Question

A current carrying circular loop of radius $R$ is placed in the $x-y$ plane with center at the origin. Half of the loop with $x>0$ is now bent so that it now lies in the $y-z$ plane.

• A. The magnetic moment does not change.
• B. The magnitude of $B$ at $\left(\mathrm{0.0.}z\right),z>>R$ increases.
• C. The magnitude of magnetic moment now diminishes.
• D. The magnitude of $B$ at $\left(\mathrm{0.0.}z\right),z>>R$ is unchanged.

Answer 1

The correct option is C The magnitude of magnetic moment now diminishes.


For a circular loop of radius $R$ carrying current, $I$ placed in $x-y$ plane, with centre at the origin the magnetic field is along its axis i.e., along $z-axis$.

Therefore, the magnetic moment is given by,
$M=IA$
Where,

$I=$ Moment of Inertia
$A=$ Area of circular loop$=\pi R^2$

So,
$M=I\left(\pi R^2\right)$

When half of the current loop is bent in $y-z$ plane, the magnetic moment due to half current loop in $x-y$ plane is given by,
$M_1=I\left(\frac{\pi R^2}{2}\right)$ acting along $z-$ direction.

Magnetic moment due to remaining half current loop in $y-z$ plane is given by,
$M_2=I\left(\frac{\pi R^2}{2}\right)$ along $x-$direction.

Here,
$M_1=M_2=M$

Therefore, effective magnetic moment$\left(M^{\prime }\right)$ due to entire bent current loop,

$M^{\prime }=\sqrt{M_1^2+M_2^2}=\sqrt{{\left(\frac{I\pi R^2}{2}\right)}^2+{\left(\frac{I\pi R^2}{2}\right)}^2}$

$M^{\prime }=\frac{I\pi R^2\sqrt{2}}{2}=\frac{M}{\sqrt{2}}$

$M^{\prime }<M$

This shows, there will be two equal magnetic moments but perpendicular to each other which will add vectorially to give some non-zero value.

Magnetic field on the axis of a current carrying loop is given by,

$B=\frac{{\mu }_o}{4\pi }\frac{2\pi R^{2}I}{(z^2+R^2{)}^{3/2}}$

For $z>>R,B=\frac{{\mu }_o}{4\pi }\frac{2\pi R^{2}I}{z^3}$

Since the half of the loop is bent in the $y-z$ plane, therefore net magnetic field due to half loop lying in $z-y$ plane along $z-$ axis will decrease. Hence last two options are incorrect.

$\text{Final Answer}:$ (a)