Physics · Moving Charges and Magnetic Field

JEE Main 2025 — 22 January, Evening Shift — Question 62

Two long parallel wires XX and YY, separated by a distance of 6 cm , carry

currents of 5 A and 4 A , respectively, in opposite directions as shown in the figure.

Magnitude of the resultant magnetic field at point P at a distance of 4 cm from

wire Y is x×10−5 T\mathrm{x} \times 10^{-5} \mathrm{~T}. The value of xx is \qquad .

Take permeability of free space as μ0=4π×10−7\mu_{0}=4 \pi \times 10^{-7} SI units.

Question figure

Answer: 1

Numerical answer — enter this value.

Step-by-step solution

Conceptual

Answer key and solution verified before publishing.

Practise Moving Charges and Magnetic Field

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Exam
JEE Main 2025
Subject
Physics
Chapter
Moving Charges and Magnetic Field
Topic
Magnetic Field Due to Current-Carrying Wire - Biot-Savart Law
Two long parallel wires X and Y , separated by a distance of 6 cm … | JEE Main 2025 PYQ with Solution · DhiX AI