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(B) The magnetic field at the center $C$ is the resultant of the magnetic fields produced by the two quarter-circular arcs of radii $r$ and $R$.The ma

Vedclass · Accepted Answer

$\frac{\mu_{0} I}{8} \left(\frac{1}{r} - \frac{1}{R}\right)$ out of the page

Vedclass · Answer

(B) The magnetic field at the center $C$ is the resultant of the magnetic fields produced by the two quarter-circular arcs of radii $r$ and $R$.The magnetic field due to a full circular loop of radius $a$ carrying current $I$ at its center is $B = \frac{\mu_0 I}{2a}$.For a quarter-circular arc, the magnetic field is $B_{arc} = \frac{1}{4} \left(\frac{\mu_0 I}{2a}\right) = \frac{\mu_0 I}{8a}$.Using the right-hand rule:$1$. For the inner arc of radius $r$, the current flows in a counter-clockwise direction, so the magnetic field at $C$ is directed out of the page $(\odot)$.$2$. For the outer arc of radius $R$, the current flows in a clockwise direction, so the magnetic field at $C$ is directed into the page $(\otimes)$.The net magnetic field at $C$ is $B_{net} = B_r - B_R = \frac{\mu_0 I}{8r} - \frac{\mu_0 I}{8R} = \frac{\mu_0 I}{8} \left(\frac{1}{r} - \frac{1}{R}\right)$.Since $r  \frac{1}{R}$, the net magnetic field is directed out of the page.

An arrangement with a pair of quarter circular coils of radii $r$ and $R$ with a common centre $C$ and carrying a current $I$ is shown in the figure. The permeability of free space is $\mu_0$. The magnetic field at $C$ is

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