1)

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 costant, the magnetic field at the centre of the loop is B2. The ratio B1/B2


A) 2

B) $\sqrt{2}$

C) $\sqrt{3}$

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

Answer:

Option B

Explanation:

Key Idea As m=IA, so to change dipole moment (current is kept constant), we have to change radius of loop

initially,$m=I\pi R^{2} $ and $B_{1}=\frac{\mu_{0}I}{2R_{1}}$

Finally,$m^{'}=2m = I\pi R_2^2$

$2I\pi R_2^2=I\pi R_2^2$

or,  $R_{2}=\sqrt{2}R_{1}$

So, $B_{2}=\frac{\mu_{0}I}{2(R_{2})}=\frac{\mu_{0}I}{2\sqrt{2}(R_{1})}$

Hence, $ratio\frac{B_{1}}{B_{2}}= \frac{\frac{\mu_{0}I}{2R_{2}}}{\frac{\mu_{0}I}{2\sqrt{2}R_{1}}}$

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