Question

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

• A. $18\mu T$
• B. $4\mu T$
• C. $1\mu T$
• D. $9\mu T$

Answer 1

The correct option is A $18\mu T$

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