Question

A circular 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 filed (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. $5\times {10}^{-25}N$
• B. $2.5\times {10}^{-25}N$
• C. $7.5\times {10}^{-25}N$
• D. ${10}^{-25}N$

Answer 1

The correct option is A $5\times {10}^{-25}N$

Number density of electrons $N=\left\{\frac{{10}^{29}}{m^3}\right\}$

Area of cross -section of copper wire, $A=\left\{\frac{{10}^{-5}}{m^2}\right\}$ magnitude of magnetic force

$\begin{array}{c}F=e\left(\frac{\left\{vd\times B\right\}}{xB}\right)=neA\left\{V\right\}\left\{d\right\}\left\{V\right\}\left\{d\right\}\\ =\frac{I}{neAF}=\frac{e.I}{NeAF}.B\sin {90}^0\\ =\frac{0.1\times 5}{{10}^{-5}\times {10}^{29}}N\\ =5\times 10-25N\end{array}$