Question

A square loop $EFGH$ of side $a$ , mass $m$ and total resistance $R$ is falling under gravity in a region of transverse non-uniform magnetic field given by $B=B_0\frac{y}{a}$ , where $B_0$ is a positive constant and $y$ is the position of side $EF$ of the loop. If at some instant the speed of the loop is $v$ , then the total Lorentz force acting on the loop is

• A. $F=\frac{B_0^2{a}^{2}v}{2R}$
• B. $F=\frac{2B_0^2{a}^{2}v}{R}$
• C. $F=\frac{B_0^2{a}^{2}v}{R}$
• D. zero

Answer 1

Answer: (C)

Motional emf in EH and FG = 0 as $\overrightarrow{v}||\overrightarrow{I}$
Motional emf in EF is $e_1=\left(\frac{{\left(\text{B}\right)}_0\text{y}}{\text{a}}\right)\left(\text{a}\right)v={\left(\text{B}\right)}_0\text{y}v$
Similarly, motional emf in GH will be
${\left(\text{e}\right)}_2=\left\{\frac{{\left(\text{B}\right)}_0\left(\text{y}+\text{a}\right)}{\text{a}}\right\}\left(\text{a}\right)\left(v\right)={\left(\text{B}\right)}_0\left(\text{y}+\text{a}\right)v$
Polarities of e₁and e₂ are shown in adjoining figures.

So the net emf is
$e=e_2-e_{1}e=B_{0}av$
$\text{i}=\frac{\text{e}}{\text{R}}=\frac{{\text{B}}_0\text{a}v}{\text{R}}$

${\left(\text{F}\right)}_{\text{EF}}=\left(\frac{B_{0}av}{\text{R}}\right)\left(\text{a}\right)\left(\frac{B_0\text{y}}{\text{a}}\right)$ (downwards)
${\left(\text{F}\right)}_{\text{GH}}=\left(\frac{{\left(\text{B}\right)}_0\text{a}v}{\text{R}}\right)\left(\text{a}\right)\left(\frac{{\left(\text{B}\right)}_0\left(\text{y}+\text{a}\right)}{\text{a}}\right)$
${\left(\text{F}\right)}_{\text{GH}}=\left(\frac{{\text{B}}_0^2\text{a}v}{\text{R}}\right)\left(\text{y}+\text{a}\right)$ (upwards)
Net Lorentz force on the loop
$F_{net}={\text{F}}_{\text{GH}}-{\text{F}}_{\text{EF}}=\frac{{\text{B}}_0^2{\text{a}}^{2}v}{\text{R}}$
$F=\frac{B_0^2{a}^{2}v}{R}$