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A ciruclar coil of 20 turns and radius 1...

A ciruclar coil of 20 turns and radius `10cm` is placed in a uniform magnetic field of `0.1T` normal to the plane of the coil . If the current in the coil is `5.0A` 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))`.

A

`2.5 xx 10^(-25) N`

B

`7.5 xx 10^(-25) N`

C

`5 xx 10^(-25) N`

D

`10^(-25) N`

Text Solution

Verified by Experts

The correct Answer is:
C
Force acting on each electron , i.e.,
Lorentz force , ` F = ev_(d) B`
or `F_m = e ( (1)/("Ane") ) B = (IB)/(An) " " (as I = "neA"v_d)`
or `F_m = (5 xx 0.10)/(10^(-5) xx 10^29) N = 5 x 10^(-25) N `
(as A = cross sectional area of the wire ` = 10^(-5) m^2 ` , n = free electron density = `10^(-29) m^(-3) ` )
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