Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the center of the loop is $B$ . It is then bent into a circular coil of $n$ turns. The magnetic field at the centre of this coil of $n$ turns will be:

$nB$

$n^2B$

$2nB$

$2n^2B$

Step-by-Step Solution

Identify the initial magnetic field: The magnetic field $B$ at the centre of a circular loop of radius $R$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2R}$ .
Relate radius to the number of turns: Let the total length of the wire be $L$ . For one turn, $L = 2\pi R$ . For $n$ turns of radius $r$ , the same wire length is used, so $L = n(2\pi r)$ . Equating the two, we get $2\pi R = n(2\pi r)$ , which implies $r = \frac{R}{n}$ .
Calculate the new magnetic field ( $B'$ ): The magnetic field at the centre of a coil with $n$ turns is $B' = \frac{\mu_0 n I}{2r}$ .
Substitute the new radius: $B' = \frac{\mu_0 n I}{2(R/n)} = n^2 \left( \frac{\mu_0 I}{2R} \right)$
Conclusion: Since $B = \frac{\mu_0 I}{2R}$ , we find that $B' = n^2 B$ .

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