JEE Main 2023 — Physics Electromagnetism

(A)

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

So, ${B}_{\mathrm{ab}}=\frac{{\mu }_{0}}{4\pi }\frac{I}{r}={B}_{\mathrm{de}}$Bab​=4πμ0​​rI​=Bde​ (directed out of the plane in both cases)

${B}_{bcd}=\frac{{\mu }_{0}}{4\pi }\frac{I}{r}(2\pi )$Bbcd​=4πμ0​​rI​(2π) (in the plane)

Thus, magnetic field at $O$O is

$${B}_{O}=-\frac{{\mu }_{0}}{4\pi }\frac{I}{r}+\frac{{\mu }_{0}}{4\pi }\frac{I}{r}(2\pi )-\frac{{\mu }_{0}}{4\pi }\frac{I}{r}$$BO​=−4πμ0​​rI​+4πμ0​​rI​(2π)−4πμ0​​rI​

${B}_{O}=\frac{{\mu }_{0}}{2\pi }\frac{I}{r}(\pi -1)$BO​=2πμ0​​rI​(π−1) (III)

(B)

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

${B}_{\mathrm{ab}}=\frac{{\mu }_{0}}{4\pi }\frac{I}{r}={B}_{\mathrm{de}}$Bab​=4πμ0​​rI​=Bde​ (directed out of the plane in both cases)

${B}_{\mathrm{bcd}}=\frac{{\mu }_{0}}{4\pi }\frac{I}{r}(\pi )$Bbcd​=4πμ0​​rI​(π) (out of the plane)

Thus, net magnetic field at $O$O is

$${B}_{O}=\frac{{\mu }_{0}}{4\pi }\frac{I}{r}+\frac{{\mu }_{0}}{4\pi }\frac{I}{r}(\pi )+\frac{{\mu }_{0}}{4\pi }\frac{I}{r}$$BO​=4πμ0​​rI​+4πμ0​​rI​(π)+4πμ0​​rI​

${B}_{0}=\frac{{\mu }_{0}}{4\pi }\frac{I}{r}(\pi +2)$B0​=4πμ0​​rI​(π+2) (I)

(C)

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

${B}_{\mathrm{ab}}=\frac{{\mu }_{0}}{4\pi }\frac{I}{r}$Bab​=4πμ0​​rI​ (directed in the plane)

${B}_{\mathrm{bcd}}=\frac{{\mu }_{0}}{4\pi }\frac{I}{r}(\pi )$Bbcd​=4πμ0​​rI​(π) (in the plane)

${B}_{\mathrm{de}}=0$Bde​=0 (passing through the axis)

Thus, magnetic field at $O$O is

${B}_{O}=\frac{{\mu }_{0}}{4\pi }\frac{I}{r}(\pi +1)$BO​=4πμ0​​rI​(π+1) (IV)

(D)

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

${B}_{\mathrm{ab}}=0={B}_{\mathrm{de}}$Bab​=0=Bde​ (passing through the axis)

${B}_{\mathrm{bcd}}=\frac{{\mu }_{0}}{4\pi }\frac{I}{r}(\pi )$Bbcd​=4πμ0​​rI​(π) (out of the plane)

Thus, magnetic field at $O$O is

${B}_{O}=\frac{{\mu }_{0}I}{4r}$BO​=4rμ0​I​ (II).

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