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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

`2.5xx10^(-25)N`

B

`5xx10^(-25)N`

C

`4xx10^(-25)N`

D

`3xx10^(-25)N`

Text Solution

Verified by Experts

The correct Answer is:
B
Force on each electron,
`ev_(d)B=(IB)/(nA)" "[{:(because" "I="neAv"_(d)),(therefore" ""ev"_(d)=(I)/(nA)):}]`
Here, `I=5A, B=0.1T, n=10^(29)m^(-3),`
`A=10^(-5)m^(2)`
So, `F=(5xx0.1)/(10^(29)xx10^(-5))=5xx10^(-25)N`
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