Question

A circular loop of wire having a radius of $8.0\text{cm}$ carries a current of $0.2\text{A}$. A vector of unit length and parallel to the dipole moment $\overrightarrow{\mu }$ of the loop is given by $0.60\hat{i}-0.80\hat{j}$. If the loop is located in uniform magnetic field given by $\overrightarrow{B}=\left(0.25\right)\hat{i}+\left(0.30\right)\hat{k}$. Find the torque acting on the loop.

• A. $(-9.6\hat{i}-7.2\hat{j}+8.0\hat{k})\times {10}^{-3}\text{Nm}$
• B. $(-9.6\hat{i}-7.2\hat{j}-8.0\hat{k})\times {10}^{-4}\text{Nm}$
• C. $(9.6\hat{i}+7.2\hat{j}+8.0\hat{k})\times {10}^{-4}\text{Nm}$
• D. $(-9.6\hat{i}-7.2\hat{j}+8.0\hat{k})\times {10}^{-4}\text{Nm}$

Answer 1

The correct option is D $(-9.6\hat{i}-7.2\hat{j}+8.0\hat{k})\times {10}^{-4}\text{Nm}$

Torque acting on a loop is given by$$\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}$$From the data given in the question,

$\overrightarrow{\mu }=I\overrightarrow{A}=0.2\times (\pi \times (0.08{)}^2)(0.6\hat{i}-0.8\hat{j})=(2.4\hat{i}-3.2\hat{j})\times {10}^{-3}{\text{Am}}^2$

$\overrightarrow{\tau }=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0.0024& -0.0032& 0.0\\ 0.25& 0.0& 0.30\end{array}\right|=\hat{i}(-0.0032\times 0.3)-\hat{j}(0.0024\times 0.3)+\hat{k}(0.0032\times 0.25)=(-9.6\hat{i}-7.2\hat{j}+8.0\hat{k})\times {10}^{-4}\text{Nm}$

Hence, option (d) is the correct answer.