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 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.

• 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

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

Given:

Number of turns on the circular coil, $n=20$

Radius of the coil, $r=10cm=0.1m$

Magnetic field strength, $B=0.10T$

Current in the coil, $I=5.0A$

The total torque on the coil is zero because the field is uniform.

Cross-sectional area of copper coil, $A={10}^{-5}{m}^2$

Number of free electrons per cubic meter in copper, $N={10}^{29}/m^3$

Charge on the electron, $e=1.6\times {10}^{-19}C$

Magnetic force, $F=Be{v}_d$

Where,

$v_d=$ Drift velocity of electrons

$v_d=\frac{I}{NeA}$

$F=\frac{BeI}{NeA}=\frac{BI}{NA}=\frac{0.1\times 5}{{10}^{29}\times {10}^{-5}}=5\times {10}^{-25}N$

Hence, the average force on each electron is $5\times {10}^{-25}N$.