The magnetic field at the centre of a circula | vedclass

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(A) Let the magnetic field due to a long straight wire at distance $r$ be $B_0 = \frac{\mu_0 i}{2\pi r}$. The field due to a full circular loop of rad

Vedclass · Accepted Answer

$\left( -\frac{\pi}{2} \right) : \left( \frac{\pi}{2} \right) : \left( \frac{3\pi}{4} - \frac{1}{2} \right)$

Vedclass · Answer

(A) Let the magnetic field due to a long straight wire at distance $r$ be $B_0 = \frac{\mu_0 i}{2\pi r}$. The field due to a full circular loop of radius $r$ is $B_{loop} = \frac{\mu_0 i}{2r} = \pi B_0$.Case $1$: The field is due to a semi-circular arc and two semi-infinite wires. The wires produce fields in opposite directions. The net field is $B_1 = \frac{1}{2} \left( \frac{\mu_0 i}{2r} \right) - \frac{\mu_0 i}{4\pi r} - \frac{\mu_0 i}{4\pi r} = \frac{\mu_0 i}{4r} - \frac{\mu_0 i}{2\pi r} = \frac{\mu_0 i}{4\pi r} (\pi - 2)$.Case $2$: The field is due to a semi-circular arc and two semi-infinite wires. The fields add up: $B_2 = \frac{1}{2} \left( \frac{\mu_0 i}{2r} \right) + \frac{\mu_0 i}{4\pi r} + \frac{\mu_0 i}{4\pi r} = \frac{\mu_0 i}{4r} + \frac{\mu_0 i}{2\pi r} = \frac{\mu_0 i}{4\pi r} (\pi + 2)$.Case $3$: The field is due to a $3/4$ circular arc and two semi-infinite wires. $B_3 = \frac{3}{4} \left( \frac{\mu_0 i}{2r} \right) + \frac{\mu_0 i}{4\pi r} + \frac{\mu_0 i}{4\pi r} = \frac{3\mu_0 i}{8r} + \frac{\mu_0 i}{2\pi r} = \frac{\mu_0 i}{4\pi r} (\frac{3\pi}{2} + 2)$.Comparing the magnitudes and directions,the ratio is $\left( -\frac{\pi}{2} \right) : \left( \frac{\pi}{2} \right) : \left( \frac{3\pi}{4} - \frac{1}{2} \right)$.

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