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 $\frac{{\mu }_{0}I}{R}$ times
$(MA=R,MB=2R,\angle DMA={90}^{\circ })$.

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

Answer 1

Answer: (D)

$\frac{7}{16}$, but into the plane of the paper

Magnetic field at the centre $M$ due to current through the curved portion $DA$ is
$\overrightarrow{B_1}=\frac{{\mu }_{0}I}{4\pi R}\times \left(\frac{3\pi }{2}\right)$
$=\frac{3{\mu }_{0}I}{8R}⨂$
Magnetic field at the centre $M$ due to current through the straight portion $AB$ is $B_2=0$, since point $M$ lies on the axis of the straight portion $AB$.
Magnetic field at the centre $M$ due to current through the curved portion $BC$ is
$\overrightarrow{B_3}=\frac{{\mu }_{0}I}{4\pi 2R}\times \frac{\pi }{2}=\frac{{\mu }_{0}I}{16R}⨂$
Magnetic field at the centre $M$ due to current through the straight portion $CD$ is $B_4=0$, since point $M$ lies on the axis of the straight portion $CD$.
The resultant magnetic field at the point $M$ is
$\overrightarrow{B}=\overrightarrow{B_1}+\overrightarrow{B_2}+\overrightarrow{B_3}+\overrightarrow{B_4}$
$=(\frac{3{\mu }_{0}I}{8R}+0+\frac{{\mu }_{0}I}{16R}+0)⨂=\frac{7{\mu }_{0}I}{16R}⨂$.