The magnetic field at the centre of a wire loop formed by two semicircular wires of radii R 1 =2 π m and R 2 =4 π m carrying current I =4 A as per figure given below is α × 10 -7 T . The value of α is (Centre O is common for all segments)

Physics · Moving Charges and Magnetic Field

JEE Main 2024 — 27 January, Shift 2 — Question 50

The magnetic field at the centre of a wire loop formed by two semicircular wires of radii $\mathrm{R}_{1}=2 \pi \mathrm{~m}$ and $R_{2}=4 \pi \mathrm{~m}$ carrying current $\mathrm{I}=4 \mathrm{~A}$ as per figure given below is $\alpha \times 10^{-7} \mathrm{~T}$ . The value of $\alpha$ is (Centre O is common for all segments)

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

$\frac{\mu_{0} \mathrm{i}}{2 \mathrm{R}_{2}}\left(\frac{\pi}{2 \pi}\right) \otimes+\frac{\mu_{0} \mathrm{i}}{2 \mathrm{R}_{1}}\left(\frac{\pi}{2 \pi}\right) \otimes$

$\left(\frac{\mu_{0} \mathrm{i}}{4 \mathrm{R}_{2}}+\frac{\mu_{0} \mathrm{i}}{4 \mathrm{R}_{1}}\right) \otimes$

$\frac{4 \pi \times 10^{-7} \times 4}{4 \times 4 \pi}+\frac{4 \pi \times 10^{-7} \times 4}{4 \times 2 \pi}$

$=3 \times 10^{-7}=\alpha \times 10^{-7}$

$\alpha=3$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 27 January, Shift 2
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Magnetic Field Due to Current-Carrying Wire - Biot-Savart Law