Question

The magnetic field at the centre of an equilateral triangular loop of side $2L$ and carrying a current $i$ is

• A. $\frac{9{\mu }_{0}i}{4\pi L}$
• B. $\frac{3\sqrt{3}{\mu }_{0}i}{4\pi L}$
• C. $\frac{2\sqrt{3}{\mu }_{0}i}{\pi L}$
• D. $\frac{3{\mu }_{0}i}{4\pi L}$

Answer 1

The correct option is A $\frac{9{\mu }_{0}i}{4\pi L}$

Magnetic field due to all 3 wires will be in same direction i.e. into the plane.

$\therefore B_{net}=3B_{AC}$

We know that magnetic field due to current carrying wire is
$B=\frac{{\mu }_{0}i}{4\pi r}(\sin {\theta }_1+\sin {\theta }_2)$

$\therefore$ for wire AC, ${\theta }_1={60}^{\circ }$ and ${\theta }_2={60}^{\circ }$

$B_{AC}=\frac{{\mu }_{0}i\sqrt{3}}{4\pi L}(\sin {60}^{\circ }+\sin {60}^{\circ })$

$B_{AC}=\frac{3{\mu }_{0}i}{4\pi L}$

$B_{net}=\frac{3\times 3{\mu }_{0}i}{4\pi L}$

$B_{net}=\frac{9{\mu }_{0}i}{4\pi L}$