Question

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

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

Answer 1

The correct option is C $3\times {10}^{-5}\text{T},\text{inside the plane of triangle}$

Magnetic field at a point which lies on perpendicular bisector of finite length wire is $B=\frac{{\mu }_{0}I}{4\pi a}(\sin {\theta }_1+\sin {\theta }_2)$


Magnetic field at centroid due to each side of triangle are equal both in magnitude and direction.
Net Magnetic field at centroid is $B_{net}=3B=\frac{3{\mu }_{0}I}{4\pi a}(\sin {\theta }_1+\sin {\theta }_2)$
As given triangle is equilateral ${\theta }_1={\theta }_2={60}^{\circ }$
From $△AOD$
$\tan {\theta }_1=\frac{AD}{OD}$
$\tan 60=\frac{9\times {10}^{-2}}{2\left(OD\right)}$
$OD=a=\frac{9\times {10}^{-2}}{2\sqrt{3}}$
On substitution,
$B_{net}=\frac{3{\mu }_{0}I}{4\pi a}(\sin {60}^{\circ }+\sin {60}^{\circ })$
$B_{net}=\frac{3\times {10}^{-7}\left(1.5\right)}{\frac{9\times {10}^{-2}}{2\sqrt{3}}}\left(\sqrt{3}\right)$
$B_{net}=\frac{3\times {10}^{-5}\left(1.5\right)}{9}\left(6\right)=3\times {10}^{-5}\text{T}$

When current in the loop is in clockwise direction, from right hand thumb rule magnetic field generated is inside the plane of triangle.