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 B₂. The ratio B₁/B₂ is:

$\frac{1}{\sqrt{2}}$

2

$\sqrt{3}$

$\sqrt{2}$

Solution

$\sqrt{2}$

Dipole moment of circular loop is m = IA

m₁ = I.A = I.πR² { R = Radius of the loop}

If moment is doubled (keeping current constant)

R becomes $\sqrt{2 R}$

$m_{2} = I . π (\sqrt{2 R})^{2} = 2 . I π R^{2} = 2 m_{1} B_{2} = \frac{μ_{0} I}{2 (\sqrt{2 R)}} ∴ \frac{B_{1}}{B_{2}} = - \frac{\frac{μ_{0} I}{2 R}}{\frac{μ_{0} I}{2 (\sqrt{2 R)}}} = \sqrt{2}$

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