Question

A current $i$ is flowing through the loop. The direction of the current and the shape of the loop are as shown in the figure. The magnetic field at the centre of the loop is
$\left( MA =R, MB =2 R, \angle DMA =90^{\circ}\right)$

• A. $\frac{5}{16}\frac{\mu _{o} i}{R},$ but out of the plane of the paper
• B. $\frac{5}{16} \frac{\mu _{o} i}{R} ,$ but into the plane of the paper
• C. $\frac{7}{16}\frac{\mu _{o} i}{R},$ but out of the plane of the paper
• D. $\frac{7}{16}\frac{\mu _{o} i}{R},$ but into the plane of the paper

Answer 1

Answer: (D)

(i) The magnetic field at the centre due to the curved portion
$D A=\frac{(\mu)_{0} i}{4 \pi R}\left(\frac{3 \pi}{2}\right)$
According to the right-hand screw rule, the magnetic field will be into the plane of the paper.


(ii) The magnetic field at $M$ due to $A B$ is zero.
(iii) The magnetic field at the centre due to the curved portion $B C$ is $\frac{(\mu)_{0} i}{4 \pi 2 R}\left(\frac{\pi}{2}\right)$.
According to the right-hand screw rule, the magnetic field will be into the plane of the paper.
(iv) The magnetic field at $M$ due to $D C$ is zero.
Hence, the resultant magnetic field at $M$
$B_{M}=B_{\text {small loop }}+B_{\text {big loop }}$
Both are perpendicular downward
$B_{M} =\frac{(\mu)_{o^{i}}}{4 \pi R}\left(\frac{3}{2} \pi\right)+\frac{(\mu)_{o} i}{4 \pi(2 R)}\left(\frac{\pi}{2}\right)$
$B_{M} =\frac{7}{16} \frac{\mu_{O} i}{R} \otimes$