Question

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 double by keeping the current constant, the magnetic field at the centre of loop is $B_2$. The ratio $\frac{B_1}{B_2}$ is

• A. $\sqrt{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. 2
• D. $\sqrt{3}$

Answer 1

The correct option is A $\sqrt{2}$


Dipole moment of circular loop is $m$
$m_1=I.A=I.\pi R^2\{R=\text{radius of the loop}\}$
$B_1=\frac{{\mu }_{0}I}{2R}$
$m_2=I.\pi (\sqrt{2}R{)}^2=2.I\pi R^2=2m_1$
Hence, if moment becomes double $R$ becomes $\sqrt{2}R$ (keeping current constant).
$B_2=\frac{{\mu }_{0}I}{2\left(\sqrt{2}R\right)}=\frac{B_1}{\sqrt{2}}$
$\frac{B_1}{B_2}=\sqrt{2}$