Single Choice

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 $$\displaystyle\frac{B_1}{B_2}$$ is?

A$$\sqrt{2}$$
Correct Answer
B$$\displaystyle\frac{1}{\sqrt{2}}$$
C$$2$$
D$$\sqrt{3}$$

Solution

The dipole moment of circular loop is m
$$m_1= I.A = I. \pi R^2$$ { R = radius of the loop}
$$B_1 = \dfrac{\mu_{o}I}{2R}$$
moment becomes double, R becomes $$\sqrt{2}R$$(keeping the current constant)
$$m_2 = I. \pi(\sqrt{2}R)^2= 2. I \pi R^2= 2 m_1$$
$$B_2 = \dfrac{\mu_o I}{2 \sqrt{2}R}=\dfrac{B_1}{\sqrt{2}}$$
$$\dfrac{B_1}{B_1}= \sqrt{2}$$

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