The magnetic field at the centre of a single- | vedclass

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(C) Let the total length of the wire be $L$. For a single-turn coil $(n_1 = 1)$,the circumference is $L = 2\pi R_1$,so $R_1 = \frac{L}{2\pi}$. The mag

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$4B$

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(C) Let the total length of the wire be $L$. For a single-turn coil $(n_1 = 1)$,the circumference is $L = 2\pi R_1$,so $R_1 = \frac{L}{2\pi}$. The magnetic field at the centre is $B_1 = \frac{\mu_0 I}{2R_1} = \frac{\mu_0 I}{2(L/2\pi)} = \frac{\mu_0 I \pi}{L} = B$. For a two-turn coil $(n_2 = 2)$,the total length $L = n_2(2\pi R_2) = 2(2\pi R_2) = 4\pi R_2$. Thus,the new radius is $R_2 = \frac{L}{4\pi} = \frac{R_1}{2}$. The magnetic field at the centre for $n$ turns is $B_n = \frac{n \mu_0 I}{2R_n}$. For $n=2$,$B_2 = \frac{2 \mu_0 I}{2R_2} = \frac{\mu_0 I}{R_2}$. Substituting $R_2 = \frac{R_1}{2}$,we get $B_2 = \frac{\mu_0 I}{R_1/2} = \frac{2 \mu_0 I}{R_1}$. Since $B = \frac{\mu_0 I}{2R_1}$,we have $\frac{\mu_0 I}{R_1} = 2B$. Therefore,$B_2 = 2(2B) = 4B$.

The magnetic field at the centre of a single-turn circular coil of a given length of wire carrying a current $I$ is $B$. If the same wire is bent into a circular coil of two turns and the same current $I$ is passed through it,the new magnetic field at the centre will be:

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