A circular coil o

Given:

Number of turns, N = 20

Radius of coil, r = 10cm

Magnetic field strength = 0.10 T

Current in the coil = 5 A

a) The torque on the coil is zero because the magnetic field is uniform along the coil.

b) The total force on the coil is Zero due to uniform magnetic field.

c) Let average force on each electron in the coil due to magnetic field = F

Cross-sectional area of the wire, A = 10^-5 m²

Free electron density or number of free electron per metre cube of copper, D = 10²⁹

Charge in each electron = 1.6 × 10¹⁹ C

We know that,

…(1)

Where,

F = force due to magnetic field

B = Magnetic field strength

e = charge on electron

V_d = drift velocity of electron

V_d = I/ (NeA) …(2)

Where, I = current through the coil

N = Density of free electron

A = cross sectional area.

Putting values in equation (2)

⇒

⇒ V_d = 3.125 × 10⁴³ ms^-1

F = B × e × V_d …(3)

By putting the values of B, e and V_d in equation 3, we get,

⇒

⇒ F = 5 × 10^-25 N

Hence, average force on each electron is 5 × 10^-25 N.