A current i is flowing through the loop. The | vedclass

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(D) The loop consists of four parts: two radial segments ($AB$ and $CD$) and two circular arcs ($DA$ and $BC$).$(i)$ For the radial segments $AB$ and

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

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(D) The loop consists of four parts: two radial segments ($AB$ and $CD$) and two circular arcs ($DA$ and $BC$).$(i)$ For the radial segments $AB$ and $CD$,the magnetic field at the center $M$ is zero because the point $M$ lies on the line of the current elements.(ii) For the circular arc $DA$ with radius $R$ and angle $	heta_1 = 270^{\circ} = \frac{3\pi}{2}$ radians,the magnetic field at $M$ is $B_1 = \frac{\mu_0 i 	heta_1}{4\pi R} = \frac{\mu_0 i (3\pi/2)}{4\pi R} = \frac{3\mu_0 i}{8R}$. By the right-hand rule,the direction is into the plane of the paper.(iii) For the circular arc $BC$ with radius $2R$ and angle $	heta_2 = 90^{\circ} = \frac{\pi}{2}$ radians,the magnetic field at $M$ is $B_2 = \frac{\mu_0 i 	heta_2}{4\pi (2R)} = \frac{\mu_0 i (\pi/2)}{8\pi R} = \frac{\mu_0 i}{16R}$. By the right-hand rule,the direction is into the plane of the paper.(iv) The total magnetic field at $M$ is $B = B_1 + B_2 = \frac{3\mu_0 i}{8R} + \frac{\mu_0 i}{16R} = \frac{6\mu_0 i + \mu_0 i}{16R} = \frac{7\mu_0 i}{16R}$.Since both fields are directed into the plane of the paper,the resultant field is $\frac{7}{16} \frac{\mu_0 i}{R}$ into the plane of the paper.

$A$ 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 $M$ of the loop is $\frac{\mu_{0} i}{R}$ times
$(MA=R, MB=2 R, \angle DMA=90^{\circ})$

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$A$ 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 $M$ of the loop is $\frac{\mu_{0} i}{R}$ times$(MA=R, MB=2 R, \angle DMA=90^{\circ})$