phy magnetic torque

(a) There is no force acting on the sides AD and BC as current and field lines are in the same direction.

Force on AB = (50x10^-3) x 0.2 x 0.5 N = 5 x 10^-3 N
The force is of direction into the paper .
Force on CD = force on AB = 5 x 10^-3 N
but in direction out of the paper

Hece, torque = (5 x 10^-3) x 0.2 N.m = 10^-3 N.m

(b) Resolve the magnetic field lines in directions perpendicular to AB and parallel to AB.

The component perpendicular to AB gives rise to a magentic force, but not the one parallel to AB.
Force acting on AB = (50x10^-3).cos(45) x 0.2 x 0.5 N = 3.54 x 10^-3 N (in direction out of paper)
Force acting on CD = force acting on AB = 3.54 x 10^-3 N (in direction into paper)

For side BC, again resolve the field lines in directions perpendicular to BC and parallel to BC.
The component perpendicular to BC gives rise to a magentic force, but not the one parallel to BC.
Force acting on BC = (50x10^-3).cos(45) x 0.3 x 0.5 N = 5.3 x 10^-3 N (in direction into paper)
Force acting on AD = force acting on BC = 5.3 x 10^-3 N (in direction out of paper)

Torque given by forces on AB and CD = 3.54x10^-3 x [0.2cos(45)] N.m = 5x10^-4 N.m
Torque given by forces on BC and AD = 5.3x10^-3 x [0.3cos(45)] N.m = 1.12x10^-3 N.m
Hence, total torque = (5x10^-4 + 1.12x10^-3) N.m = 1.62 x 10^-3 N.m

The coil rotates with BD as the axis of rotation.