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A circular coil of 20 turns and radius $10 \mathrm{~cm}$ is placed in a uniform magnetic field of $0.10 \mathrm{~T}$ normal to the plane of the coil. If the current in the coil is $5 \mathrm{~A}$, then the average force on each electron in the coil due to the magnetic field is
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Verified Answer
The correct answer is:
$5 \times 10^{-25} \mathrm{~N}$
Magnetic Lorenz force acting on each electron, $F^*=e v B$
As, current, $i=n e A v$
$$
e v=\frac{i}{n A}
$$
Therefore, substituting value of $e v$ into Eq. (i),
$$
F=\frac{i}{n A} B
$$
Given, $i=5 \mathrm{~A}, n=20$ turns, $B=0.1 \mathrm{~T}$
$$
F=e v B=\frac{i B}{n A}=\frac{5 \times 0.1}{10^{29} \times 10^{-5}}=5 \times 10^{-25} \mathrm{~N}
$$
As, current, $i=n e A v$
$$
e v=\frac{i}{n A}
$$
Therefore, substituting value of $e v$ into Eq. (i),
$$
F=\frac{i}{n A} B
$$
Given, $i=5 \mathrm{~A}, n=20$ turns, $B=0.1 \mathrm{~T}$
$$
F=e v B=\frac{i B}{n A}=\frac{5 \times 0.1}{10^{29} \times 10^{-5}}=5 \times 10^{-25} \mathrm{~N}
$$
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