Question

A rectangular loop of sides $10$cm and $5$cm carrying a current I of $12$A is placed in different orientations as shown in the figure.
If there is a uniform magnetic field of $0.3$T in the positive z-direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium.

• A. $\left(2\right)$ and $\left(4\right)$, respectively
• B. $\left(2\right)$ and $\left(3\right)$, respectively
• C. $\left(1\right)$ and $\left(2\right)$, respectively
• D. $\left(1\right)$ and $\left(3\right)$, respectively

Answer 1

The correct option is A $\left(2\right)$ and $\left(4\right)$, respectively

$I=12$A, $\overline{B}=0.3\hat{k}T$,
$A=10\times 5c{m}^2=50\times {10}^{-4}{m}^2$
$\overrightarrow{M}=IA\hat{n}=12\times 50\times {10}^{-4}\hat{n}A{m}^2$
$=6\times {10}^{-2}\hat{n}A{m}^2$

Here, $\overrightarrow{M_1}=6\times {10}^{-2}\hat{i}A{m}^2$, $\overrightarrow{M_2}=6\times {10}^{-2}\hat{k}$A $m^2$

$\overrightarrow{M_3}=-6\times {10}^{-2}\hat{j}$A $m^2,\overrightarrow{M_4}=-6\times {10}^{-2}\hat{k}$A $m^2$

$\overrightarrow{M_2}$ is parallel to $\overrightarrow{B}$, it means potential energy is minimum, therefore in orientation $\left(2\right)$ the loop is in stable equilibrium.

$\overrightarrow{M_4}$ is antiparallel to $\overrightarrow{B}$, it means potential energy is maximum, therefore in orientation $\left(4\right)$ the loop is in unstable equilibrium.