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)` .)

Explanation

N = 20, R = 0.10m, B = 0.10 T, I = 5. A = `10^(-5)m^(2), n=10^(29)m^(-3)`.
Torque `=BAN I sintheta`.
(a) Torque on the coil = 0, since there will be no rotatory effect of the coil.
(b) Total force on the coil = 0 because the field will be uniform.
(c) magnetic force on each electron `F=Bev=(BI)/(nA)`
i.e., `" "F=(0.10 times 5)/(10^(29) times 10^(-5))=5 times 10^(-25)N.`