Question

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

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

Answer 1

The correct option is A $18\mu \text{T}$

Given:
Side of equilateral triangle $=1\text{m}$
Current in the loop $=10\text{A}$

From the figure,
$r=\left(\frac{1}{3}\right)\left(a\sin {60}^{\circ }\right)$

$r=\frac{a}{3}\times \frac{\sqrt{3}}{2}=\left(\frac{a}{2\sqrt{3}}\right)$

Magnetic field at the centre of the equilateral triangle

$B_0=3\left[\frac{{\mu }_{0}I}{4\pi r}\right(\sin {60}^{\circ }+\sin {60}^{\circ }\left)\right]=\frac{3{\mu }_{0}I}{4\pi }\left(\frac{2\sqrt{3}}{a}\right)\times \left(2\right)\left(\frac{\sqrt{3}}{2}\right)=\frac{9}{2}\left(\frac{{\mu }_{0}I}{\pi a}\right)$

$B_0=\frac{9\times 4\times {10}^{-7}\times 10}{2\times 1}$

$B_0=18\mu \text{T}$.

Hence option $\left(A\right)$ is correct.