Question

As shown in figure, a metal rod makes contact with a partial circuit and completes the circuit. The circuit area is perpendicular to a magnetic field with B = 0.15 T. If the resistance of the total circuit is $3\Omega$ , the force needed to move the rod as indicated with a constant speed of $2m{s}^{-1}$ will be equal to

• A. $3.75\times {10}^{-3}N$
• B. $2.75\times {10}^{-3}N$
• C. $6.75\times {10}^{-4}N$
• D. $4.36\times {10}^{-4}N$

Answer 1

The correct option is A

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

$|ϵ|=Blv=\left(0.15T\right)\left(0.5m\right)\left(2m{s}^{-1}\right)=0.15V$

Current induced in the rod is $I=\frac{|ϵ|}{r}=\frac{0.15V}{3\Omega }=0.05A$

$\therefore F=IlBsin{90}^{\circ }=\left(0.05A\right)\left(0.5m\right)\left(0.15T\right)$

$=3.75\times {10}^{-3}N$