Question 224: Magnetism and Matter

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

Accepted Answer

Dipole moment, M = IA
Let radius of circular loop = R
$$	herefore\,\,\,$$ M  =  I $$ 	imes $$ $$\pi $$R
2
Later,  we keep current constant ,
But dipole moment becomes double, let new radius = R
1
$$	herefore\,\,\,$$ 2M  =  I $$ 	imes $$ $$\pi $$R$$_1^2$$
$$ \Rightarrow $$$$\,\,\,$$ 2I$$\pi $$R
2
=  I$$\pi $$R$$_1^2$$
$$ \Rightarrow $$$$\,\,\,$$ R
1
=  $$\sqrt 2 $$ R
At the center of circular ring, the magnetic field,
B  =  $${{{\mu _0}I} \over {2R}}$$
$$	herefore\,\,\,$$ B
1
=  $${{{\mu _0}I} \over {2R}}$$   and    B
2
= $${{{\mu _0}I} \over {2 	imes \left( {\sqrt 2 R} ight)}}$$
$$	herefore\,\,\,$$ $${{{B_1}} \over {{B_2}}}$$  =   $${{{{{\mu _0}I} \over {2R}}} \over {{{{\mu _0}I} \over {2\sqrt 2 \,R}}}}$$
= $${{2\sqrt 2 } \over 2}$$
= $$  \sqrt 2 $$

Answer

2

Answer

$$\sqrt 3 $$

Answer

$$\sqrt 2 $$

Answer

$$1 \over \sqrt 2 $$

Magnetism and Matter

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 ${{B_2}}. The ratio {{{B_1}} \over {{B_2}}}$ is:

Choose the correct answer: