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A circular coil of $20$ turns and $10$ cm radius 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 $5 A$ , cross-sectional area is $10^{- 5} m^{2}$ and coil is made up of copper wire having free electron density about $10^{29} m^{- 3}$ , then the average force on each electron in the coil due to magnetic field is:

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$3 \times 10^{- 25} N$

$2.5 \times 10^{- 25} N$

$5 \times 10^{- 25} N$

$4 \times 10^{- 25} N$

answer is B.

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Detailed Solution

Force on each electron, $e ν_{d} B = \frac{I B}{n A} [∵ I = n e A ν_{d} ∴ e ν_{d} = \frac{I}{n A}]$

Here, $I = 5 A, B = 0.1 T, n = 10^{29} m^{- 3}, A = 10^{- 5} m^{2}$

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

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