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 m², and the free electron density in copper is given to be about 10^–29 m^–3.)

Given,
Number of turns, N = 20 
Radius of the coil, r = 10 cm = 10 x 10^–2 m
Magnetic field, B = 0.10 T
Current carried, I = 5.0 A
θ = 0° (angle between field and normal to the coil)
Area of the coil, A = $\mathrm{\pi }$r² 
                          = $\mathrm{\pi }$ x (10 x 10^–2)²
                          = $\mathrm{\pi }$ x 10^–2 m²

(a) Torque, τ = NIBA sin θ
                   = 20 x 5.0 x 0.10 x $\mathrm{\pi }$ x 10^–2 sin 0°
                   = 20 x 5.0 x 0.10 x $\mathrm{\pi }$ x 10^–2 x 0 = 0

(b) Net force on a planer current loop in a magnetic field is always zero, as net force due to couple of force is zero. 

(c) If v_d is the drift velocity of electron

                       F = qv x B
                         = ev_d. B sin 90°

Force on one electron = Be  ${\mathrm{v}}_{\mathrm{d}} = \mathrm{Be}\frac{\mathrm{I}}{\mathrm{neA}} = \frac{\mathrm{BI}}{\mathrm{nA}}$
Here,  

                    $\mathrm{n} = {10}^{29} {\mathrm{m}}^{-3}, \mathrm{A} ={10}^{-5}{\mathrm{m}}^2$

$\therefore$ Force on one electron = $\frac{0.10 \times 5.0}{{10}^{29} \times {10}^{-5}} = 5 \times {10}^{-25}\mathrm{N}.$