The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is B1. 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 B_1 \over B_2 is:

Magnetics PYQ — JEE Main 2018 · Shift 0 · Q49 with Solution

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Correct answer: C

For a circular loop of radius $r$ carrying current $I$ ,

m = I A = I \pi r^2

For the same loop,

B = \frac{\mu_0 I}{2r}

So initially,

m = I\pi r^2, \qquad B_1 = \frac{\mu_0 I}{2r}

Since current $I$ is constant,

m \propto r^2

If the dipole moment is doubled,

m_2 = 2m_1

So,

I\pi r_2^2 = 2I\pi r_1^2

r_2^2 = 2r_1^2

r_2 = \sqrt{2}\, r_1

Using

B = \frac{\mu_0 I}{2r}

we get

B_2 = \frac{\mu_0 I}{2r_2} = \frac{\mu_0 I}{2\sqrt{2}r_1}

Thus,

\frac{B_1}{B_2} = \frac{\mu_0 I/(2r_1)}{\mu_0 I/(2\sqrt{2}r_1)} = \sqrt{2}

Therefore, the correct answer is Option C.