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 y-axis. A uniform magnetic field $\overrightarrow{B}=(3\hat{i}+4\hat{k})B_0$ 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 I in PQ?
(b) Find the magnetic force on the arm RS.
(c) Find the expression for I in terms of $B_0$, a b and m.

Answer 1

$B=(3\overrightarrow{i}+4\overrightarrow{k})B_0$

Force due to mag field on a current carrying conductor $=Idl\times B$

$\to$ Torque due to gravity about $PQ=\frac{a}{2}\overrightarrow{i}\times (-mg\overrightarrow{k})=\frac{mga}{2}\overrightarrow{J}$

$\to$ For any direction of current the force due to mag field will be equal and cancel each other on the wires PS and QR

$\to$ Direction of current should be clockwise (from P to Q) so mthat force on RS due to B

$F_B=-Ib\overrightarrow{J}\times (3\overrightarrow{i}+4\overrightarrow{J})B_0=Ib{B}_0(-4\overrightarrow{i}-3\overrightarrow{k})$

$\to$ As only degree of freedom is rotation about PQ. neglect $X$ component of $F_B$ (gets canceled by reaction forces from PS, PQ)

$\to$ Torque due to $F_B=-3Ib{B}_{0}a\overrightarrow{j}$

$\to$ For eqb, torque about $PQ=0$, so

$I=\frac{mg}{6b{B}_0}$