Question

The dipole moment of a circular loop carrying a current $I$, is $m$ and 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

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

Answer 1

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

Magnetic dipole moment of a coil, $m=iA$
$(current\times area)$

Given current is constant and dipole moment is doubled
$m\propto A$

So, the area will be doubled

$\frac{m_1}{m_2}=\frac{\pi {r_1}^2}{\pi {r_2}^2}=\frac{1}{2}$

$\frac{r_2}{r_1}=\sqrt{2}$

$\overrightarrow{B}$ at the centre of the coil

$\frac{{\mu }_{0}i}{2r}$

($r=$ radius of coil)

$\Rightarrow B\propto \frac{1}{r}(i=constant)$

$\frac{B_1}{B_2}=\frac{r_2}{r_1}$

$\frac{B_1}{B_2}=\frac{\sqrt{2}}{1}$

$\frac{B_1}{B_2}=\sqrt{2}$