Physics - Magnetism Question with Solution | TestHub

(A)

Here net magnetic field will be sum of magnetic field due to straight wires $\mathrm{ab}, \mathrm{de}$ and due to loop $\mathrm{bcd}$.

So, $B_{\mathrm{ab}}=\frac{{\mu }_0}{4\pi }\frac{I}{r}=B_{\mathrm{de}}$ (directed out of the plane in both cases)

$B_{\mathit{b}\mathit{c}\mathit{d}}=\frac{{\mu }_0}{4\pi }\frac{I}{r}\left(2\pi \right)$ (in the plane)

Thus, magnetic field at $O$ is

$B_O=-\frac{{\mu }_0}{4\pi }\frac{I}{r}+\frac{{\mu }_0}{4\pi }\frac{I}{r}\left(2\pi \right)-\frac{{\mu }_0}{4\pi }\frac{I}{r}$

$B_O=\frac{{\mu }_0}{2\pi }\frac{I}{r}(\pi -1)$  (III)

(B)

Here net magnetic field will be sum of magnetic field due to straight wires $\mathrm{ab}, \mathrm{de}$ and semicircular arc $\mathrm{bcd}$.

$B_{\mathrm{ab}}=\frac{{\mu }_0}{4\pi }\frac{I}{r}=B_{\mathrm{de}}$ (directed out of the plane in both cases)

$B_{\mathrm{bcd}}=\frac{{\mu }_0}{4\pi }\frac{I}{r}\left(\pi \right)$ (out of the plane)

Thus, net magnetic field at $O$ is

$B_O=\frac{{\mu }_0}{4\pi }\frac{I}{r}+\frac{{\mu }_0}{4\pi }\frac{I}{r}\left(\pi \right)+\frac{{\mu }_0}{4\pi }\frac{I}{r}$

$B_0=\frac{{\mu }_0}{4\pi }\frac{\mathit{I}}{\mathit{r}}(\pi +2)$ (I)

(C)

Here net magnetic field will be sum of magnetic field due to straight wires $\mathrm{ab}, \mathrm{de}$ and due to loop $\mathrm{bcd}$.

$B_{\mathrm{ab}}=\frac{{\mu }_0}{4\pi }\frac{I}{r}$ (directed in the plane)

$B_{\mathrm{bcd}}=\frac{{\mu }_0}{4\pi }\frac{I}{r}\left(\pi \right)$ (in the plane)

$B_{\mathrm{de}}=0$ (passing through the axis)

Thus, magnetic field at $O$ is

$B_O=\frac{{\mu }_0}{4\pi }\frac{I}{r}(\pi +1)$  (IV)

(D)

Here net magnetic field will be sum of magnetic field due to straight wires $\mathrm{ab}, \mathrm{de}$ and due to loop $\mathrm{bcd}$.

$B_{\mathrm{ab}}=0=B_{\mathrm{de}}$ (passing through the axis)

$B_{\mathrm{bcd}}=\frac{{\mu }_0}{4\pi }\frac{I}{r}\left(\pi \right)$ (out of the plane)

Thus, magnetic field at $O$ is

$B_O=\frac{{\mu }_{0}I}{4r}$ (II).

A-III, B-I, C-IV, D-II