Question

A symmetric star shaped conducting wire loop is carrying a steady state current $I$ as shown in 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. $\frac{3{\mu }_{0}I}{4\pi a}[2-\sqrt{3}]$
• B. $\frac{3{\mu }_{0}I}{2\pi a}[\sqrt{3}-1]$
• C. $\frac{3{\mu }_{0}I}{4\pi a}[\sqrt{3}-1]$
• D. $\frac{6{\mu }_{0}I}{4\pi a}[\sqrt{3}+1]$

Answer 1

The correct option is B $\frac{3{\mu }_{0}I}{2\pi a}[\sqrt{3}-1]$


Magnitude of magnetic field at a particular distance $d$ from a current carrying wire :-
$\left|{\overrightarrow{B}}_P\right|=\frac{{\mu }_{0}I}{4\pi d}[\sin {\theta }_1-\sin {\theta }_2]$
Magnitude of magnetic field at the centre due to wire $AB$ :-
${\left|{\overrightarrow{B}}_C\right|}_{\text{due to}AB}=\frac{{\mu }_{0}I}{4\pi a}[\sin {60}^{\circ }-\sin {30}^{\circ }]$
Net magnetic field due to star at the centre is equal to $12$ times the magnetic field due to one section of star side.
${\left|{\overrightarrow{B}}_C\right|}_{\text{net}}=12{\left|{\overrightarrow{B}}_C\right|}_{\text{due to}AB}$
$=12\times \frac{{\mu }_{0}I}{4\pi a}\left[\frac{\sqrt{3}-1}{2}\right]$
$=\frac{6{\mu }_{0}I}{4\pi a}[\sqrt{3}-1]=\frac{3{\mu }_{0}I}{2\pi a}[\sqrt{3}-1]$