Physics · Moving Charges and Magnetic Field

JEE Main 2024 — 4 April, Shift 2 — Question 58

Two parallel long current carrying wire separated by a distance 2 r are shown in the figure. The ratio of magnetic field at $A$ to the magnetic field produced at $C$ is $\frac{x}{7}$ . The value of $x$ is $\qquad$

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

$\quad \mathrm{B}_{\mathrm{A}}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}+\frac{\mu_{0}(2 \mathrm{i})}{2 \pi(3 \mathrm{r})}=\frac{5 \mu_{0} \mathrm{i}}{6 \pi \mathrm{r}}$

$B_{C}=\frac{\mu_{0}(2 i)}{2 \pi r}+\frac{\mu_{0} i}{2 \pi(3 r)}=\frac{7 \mu_{0} i}{6 \pi r}$ $\therefore \frac{\mathrm{B}_{\mathrm{A}}}{\mathrm{B}_{\mathrm{C}}}=\frac{5}{7}$

$\therefore \mathrm{x}=5$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 2
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Magnetic Field Due to Current-Carrying Wire - Biot-Savart Law