Question

A symmetric star shaped conducting wire loop is carrying a steady state current $I$ as shown the figure. The distance between the diametrically opposite vertices of the star is $4a$. The magnitude of the magnetic field at the center of the loop is

• A. $\frac{{\mu }_{0}I}{4\pi a}6[\sqrt{3}-1]$
• B. $\frac{{\mu }_{0}I}{4\pi a}3[\sqrt{3}-1]$
• C. $\frac{{\mu }_{0}I}{4\pi a}6[\sqrt{3}+1]$
• D. $\frac{{\mu }_{0}I}{4\pi a}3[2-\sqrt{3}]$

Answer 1

The correct option is A

$\frac{{\mu }_{0}I}{4\pi a}6[\sqrt{3}-1]$

In $\Delta OACcos{60}^o=\frac{OC}{OA}$
$\therefore$ $OC=2a\times \frac{1}{2}=a$

The magnitude field at ${}^{\prime }{O}^{\prime }$ due to
$AB=\frac{{\mu }_{0}I}{4\pi a}[sin{60}^0-sin{30}^0]$

$=\frac{{\mu }_{0}I}{4\pi a}[\frac{\sqrt{3}}{2}-\frac{1}{2}]=\frac{{\mu }_{0}I}{4\pi a}\frac{1}{2}[\sqrt{3}-1]$

The total magnetic field due to all the straight segments of the star is $\frac{{\mu }_{0}I}{4\pi a}\frac{1}{2}[\sqrt{3}-1]\times 12=$ $\frac{{\mu }_{0}I}{4\pi a}\times 6[\sqrt{3}-1]$
$\therefore$ Option A is correct answer