Question

A circular coil of $20$ turns and $10cm$ radius is placed in a uniform magnetic field of $0.10T$ normal to the plane of the coil. If the current in the coil is $5A,$ cross-sectional area is $10^{- 5}m^{2}$ and coil is made up of copper having free electron density about $10^{29}m^{- 3},$ then the average force on each electron in the coil due to magnetic field is:

• A. $2.5\times 10^{- 25}N$
• B. $5\times 10^{- 25}N$
• C. $4\times 10^{- 25}N$
• D. $3\times 10^{- 25}N$

Answer 1

Answer: (B)

Force on each electron,
$F=ev_{d}B=\frac{I B}{n A}\begin{bmatrix} \because & I=neAv_{d} \\ \therefore & eV_{d}=\frac{I}{n A} \end{bmatrix}$
Here, $I=5A,B=0.1T,n=10^{29}m^{- 3}$
$A=10^{- 5}m^{2}$
so, $F=\frac{5 \times 0 . 1}{10^{29} \times 10^{- 5}}=5\times 10^{- 25}N$