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. $(MA = R,\ MB = 2R,\ \angle DMA = 90{^\circ}$ [image:4_image44.jpeg]

Question

Dr. Abubacker · Accepted Answer

(d) (i) Magnetic field at the centre due to the curved portion $DA = \frac{\mu_{0}i}{4\pi R}\ \left( \frac{3\pi}{2} \right)$
According to right hand screw rule, the magnetic field will be into the plane of paper.
[image:4_image122.emf]
(ii) Magnetic field at $M$ due to $AB$ is zero.
(iii) Magnetic field at the centre due to the curved portion $BC$ is $\frac{\mu_{0}i}{4\pi 2R}\ \left( \frac{\pi}{2} \right).$ According to
right hand screw rule, the magnetic field will be into the plane of paper.
(iv) Magnetic field at $M$ due to $DC$ is zero.
Hence, the resultant magnetic field at $M$
$\ \ \ \  = \frac{3\mu_{0}i}{8R} + 0 + \frac{\mu_{0}i}{16R} + 0 = \frac{7\mu_{0}i}{16R}\ $

Q. 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. $(MA = R,\ MB = 2R,\ \angle DMA = 90{^\circ}$

Explanation

Reviewed by Dr. Abubacker

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