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 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.
$$2$$
B.
$$\sqrt 3 $$
C.
$$\sqrt 2 $$
D.
$$\frac{1}{{\sqrt 2 }}$$
Answer :
$$\sqrt 2 $$
Solution :
Magnetic field at the centre of loop, $${B_1} = \frac{{{\mu _0}I}}{{2R}}$$
Dipole moment of circular loop is $$m = IA$$
$${m_1} = I.A = I.\pi {R^2}\,\,\,\,\left\{ {R = {\text{Radius of the loop}}} \right\}$$
If moment is doubled (keeping current constant) $$R$$ becomes $$\sqrt 2 R$$
$$\eqalign{
& {m_2} = I.\pi {\left( {\sqrt 2 R} \right)^2} = 2.I\pi {R^2} = 2{m_1} \cr
& {B_2} = \frac{{{\mu _0}I}}{{2\left( {\sqrt 2 R} \right)}} \cr
& \therefore \frac{{{B_1}}}{{{B_2}}} = \frac{{\frac{{{\mu _0}I}}{{2R}}}}{{\frac{{{\mu _0}I}}{{2\left( {\sqrt 2 R} \right)}}}} = \sqrt 2 \cr} $$
Releted MCQ Question on Electrostatics and Magnetism >> Magnetic Effect of Current
Releted Question 1
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