Question

A current $I$ is flowing through a 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 (Here, $MA=R,MB=2R,\angle DMA=90^{\circ })$

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

Answer 1

Answer: (D)

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