Magnetic Moment of Planar Current Loop with Semi Circular Arcs

To find the magnetic moment vector $\vec{M}$ of the planar current-carrying loop, we use the definition of the magnetic dipole moment of a closed planar loop:

$\vec{M} = I \vec{A} = I A \hat{n}$

where $I$ is the current flowing in the loop, $A$ is the total area enclosed by the loop, and $\hat{n}$ is the unit normal vector determined by the right-hand rule.

1. Determination of Direction ($\hat{n}$)

The current $I$ flows in a counter-clockwise direction around the loop in the $x$-$y$ plane. By applying the right-hand thumb rule (curling the fingers of the right hand along the direction of current flow), the thumb points out of the plane of the paper along the positive $z$-axis. Therefore:

$\hat{n} = \hat{k}$

2. Calculation of Enclosed Area ($A$)

The geometry of the loop consists of:

1. A central square region of side length $a$.
2. Four identical semicircular arcs, each attached outward onto one of the four sides of the square.

• Area of the central square ($A_{\text{square}}$): $A_{\text{square}} = a^2$

• Area of one semicircular arc ($A_{\text{semi}}$): Since the diameter of each semicircular arc is equal to the side length $a$ of the square, the radius of each arc is $r = \frac{a}{2}$. $A_{\text{semi}} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi \left(\frac{a}{2}\right)^2 = \frac{\pi a^2}{8}$

• Total area of the four semicircular arcs ($A_{\text{arcs}}$): $A_{\text{arcs}} = 4 \times A_{\text{semi}} = 4 \times \left(\frac{\pi a^2}{8}\right) = \frac{\pi a^2}{2}$

• Total enclosed area ($A$): $A = A_{\text{square}} + A_{\text{arcs}} = a^2 + \frac{\pi a^2}{2} = \left(\frac{\pi}{2} + 1\right) a^2$

3. Magnetic Moment Vector

Substituting the total area $A$ and direction $\hat{n}$ into the magnetic moment expression:

$\vec{M} = I \left(\frac{\pi}{2} + 1\right) a^2 \hat{k} = \left(\frac{\pi}{2} + 1\right) a^2 I \hat{k}$

Hence, the correct option is B.