Question

A current of $1.5 A$ is flowing through a triangle, of side $9 cm$ 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.