Question

A rectangular loop PQRS made from a uniform wire has length a, width b and mass m. It is free to rotate about the arm PQ, which remains hinged along a horizontal line taken as the v-axis (see figure). Take the vertically upward direction as the z-axis. A uniform magnetic field$B=(3\hat{i}+4\hat{k})B_0$ exists in the region. The loop is held in the x-y plane and a current I is passed through it. The loop is now released and is found to stay in the horizontal position in equilibrium. (a) What is the direction of the current / in PQ ? (b) Find the magnetic force on the arm RS. (c) Find the expression for / in terms of $B_0$,a ,b and m.

Answer 1

Let the direction of current in wire PQ is from P to Q and its
magnitude be I
The magnetic moment of the given loop is
$M=-Iab\hat{k}$
Torque on the loop due to magnetic forces is
$\tau =r\times F=\left(\frac{a}{2}\hat{i}\right)\times (-mg\hat{K})=\frac{mga}{2}\hat{j}$
We see that when the current in the wire PQ is from P to Q,
${\tau }_1$ and ${\tau }_2$ are in opposite directions, so they can cancel each
other and the loop may remain in equilibrium. So, the
direction of current I in wire PQ is from P to Q. Further for
equilibrium of the loop
$|{\tau }_1|=|{\tau }_2|$
or $3Iab{B}_0=\frac{mg}{6b{B}_0}$
(b) Magnetic force on wire RS is
$F=I(I\times B)=I[(-b\hat{j}\times (3\hat{i}+4\hat{k})B_0]$
$F=Ib{B}_0(3\hat{k}-4\hat{i})$