Question

A rectangular loop of sides $10cm$ and $5cm$ carrying a current $I$ of $12A$ is placed in different orientations as shown in the figure below.

If there is a uniform magnetic field of $0.3T$ in the positive z-direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium.

• A. (2) and (4), respectively
• B. (2) and (3), respectively
• C. (1) and (2), respectively
• D. (1) and (3), respectively

Answer 1

Answer: (A)

$I=12A$,
$\vec{B}=0.3\hat{k}T$,
$A=10\times 5c{m}^2$
$=50\times 10^{-4}{m}^2$
$\overrightarrow{M}=IA\hat{n}=12\times 50\times 10^{-4}\hat{n}A{m}^2$
$=6\times 10^{-2}\hat{n}A{m}^2$
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 $\vec{B}$, it means potential energy is minimum, therefore in orientation $\left(2\right)$ the loop is in stable equilibrium.
$\overrightarrow{M_4}$ is antiparallel to $\vec{B}$, it means potential energy is maximum, therefore in orientation $\left(4\right)$ the loop is in unstable equilibrium.