Question

A rectangular loop is connected to a battery as shown in the figure. If the total resistance of the loop is $1\Omega$, the magnitude of net force acting on it is,
(Take $B=0.1\text{T}$)

• A. $0.6\text{N}$
• B. $1.2\text{N}$
• C. $2.4\text{N}$
• D. $6.0\text{N}$

Answer 1

The correct option is B $1.2\text{N}$


Resistance of $\text{ABCD}$ = resistance $\text{AFED}$ = $0.5\Omega$

$\therefore$ Effective resistance $=0.25\Omega (\because$ the two segments are connected in parallel)

Current in the circuit is given by,

$I=\frac{E}{R}=\frac{15}{0.25}=60\text{A}$
$I_{\text{ABCD}}=I_{\text{AFED}}=\frac{60}{2}=30\text{A}$


Here, $F_{\text{AB}}=F_{\text{FA}}=F_{\text{CD}}=F_{\text{ED}}=0$

and $F_{\text{BC}}=F_{\text{FE}}=BIl$

$=0.1\times 30\times 0.2=0.6\text{N}$

$\therefore F_{net}=F_{\text{BC}}+F_{\text{FE}}=0.6+0.6=1.2\text{N}$

Hence, option $\left(B\right)$ is the correct answer.