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 \widehat{i}+ 4 \widehat{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 \widehat{k}$
Torque on the loop due to magnetic forces is
$\tau = r\times F = \bigg( \frac{a}{2} \widehat{i} \bigg) \times (-mg \widehat{ K}) = \frac{mga}{2} \widehat{j}$
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 $ \, \, \, \, \, \, \, \, \, \, \, \, 3IabB_0 = \frac{ mg}{6bB_0}$
(b) Magnetic force on wire RS is
$ \, \, \, \, \, \, \, \, \, F = I( I \times B) = I[ (-b \widehat{j} \times { (3 \widehat{i} + 4 \widehat{k}) B_0 } ]$
$ \, \, \, \, \, \, \, \, F = IbB_0 (3 \widehat{ k} -4 \widehat{i})$