Question 45 of 276

Find the magnetic field at the point P in figure. The curved portion is a semicircle connected to two long straight wires.

A$\dfrac{{\mu }_{0}i}{2r}\left(1+\dfrac{2}{\pi }\right)$B$\dfrac{{\mu }_{0}i}{2r}\left(1+\dfrac{1}{\pi }\right)$C$\dfrac{{\mu }_{0}i}{2r}\left(\dfrac{1}{2}+\dfrac{1}{2\pi }\right)$D$\dfrac{{\mu }_{0}i}{2r}\left(\dfrac{1}{2}+\dfrac{1}{\pi }\right)$

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Answer: (C) $\dfrac{{\mu }_{0}i}{2r}\left(\dfrac{1}{2}+\dfrac{1}{2\pi }\right)$

Magnitude of magnetic field due to semi-infinite portion of wire AB at point C is
${B}_{1}=\dfrac{{\mu }_{0}i}{4\pi r}$
Magnitude of magnetic field due to semicircular arc BD at point C is
${B}_{2}=\dfrac{{\mu }_{0}i}{4r}$
Magnitude of magnetic field due to semi-infinite portion DE at point C is 0.
Using right-hand rule, it can be seen that directions of $\vec{{B}_{1}}$ and $\vec{{B}_{2}}$ are both outwards from the plane.
Therefore, the magnitude of net magnetic field at point C is
$B={B}_{1}+{B}_{2}$
$=\dfrac{{\mu }_{0}i}{2r}\left[\dfrac{1}{2}+\dfrac{1}{2\pi }\right]$

JEE Main Physics practice question on Magnetic Effects & Magnetism (Magnetic Field), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.