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(B) The magnetic field at the centre of a circular loop of radius $r$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.Case $(I)$: For a loop

Vedclass · Accepted Answer

$\frac{B}{4}$

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(B) The magnetic field at the centre of a circular loop of radius $r$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.Case $(I)$: For a loop of radius $r$, the length is $l_1 = 2\pi r$. The resistance is $R_1 = \rho \frac{l_1}{A} = \rho \frac{2\pi r}{A}$. The current is $I_1 = \frac{V}{R_1}$. The magnetic field is $B_1 = \frac{\mu_0 I_1}{2r}$.Case $(II)$: For a loop of radius $2r$, the length is $l_2 = 2\pi(2r) = 2l_1$. The resistance is $R_2 = \rho \frac{l_2}{A} = 2R_1$. The current is $I_2 = \frac{V}{R_2} = \frac{V}{2R_1} = \frac{I_1}{2}$.The new magnetic field is $B_2 = \frac{\mu_0 I_2}{2(2r)} = \frac{\mu_0 (I_1/2)}{4r} = \frac{1}{4} \left( \frac{\mu_0 I_1}{2r} \right) = \frac{B_1}{4}$.

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