Question

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

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

Answer 1

Answer: (A)

Magnetic field due to wire $BC$ is $\Rightarrow B=\frac{{\mu }_0}{4\pi }\frac{i}{\ell }\times 2\sqrt{3}\left[\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right]⨂$
Net magnetic field at centre $=3B$
$=\frac{{\mu }_0}{4\pi }\frac{i}{\ell }\times 2\sqrt{3}\left[\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right]3⨂$
$=\frac{{\left(\mu \right)}_0}{4\pi }\frac{1}{\ell }6(1)3={\left(10\right)}^{-7}\times \frac{10}{1}\times 18=18\times {\left(10\right)}^{-6}T$