Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# A loop carrying current I lies in the x-y plane as shown in the figure. The unit vector $\widehat{k}$ is coming out of the plane of the paper. The magnetic moment of the current loop is?\begin{align} & A)\text{ }{{\text{a}}^{2}}I\widehat{k} \\ & B)\text{ (}\dfrac{\pi }{2}+1){{\text{a}}^{2}}I\widehat{k} \\ & C)\text{ -(}\dfrac{\pi }{2}+1){{\text{a}}^{2}}I\widehat{k} \\ & D)\text{ (2}\pi \text{-1)}{{\text{a}}^{2}}I\widehat{k} \\ \end{align}

Last updated date: 13th Jun 2024
Total views: 402.9k
Views today: 7.02k
Verified
402.9k+ views
Hint: The current carrying coil develops a magnetic field around the coil. The magnetic field is always perpendicular to the direction of current flow from the right-hand thumb rule. The area of the coil determines the magnetic field strength at a point.

Let us consider the given coil as the source of the magnetic field. The magnetic moment is a quantity which determines the magnetic strength and its orientation with respect to the object that develops the magnetic field, here the coil. The magnetic moment is given by: $\mu =N(I\times A)$, where N is the number of turns in the coil, I is the current through the coil and A is the area of the coil.

So, let us find the area of the given coil. It consists of four semicircles each of radius, $r=\dfrac{a}{2}$ and a square of side a.
i.e., The area of the coil is given by,
\begin{align} & A=4\times \dfrac{\pi {{\left( \dfrac{a}{2} \right)}^{2}}}{2}+{{a}^{2}} \\ & A=2\pi \dfrac{{{a}^{2}}}{4}+{{a}^{2}} \\ & A={{a}^{2}}(\dfrac{\pi }{2}+1) \\ \end{align}
Now, let us find the magnetic moment of the coil,
\begin{align} & \mu =N(I\times A) \\ & \mu =NIA\sin \theta \\ & but\text{ }\theta ={{90}^{0}} \\ \end{align}
The current and area are perpendicular to each other,
So, $\sin \theta =1$
The magnetic moment is,
\begin{align} & \mu =NIA \\ & \mu =1\times I{{a}^{2}}(\dfrac{\pi }{2}+1) \\ & \therefore \mu =I{{a}^{2}}(\dfrac{\pi }{2}+1) \\ \end{align}
The direction of moment is perpendicular to both current and area, i.e., it is along $\widehat{k}$
Therefore, the magnetic moment is given by:
$\mu \text{=(}\dfrac{\pi }{2}+1){{\text{a}}^{2}}I\widehat{k}$

The correct answer is option B.

Note:
Magnetic dipole moment is the elaborated name given to the magnetic moments. In electrostatics, a pair of equal and opposite charges constitute a dipole, whereas in magnetostatics, there is no chance of monopoles. Magnetism is always dipole in nature.
The magnetic moment of an electron has a special unit known as Bohr Magneton (BM).