Electromagnetic Waves

JEE Main level, single correct, medium

71 of 165

Question 71 of 165

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 conductors is 1:11:1. The magnetic field at point PP is_________.

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Answer: (C) μ0I4πxy[x2+y2+(x+y)]\dfrac{{\mu }_{0}I}{4\pi xy}\left[\sqrt{{x}^{2}+{y}^{2}}+(x+y)\right]

B=μ0I4πy(1+sin⁡θ1)+μ0I4πx(1+sin⁡θ2)B=\dfrac{{\mu }_{0}I}{4\pi y}\left(1+\sin {\theta }_{1}\right)+\dfrac{{\mu }_{0}I}{4\pi x}\left(1+\sin {\theta }_{2}\right)
B=μ0I4πy(1+xx2+y2)+μ0I4πx(1+yx2+y2)B=\dfrac{{\mu }_{0}I}{4\pi y}\left(1+\dfrac{x}{\sqrt{{x}^{2}+{y}^{2}}}\right)+\dfrac{{\mu }_{0}I}{4\pi x}\left(1+\dfrac{y}{\sqrt{{x}^{2}+{y}^{2}}}\right)
B=μ0I4πx+μ0I4πy+μ0I4π[xyx2+y2+yxx2+y2]⇒B=μ0I4π[1x+1y]+μ0I4πx2+y2xy=μ0I4π[x+y+x2+y2xy]B=\dfrac{{\mu }_{0}I}{4\pi x}+\dfrac{{\mu }_{0}I}{4\pi y}+\dfrac{{\mu }_{0}I}{4\pi }\left[\dfrac{x}{y\sqrt{{x}^{2}+{y}^{2}}}+\dfrac{y}{x\sqrt{{x}^{2}+{y}^{2}}}\right] \Rightarrow B=\dfrac{{\mu }_{0}I}{4\pi }\left[\dfrac{1}{x}+\dfrac{1}{y}\right]+\dfrac{{\mu }_{0}I}{4\pi }\dfrac{\sqrt{{x}^{2}+{y}^{2}}}{xy}=\dfrac{{\mu }_{0}I}{4\pi }\left[\dfrac{x+y+\sqrt{{x}^{2}+{y}^{2}}}{xy}\right]

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