An element Δ l=Δ x hat i is placed at the origin and carries a large current I=10 A . The magnetic field on the y -axis at a distance of 0.5 m from the elements Δ x of 1 cm length is :

Physics · Moving Charges and Magnetic Field

JEE Main 2024 — 6 April, Shift 1 — Question 48

An element $\Delta l=\Delta x \hat{i}$ is placed at the origin and carries a large current $I=10 \mathrm{~A}$ . The magnetic field on the y -axis at a distance of 0.5 m from the elements $\Delta \mathrm{x}$ of 1 cm length is :

Option A:
$4 \times 10^{-8} \mathrm{~T}$
Correct
Option B:
$8 \times 10^{-8} \mathrm{~T}$
Option C:
$12 \times 10^{-8} \mathrm{~T}$
Option D:
$10 \times 10^{-8} \mathrm{~T}$

Answer: A

Step-by-step solution

$\overrightarrow{\mathrm{dB}}=\frac{\mu_{0} \mathrm{I}}{4 \pi} \frac{(\overrightarrow{\mathrm{~d} l} \times \overrightarrow{\mathrm{r}})}{\mathrm{r}^{3}}$ (Tesla) $=\frac{10^{-7} \times 10 \times\left(\frac{1}{2} \times \frac{1}{100}\right)(+\hat{\mathrm{k}})}{\left(\frac{1}{2}\right)^{3}}=4 \times 10^{-8} \mathrm{~T}(+\hat{\mathrm{k}})$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 1
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Magnetic Field Due to Current-Carrying Wire - Biot-Savart Law