Question 14 of 276

The magnitude of the magnetic field $(B)$ due to loop $A B C D$ at the origin $(O)$ is

AzeroB$\frac{\mu_0(b-a)}{24 a b}$C$\frac{\mu_0 \mathrm{I}}{4 \pi}\left[\frac{b-a}{a b}\right]$D$\frac{\mu_o \mid}{4 \pi}\left[2(b-a)+\frac{\pi}{3}(a+b)\right]$

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Answer: (B) $\frac{\mu_0(b-a)}{24 a b}$

Net magnetic field due to loop $A B C D$ at $O$ is
$$\begin{aligned} & B=B_{A B}+B_{B C}+B_{C D}+B_{D A} \\ & =0+\frac{\mu_o I}{4 \pi a} \times \frac{\pi}{6}+0-\frac{\mu_o I}{4 \pi b} \times \frac{\pi}{6}=\frac{\mu_o I}{24 a}-\frac{\mu_o I}{24 b}=\frac{\mu_o I}{24 a b}(b-a) \end{aligned}$$

JEE Main Physics practice question on Magnetic Effects & Magnetism (Biot-Savart's Law and Ampere's Circuital Law), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.