$A$ 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 $10^{29}\; m ^{-3}.$)

(A) Given:
Number of turns,$n = 20$
Radius,$r = 10 \; cm = 0.1 \; m$
Magnetic field,$B = 0.10 \; T$
Current,$I = 5.0 \; A$
Cross-sectional area,$A = 10^{-5} \; m^2$
Electron density,$n_e = 10^{29} \; m^{-3}$
Charge of electron,$e = 1.6 \times 10^{-19} \; C$
$(a)$ The torque $\tau$ on a current loop is given by $\tau = N I A B \sin \theta$. Since the magnetic field is normal to the plane of the coil,the angle $\theta$ between the magnetic field and the area vector is $0^\circ$. Thus,$\tau = N I A B \sin(0^\circ) = 0$.
$(b)$ The total force on a closed current loop in a uniform magnetic field is always zero,as the forces on opposite segments cancel each other out.
$(c)$ The magnetic force on an electron is $F = B e v_d$,where $v_d$ is the drift velocity. The drift velocity is given by $v_d = \frac{I}{n_e e A}$. Substituting this into the force equation:
$F = B e \left( \frac{I}{n_e e A} \right) = \frac{B I}{n_e A}$
$F = \frac{0.10 \times 5.0}{10^{29} \times 10^{-5}} = \frac{0.5}{10^{24}} = 5 \times 10^{-25} \; N$.
Thus,the average force on each electron is $5 \times 10^{-25} \; N$.