Question

As shown in the figure a metal rod makes contact and complete the circuit. The circuit is perpendicular to the magnetic field with B=0.15 tesla If the resistance is $3\Omega$ , force needed to move the rod as indicated with a constant speed of 2m/sec is

• A. $3.75\times {10}^{-3}N$
• B. $3.75\times {10}^{-2}N$
• C. $3.75\times {10}^{2}N$
• D. $3.75\times {10}^{-4}N$

Answer 1

The correct option is A $3.75\times {10}^{-3}N$

Induced current in the circuit $i=\frac{Bvl}{R}$
Magnetic force acting on the wire $F_m=Bil=B\left(\frac{Bvl}{R}\right)l$
$\Rightarrow F_m=\frac{B^{2}v{l}^2}{R}$ External force needed to move the rod with constant velocity
$\left(F_m\right)=\frac{B^{2}v{l}^2}{R}=\frac{(0.15{)}^2\times (2)\times (0.5{)}^2}{3}=3.75\times {10}^{-3}N$