Magnetic Field at Center of Star Shaped Wire Loop

To find the magnitude of the magnetic field at the geometric center of the symmetric, 6-pointed star-shaped wire loop, we analyze its geometric properties and apply the Biot-Savart Law.

1. Geometry of the Star Loop

1. Outer Radius ($R_{\text{out}}$): The distance between two diametrically opposite outer vertices of the star is given as $4a$. Thus, the distance from the geometric center $O$ to any outer vertex $V_{\text{out}}$ is: $R_{\text{out}} = \frac{4a}{2} = 2a$

2. Structure of the Loop: A regular 6-pointed star (hexagram) consists of 12 identical straight wire segments, formed by the outer edges of two overlapping equilateral triangles inscribed in a circle of radius $R_{\text{out}} = 2a$.

3. Perpendicular Distance ($h$): Each wire segment lies along a side of one of these equilateral triangles. The perpendicular distance $h$ from the center $O$ to the line containing any wire segment is: $h = R_{\text{out}} \cos(60^\circ) = 2a \times \frac{1}{2} = a$

4. Angles Subtended by a Single Segment: Let $P$ be the foot of the perpendicular from the center $O$ to the extended line of a wire segment.

   • The outer vertex $V_{\text{out}}$ subtends an angle $\phi_1$ with respect to $OP$ such that: $\tan \phi_1 = \frac{\text{Distance from } P \text{ to } V_{\text{out}}}{h} = \sqrt{3} \implies \phi_1 = 60^\circ$
   • The adjacent inner vertex $V_{\text{in}}$ subtends an angle $\phi_2$ with respect to $OP$ such that: $\tan \phi_2 = \frac{\text{Distance from } P \text{ to } V_{\text{in}}}{h} = \frac{1}{\sqrt{3}} \implies \phi_2 = 30^\circ$

Since both endpoints of the segment ($V_{\text{out}}$ and $V_{\text{in}}$) lie on the same side of the foot of the perpendicular $P$, the field from this single segment is determined by the difference of the sines of these angles.

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2. Magnetic Field Due to a Single Segment

Using the Biot-Savart Law for a finite straight conductor, the magnitude of the magnetic field $B_1$ due to one segment at distance $h = a$ is: $B_1 = \frac{\mu_0 I}{4\pi h} (\sin \phi_1 - \sin \phi_2)$

Substituting $h = a$, $\phi_1 = 60^\circ$, and $\phi_2 = 30^\circ$: $B_1 = \frac{\mu_0 I}{4\pi a} \left( \sin 60^\circ - \sin 30^\circ \right) = \frac{\mu_0 I}{4\pi a} \left( \frac{\sqrt{3}}{2} - \frac{1}{2} \right) = \frac{\mu_0 I}{4\pi a} \left[ \frac{\sqrt{3} - 1}{2} \right]$

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3. Total Magnetic Field at the Center

By symmetry, the star consists of 12 such identical wire segments. The current flows in a direction such that the magnetic fields created by all 12 segments at the geometric center $O$ point in the same direction (perpendicular to the plane of the loop).

Thus, the total magnetic field $B_{\text{net}}$ is: $B_{\text{net}} = 12 \times B_1 = 12 \times \frac{\mu_0 I}{4\pi a} \left[ \frac{\sqrt{3} - 1}{2} \right] = \frac{\mu_0 I}{4\pi a} 6[\sqrt{3} - 1]$

Correct Answer:

A $\frac{\mu_0 I}{4\pi a} 6[\sqrt{3} - 1]$