Question

The square loop in the figure has sides of length $20\text{cm}$. It has $5$ turns and carries a current of $2\text{A}$. The normal to the loop is at ${37}^{\circ }$ to a uniform field, $B=0.5\hat{j}\text{T}$. Find the work needed to rotate the loop from its position of minimum energy to the given orientation.

• A. $-0.04\text{J}$
• B. $+0.04\text{J}$
• C. $-0.02\text{J}$
• D. $+0.02\text{J}$

Answer 1

The correct option is B $+0.04\text{J}$



The potential energy of the loop is,

$U=-\mu B\cos \theta$

Where,

$\mu =NIA=5\times 2\times (0.2{)}^2=0.4{\text{Am}}^2$

And the position of minimum energy is $\theta =0^{\circ }$.

Thus, the external work, $W_{ext}$ needed to rotate it to the given orientation, is given by

$W_{ext}=U_f-U_i$

$=(-\mu B\cos {37}^{\circ })-(-\mu B\cos 0^{\circ })$

$=\left(0.4\right)\left(0.5\right)(1-0.8)$

$=0.04\text{J}$

The external work is positive since the dipole is rotated away from alignment with the field.

Hence, option (b) is the correct answer.