25.(c) 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 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} .)

Answers (1)
S Sayak

The average force on each electron in the coil due to the magnetic field will be eVdB where Vd is the drift velocity of the electrons.

The current is given by

I=neAV_{d}

where n is the free electron density and A is the cross-sectional area.

\\V_{d}=\frac{I}{neA}\\ \\V_{d}=\frac{5}{10^{29}\times 1.6\times 10^{-19}\times 10^{-5}}

Vd=3.125\times10^{-5} ms^{-1}

The average force on each electron is

F=eV_dB

F=1.6\times10^{-19}\times3.125\times10^{-5}\times00.1

F=5\times10^{-25} N

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