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QJEEM25Ph0bbe: JEE Main 2025 Physics - Class 12 JEEM25Ph0bbe Solution: Two long parallel wires and , separated by a distance of 6 cm , …

Two long parallel wires XX and YY , separated by a distance of 6 cm , carry currents of 5 A and 4A, 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 Tx \times 10^{-5} \mathrm{~T} . The value of xx is _________ . Take permeability of free space as μ0=4π×10−7\mu_0=4 \pi \times 10^{-7} SI units.

JEE Main 2025 (Online) 22nd January Evening Shift Physics - Magnetic Effect of Current Question 7 English

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JEE Main 2025 (Online) 22nd January Evening Shift Physics - Magnetic Effect of Current Question 7 English Explanation

amp;B=μ0(5)2π×.01−μ042π×0.04amp;=−100μ04πamp;=−100×10−7amp;=−1×10−5 T\begin{aligned}& B=\frac{\mu_0(5)}{2 \pi \times .01}-\frac{\mu_0 4}{2 \pi \times 0.04} \\& =-\frac{100 \mu_0}{4 \pi} \\& =-100 \times 10^{-7} \\& =-1 \times 10^{-5} \mathrm{~T}\end{aligned}

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