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:

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

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

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

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

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

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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

Watch 3-min video & get full concept clarity