Question

A triangular loop of side $a=10\text{m}$ carrying current of $2\text{A}$ is kept in a uniform magnetic field of $30\text{mT}$ as shown in the figures below. Find the torque about its center in all the three cases in the same order. (Assume the triangular loop is in the first quadrant of $x-y$ plane with $BC$ parallel to $x$ axis)

• A. $0;0;0$
• B. $0;-2.6\hat{i};2.6\hat{j}$
• C. $0;2.6\hat{i};2.6\hat{j}$
• D. $0;-2.6\hat{i};-2.6\hat{j}$

Answer 1

The correct option is B $0;-2.6\hat{i};2.6\hat{j}$

Torque due to a current carrying loop is given by $$\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}$$From the given figure,

Magnetic moment $\overrightarrow{\mu }=2\times \frac{\sqrt{3}}{4}(10{)}^2\hat{k}{\text{Am}}^2$

Case-I:

Magnetic field $\overrightarrow{B}=30\times {10}^{-3}\hat{k}\text{T}$

$\therefore \overrightarrow{\tau }=\frac{\sqrt{3}}{2}(10{)}^2\times 30\times {10}^{-3}(\hat{k}\times \hat{k})=\overrightarrow{0}$

Case-II:

Magnetic field $\overrightarrow{B}=30\times {10}^{-3}\hat{j}\text{T}$

$\therefore \overrightarrow{\tau }=\frac{\sqrt{3}}{2}\times (10{)}^2\times 30\times {10}^{-3}(\hat{k}\times \hat{j})=-2.6\hat{i}$

Case-III:

Magnetic field $\overrightarrow{B}=30\times {10}^{-3}\hat{i}\text{T}$

$\therefore \overrightarrow{\tau }=\frac{\sqrt{3}}{2}\times (10{)}^2\times 30\times {10}^{-3}(\hat{k}\times \hat{i})=2.6\hat{j}$

Hence, option (b) is the correct answer.