Question

A circular coil of 200 turns and radius 10 cm is placed in a uniform magnetic field of 0.5T,normal to the plane of the coil.If the current in the coil id 3.0A, calculate the total torque on the coil total force on the coil average force on each electron in the coil,due to the magnetic field .Assume the area of cross section of the wire to be 10^-5 m^2 and the free electron density is 10^29/m^3

Answer 1

F = n * i dl X B    in vector notation.   This is the Biot Savarts law.

As the circumference is perpendicular to magnetic field,
  F = n * i * dl * B.         n = 200 turns    B = 0.50 T         i = 3.0 A
  F = 300 dl  Newtons.
  Force / meter  length of the coil  = 300   N/m

The force on a small segment of coil, is along the radius (inwards or outwards).  Hence the force due to one part of the coil is cancelled by the part which is diametrically opposite.  Hence net force is 0 on the coil.

The torque is zero, as radius and force vectors are parallel.  So net torque is 0.

In 1 meter of coil, there are 10^-5 * 10^29 = 10^24 number of electrons.

So force on 1 electron = 300/10^24 = 3 * 10⁻²²  Newtons  along the radius.