A wire of length L = 1 m carries a constant | vedclass

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(B) The magnetic field at the centre of a circular loop with $n$ turns is given by $B = \frac{n \mu_0 I}{2R}$.For the first case,$n_1 = 1$. The circum

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$16 B$

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(B) The magnetic field at the centre of a circular loop with $n$ turns is given by $B = \frac{n \mu_0 I}{2R}$.For the first case,$n_1 = 1$. The circumference of the loop is $L = 2 \pi R_1$,so $R_1 = \frac{L}{2 \pi}$.The magnetic field is $B = \frac{1 \cdot \mu_0 I}{2 R_1} = \frac{\mu_0 I}{2 (L / 2 \pi)} = \frac{\mu_0 I \pi}{L}$.For the second case,the wire is bent into $n_2 = 4$ turns. The total length of the wire is $L = n_2 (2 \pi R_2)$,so $R_2 = \frac{L}{2 \pi n_2} = \frac{L}{8 \pi}$.The new magnetic field is $B' = \frac{n_2 \mu_0 I}{2 R_2} = \frac{4 \mu_0 I}{2 (L / 8 \pi)} = \frac{4 \mu_0 I \cdot 8 \pi}{2 L} = 16 \left( \frac{\mu_0 I \pi}{L} \right) = 16 B$.

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