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Physics-

General

Easy

Question

As shown in the figure a metal rod makes contact and completes the circuit. The circuit is perpendicular to the magnetic field with $B equals 0.15 blank t e s l a$ . If the resistance is $3 capital omega$ , force needed to move the rod as indicated with a constant speed of $2 m divided by s e c$ is

$3.75 \times 10^{- 3} N$
$3.75 \times 10^{- 2} N$
$3.75 \times 10^{2} N$
$3.75 \times 10^{- 4} N$

The correct answer is: $3.75 cross times 10 to the power of negative 3 end exponent blank N$

Induced current in the circuit $i equals fraction numerator B v l over denominator R end fraction$
Magnetic force acting on the wire $F subscript m end subscript equals B i l equals B open parentheses fraction numerator B v l over denominator R end fraction close parentheses l$
$rightwards double arrow F subscript m end subscript equals fraction numerator B to the power of 2 end exponent v l to the power of 2 end exponent over denominator R end fraction$ . External force needed to move the rod with constant velocity
$open parentheses F subscript m end subscript close parentheses equals fraction numerator B to the power of 2 end exponent v l to the power of 2 end exponent over denominator R end fraction equals fraction numerator open parentheses 0.15 close parentheses to the power of 2 end exponent cross times left parenthesis 2 right parenthesis cross times open parentheses 0.5 close parentheses to the power of 2 end exponent over denominator 3 end fraction equals 3.75 cross times 10 to the power of negative 3 end exponent N$

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