Question

A rectangular loop $ABCD$ is being pulled out of a magnetic field ${}^{\prime }{B}^{\prime }$ with uniform velocity ${}^{\prime }{v}^{\prime }$ by applying an external force $F$. Length $AB$ is $l$. Length $AD$ is $3l$ and total resistance of the loop is $R$. The thermal power developed in the loop and the force $F$ respectively is

• A. $\frac{B^2{v}^2{l}^2}{R},\frac{B^2{l}^{2}v}{R}$
• B. $\frac{9B^2{v}^2{l}^2}{R},\frac{9B^2{l}^{2}v}{R}$
• C. $\frac{8B^2{v}^2{l}^2}{R},\frac{8B^2{l}^{2}v}{R}$
• D. $\frac{3B^2{v}^2{l}^2}{R},\frac{3B^2{l}^{2}v}{R}$

Answer 1

The correct option is A $\frac{B^2{v}^2{l}^2}{R},\frac{B^2{l}^{2}v}{R}$

When the loop is passing out of the magnetic field, the flux through the loop decreases. Therefore, induced current will flow in such a direction as to oppose the change in flux. Since the flux into the plane is decreasing, current in the loop will flow in clockwise direction.


A rod of length $l$ moving in a uniform perpendicular magnetic field with velocity $v$ acts like a battery of emf
$ϵ=Bvl$
And current induced in the rod $i=\frac{Bvl}{R}$ where $R$ is the resistance of the loop.

Thermal power $P=i^{2}R=\frac{B^2{l}^2{v}^2}{R}$

Segment AB is a current carrying rod moving in a magnetic field. Hence it will experience the force
$\overrightarrow{F_{AB}}=i(\overrightarrow{l}\times \overrightarrow{B})$
i.e $F_{AB}=ilB$ in the direction shown


Inorder to keep the loop moving with constant velocity,
$F=F_{AB}=ilB=\left(\frac{Bvl}{R}\right)lB=\frac{B^2{l}^{2}v}{R}$