetic field of B=5× 10- 2T extending over the region PQRS and directed into the plane of the paper. The circuit is moving out of the field at a uniform speed of 0.2ms- 1 for 1.5s . During this time, the current in the 5Ω resistor is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-lem6xcmkpwmcoz9s.jpg" /> (A) 0.6mA form B texttoC (B) 0.9mA form B texttoC (C) 0.9mA form C texttoB (D) 0.6mA form C texttoB

Tardigrade Admin · Accepted Answer

I=(B ℓ v/R) I=(5 × 10- 2 × 0 .3 × 0 .2/5)A=0.6mA. Area and flux are decreasing. So, current flows to increase the flux. Clearly, current should be clockwise. So, it flows from B to C through 5Ω .

$0.6 m A$ form $B to C$

$0.9 m A$ form $B to C$

$0.9 m A$ form $C to B$

$0.6 m A$ form $C to B$

$I = \frac{B ℓ v}{R}$
$I = \frac{5 \times 1 0 ^{- 2} \times 0.3 \times 0.2}{5} A = 0.6 m A .$
Area and flux are decreasing. So, current flows to increase the flux. Clearly, current should be clockwise. So, it flows from $B$ to $C$ through $5Ω$ .

Q. A circuit ABCD is held perpendicular to the uniform magnetic field of B=5×10−2T extending over the region PQRS and directed into the plane of the paper. The circuit is moving out of the field at a uniform speed of 0.2ms−1 for 1.5s . During this time, the current in the 5Ω resistor is

0.6mA form BtoC

0.9mA form BtoC

0.9mA form CtoB

0.6mA form CtoB

I=RBℓv​ I=55×10−2×0.3×0.2​A=0.6mA. Area and flux are decreasing. So, current flows to increase the flux. Clearly, current should be clockwise. So, it flows from B to C through 5Ω .

$0.6 m A$ form $B to C$

$0.9 m A$ form $B to C$

$0.9 m A$ form $C to B$

$0.6 m A$ form $C to B$

$I = \frac{B ℓ v}{R}$
$I = \frac{5 \times 1 0 ^{- 2} \times 0.3 \times 0.2}{5} A = 0.6 m A .$
Area and flux are decreasing. So, current flows to increase the flux. Clearly, current should be clockwise. So, it flows from $B$ to $C$ through $5Ω$ .