QJEEM18Phad09: JEE Main 2018 Physics - Class 12 JEEM18Phad09 Solution: The dipole moment of a circular loop carrying a current I, is m …

Dipole moment, M = IA

Let radius of circular loop = R

$\therefore\,\,\,$ M  =  I $\times$ $\pi$ R²

Later, we keep current constant ,

But dipole moment becomes double, let new radius = R₁

$\therefore\,\,\,$ 2M  =  I $\times$ $\pi$ R $_1^2$

$\Rightarrow$ $\,\,\,$ 2I $\pi$ R²  =  I $\pi$ R $_1^2$

$\Rightarrow$ $\,\,\,$ R₁  =   $\sqrt 2$ R

At the center of circular ring, the magnetic field,

B  =   ${{{\mu _0}I} \over {2R}}$

$\therefore\,\,\,$ B₁  =  ${{{\mu _0}I} \over {2R}}$    and    B₂ = ${{{\mu _0}I} \over {2 \times \left( {\sqrt 2 R} \right)}}$

$\therefore\,\,\,$ ${{{B_1}} \over {{B_2}}}$  =   ${{{{{\mu _0}I} \over {2R}}} \over {{{{\mu _0}I} \over {2\sqrt 2 \,R}}}}$

= ${{2\sqrt 2 } \over 2}$

= $\sqrt 2$