Question

A rectangular loop of sides $10\text{cm}$ and $5\text{cm}$ carrying a current $I$ of $12\text{A}$ is placed in different orientations as shown in the figures below ;


If there is a uniform magnetic field of $0.3\text{T}$ in the positive $z$ direction, in which orientations the loop would be in $\left(i\right)$ stable equilibrium and $\left(ii\right)$ unstable equilibrium ?

• A. (b) and (d), respectively
• B. (b) and (c), respectively
• C. (a) and (b), respectively
• D. (a) and (c), respectively

Answer 1

The correct option is A (b) and (d), respectively

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

Here,
${\overrightarrow{M}}_a=6\times {10}^{-2}\hat{i}{\text{Am}}^2,{\overrightarrow{M}}_b=6\times {10}^{-2}\hat{k}{\text{Am}}^2$

${\overrightarrow{M}}_c=-6\times {10}^{-2}\hat{j}{\text{Am}}^2,{\overrightarrow{M}}_d=-6\times {10}^{-2}\hat{k}{\text{Am}}^2$
Using potential energy $U=-\overrightarrow{M}.\overrightarrow{B}$
${\overrightarrow{M}}_b$​​ is parallel to $\overrightarrow{B}$, it means potential energy is minimum, therefore in orientation (b) the loop is in stable equilibrium.
${\overrightarrow{M}}_d$ is antiparallel to $\overrightarrow{B}$, it means potential energy is maximum, therefore in orientation (d) the loop is in unstable equilibrium.