Question

Find the magnitude of the magnetic field at the center of an equilateral triangular loop of side length $1 \, m$ which is carrying a current of $10 \, A$ . (Take $\mu _{0}=4\pi \times 10^{- 7}NA^{- 2}$ )

• 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 $\Rightarrow B=\frac{\mu _{0}}{4 \pi }\frac{i}{\ell }\times 2\sqrt{3}\left[\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}\right]\bigotimes$
Net magnetic field at centre $ \, = \, 3B \, $
$=\frac{\mu _{0}}{4 \pi }\frac{i}{\ell }\times 2\sqrt{3}\left[\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}\right]3\bigotimes$
$=\frac{\left(\mu \right)_{0}}{4 \pi }\frac{1}{\ell }6\left(\right.1\left.\right)3=\left(10\right)^{- 7}\times \frac{10}{1}\times 18=18\times \left(10\right)^{- 6}T$