Question

The magnitude of the magnetic field at the center of an equilateral triangular loop of side $1m$ which is carrying a current of $10A$ is :
[Take ${\mu }_0=4\pi \times 10^{-7}N{A}^{-2}]$

• A. $18\mu T$
• B. $3\mu T$
• C. $1\mu T$
• D. $9\mu T$

Answer 1

Answer: (A)

The magnetic field at the center of an equilateral triangular loop will be given as
B=$B=3\left[\frac{{\mu }_{0}i}{4\pi r}\left(\sin 60^{\circ }+\sin 60^{\circ }\right)\right]$
Here, $r=\frac{a}{2\sqrt{3}}=\frac{1}{2\sqrt{3}}$
$B=3\left[\frac{4\pi \times 10^7\times 10\times 2\sqrt{3}}{4\pi \times 1}\left[\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right]\right]$
$B=18\times 10^{-6}=18\mu T$