- JEE Mains 2021
- Hard
- MCQ
There are two infinitely long straight current carrying conductors and they are held at right angles to each other so that their common ends meet at the origin as shown in the figure given below. The ratio of current in both conductor is \(1: 1\). The magnetic field at point \(P\) is ...... .

Answer & Solution
(A) \(\frac{\mu_{0} I}{4 \pi x y}\left[\sqrt{x^{2}+y^{2}}+(x+y)\right]\)
Step-by-step Solution
Detailed explanation
\(=\frac{\mu_{0}}{4 \pi} \frac{{I}}{{y}}\left(1+\frac{{x}}{\sqrt{{x}^{2}+{y}^{2}}}\right) \ldots \ldots(1)\)
\({B}_{\text {due to wire }(2)}=\frac{\mu_{0}}{4 \pi} \frac{{I}}{{x}}\left(\sin 90^{\circ}+\sin \theta_{2}\right)\)
\(=\frac{\mu_{0}}{4 \pi} \frac{{I}}{{x}}\left(1+\frac{{y}}{\sqrt{{x}^{2}+{y}^{2}}}\right) \ldots \ldots(2)\)
Total magnetic field
\({B}={B}_{1}+{B}_{2}\)
\({B}=\frac{\mu_{0} {I}}{4 \pi}\left[\frac{1}{{y}}+\frac{{x}}{{y} \sqrt{{x}^{2}+{y}^{2}}}+\frac{1}{{x}}+\frac{{y}}{{x} \sqrt{{x}^{2}+{y}^{2}}}\right]\)
\({B}=\frac{\mu_{0} {I}}{4 \pi}\left[\frac{{x}+{y}}{{xy}}+\frac{{x}^{2}+{y}^{2}}{{xy} \sqrt{{x}^{2}+{y}^{2}}}\right]\)
\({B}=\frac{\mu_{0} {I}}{4 \pi}\left[\frac{{x}+{y}}{{xy}}+\frac{\sqrt{{x}^{2}+{y}^{2}}}{{xy}}\right]\)
\({B}=\frac{\mu_{0} {I}}{4 \pi {xy}}\left[\sqrt{{x}^{2}+{y}^{2}}+({x}+{y})\right]\)
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