JEE MAIN - Physics (2023 - 25th January Evening Shift - No. 20)
Two long parallel wires carrying currents 8A and 15A in opposite directions are placed at a distance of 7 cm from each other. A point P is at equidistant from both the wires such that the lines joining the point P to the wires are perpendicular to each other. The magnitude of magnetic field at P is _____________ $$\times~10^{-6}$$ T.
(Given : $$\sqrt2=1.4$$)
Answer
68
Explanation
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$$ \begin{aligned} & B_1=\frac{\mu_0 i_1}{2 \pi d} \quad B_2=\frac{\mu_0 i_2}{2 \pi d} \\\\ & B_{\text {net }}=\sqrt{B_1^2+B_2^2} = \frac{\mu_0}{2 \pi d} \sqrt{i_1^2+i_2^2} \\\\ & = \frac{4 \pi \times 10^{-7}}{2 \pi \times(7 / \sqrt{2}) \times 10^{-2}} \times \sqrt{15^2+8^2}\left(\text {As }d=\frac{7}{\sqrt{2}} \mathrm{~cm}\right) \\\\ & = 68 \times 10^{-6} \mathrm{~T} \end{aligned} $$
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