Question

A current of $1.5A$ is flowing through a triangle, of side $9cm$ each. The magnetic field at the centroid of the triangle is : (Assume that the current is flowing in the clockwise direction.)

• A. $3\times 10^{-7}T$, outside the plane of triangle
• B. $2\sqrt{3}\times 10^{-7}T$, outside the plane of triangle
• C. $2\sqrt{3}\times 10^{-5}T$, inside the plane of triangle
• D. $3\times 10^{-5}T$, inside the plane of triangle

Answer 1

Answer: (D)

$B=3\left[\frac{{\mu }_{0}i}{4\pi r}\left(\sin 60^{\circ }+\sin 60^{\circ }\right)\right]$
$\tan 60^{\circ }=\frac{\ell /2}{r}$
Where $r=\frac{9\times 10^{-2}}{2\sqrt{3}}M$
$\therefore B=3\times 10^{-5}T$
Current is flowing in clockwise direction so, $\vec{B}$ is inside plane of triangle by right hand rule.