The horizontal component of earth's magnetic field at a place is 3.5 × 10 -5 T . A very long straight conductor carrying current of √(2) A in the direction from South east to North West is placed. The force per unit length experienced by the conductor is \ \ \ \ \ \ \ × 10 -6 N / m .

Physics · Moving Charges and Magnetic Field

JEE Main 2024 — 30 January, Shift 1 — Question 46

The horizontal component of earth's magnetic field at a place is $3.5 \times 10^{-5} \mathrm{~T}$ . A very long straight conductor carrying current of $\sqrt{2} A$ in the direction from South east to North West is placed. The force per unit length experienced by the conductor is $\_\_\_\_\_\_\_$ $\times 10^{-6} \mathrm{~N} / \mathrm{m}$ .

Answer: 35

Numerical answer — enter this value.

Step-by-step solution

$\quad B_{H}=3.5 \times 10^{-5} T$

$F=i \ell B \sin \theta, \quad i=\sqrt{2} A$

$\frac{F}{\ell}=i B \sin \theta=\sqrt{2} \times 3.5 \times 10^{-5} \times \frac{1}{\sqrt{2}}$

$=35 \times 10^{-6} \mathrm{~N} / \mathrm{m}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 30 January, Shift 1
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Magnetic Field Due to Current-Carrying Wire - Biot-Savart Law