A circular coil of 20 turns and a radius of $10 \mathrm{~cm}$ is placed in a uniform magnetic field of $0.10 \mathrm{~T}$ normal to the plane of the coil. If the current in the coil is $5.0 \mathrm{~A}$, what is the average force on each electron in the coil due to the magnetic field? (The coil is made of copper wire of cross-sectional area $10^{-5} \mathrm{~m}^{2}$, and the free electron density in copper is given to be about $10^{29} \mathrm{~m}^{-3} .$

Number of turns is given as $\mathrm{n}=20$ turns

Radius of the coil is given as $r=10 \mathrm{~cm}=0.1 \mathrm{~m}$

Current in the coil is given as $I=5 \mathrm{~A}$

Magnetic field strength is given as $B=0.10 \mathrm{~T}$

Cross-sectional area of the wire is given as $a=10^{-5} \mathrm{~m}^{2}$

Average force on the moving electron can be expressed as,

$F=B e v_{d}$

$\mathrm{v}_{\mathrm{d}}$ is the drift velocity of the electrons

$\mathrm{V}_{\mathrm{d}}=\mathrm{I} / \mathrm{N}$ e $\mathrm{A}$

Therefore, $F=B$ e $/ / N$ e a

$F=B \mid / n ~ a$

$=\frac{0.1 \times 5}{10^{29} \times 10^{-5}}=5 \times 10^{-25} N$

$5 \times 10^{-25} \mathrm{~N}$ is the average force on each electron.