Question

Question 4.25: A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the (a) total torque on the coil, (b) total force on the coil, (c) average force on each electron in the coil due to the magnetic field? (The coil is made of copper wire of cross-sectional area 10 −5 m 2 , and the free electron density in copper is given to be about 10 29 m −3 .)

Class 12 - Physics - Moving Charges And Magnetism Moving Charges And Magnetism Page-172

Answer 1

the magnetic field is normal to the plane of the coil, so condition of minimum torque .
(a) torque on the coil, $\tau=BINAsin\theta$
here $\theta = 0$
so, torque, $\tau=0$

(b) force on every element of the coil is cancelled by force on corresponding element as shown in figure. so, net force is zero.

(c) To calculate force on each electron,
let us find drift velocity.
I = neAV
here , I is the current , n is the number of electrons per unit volume, A is the cross section area and V is the drift velocity.
so, V = I/neA = 5/(10^29 × 1.6 × 10^-19 × 10^-5)
= 3.125 × 10^-5 m/s

hence, force = BeV
here, B is magnetic field and e is charge on electron.
so, F = 0.1 × 1.6 × 10^-19 × 3.125 × 10^-5 N
F = 5 × 10^-25 N