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(B) The magnetic field at the centre of a circular loop of radius $R$ carrying current $I$ is given by $B_1 = \frac{\mu_0 I}{2R}$.The magnetic dipole

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$\sqrt{2}$

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(B) The magnetic field at the centre of a circular loop of radius $R$ carrying current $I$ is given by $B_1 = \frac{\mu_0 I}{2R}$.The magnetic dipole moment of the loop is $m_1 = I A = I \pi R^2$.When the dipole moment is doubled $(m_2 = 2m_1)$ while keeping the current $I$ constant,the area $A$ must double. Since $A = \pi R^2$,the new radius $R'$ must satisfy $\pi (R')^2 = 2 \pi R^2$,which gives $R' = \sqrt{2} R$.The new magnetic field at the centre is $B_2 = \frac{\mu_0 I}{2R'} = \frac{\mu_0 I}{2(\sqrt{2} R)}$.Taking the ratio $\frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{2R}}{\frac{\mu_0 I}{2\sqrt{2}R}} = \sqrt{2}$.

The dipole moment of a circular loop carrying a current $I$ is $m$ and the magnetic field at the centre of the loop is $B_1$. When the dipole moment is doubled by keeping the current constant,the magnetic field at the centre of the loop is $B_2$. The ratio $\frac{B_1}{B_2}$ is

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