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JEE Main 20231 FebruaryMorning Shift
Exam
JEE Main
Year
2023
Date
1 February
Shift
Morning Shift
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Main 2023 · 1 February · Morning Shift · Physics · Magnetic Effects of Current

Find the magnetic field at the point P in figure. The curved...

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

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

Options

  1. Aμ0i2r(1+2π)\dfrac{{\mu}_{0}i}{2r}\left(1 + \dfrac{2}{\pi}\right)
  2. Bμ0i2r(1+1π)\dfrac{{\mu}_{0}i}{2r}\left(1 + \dfrac{1}{\pi}\right)
  3. Cμ0i2r(12+12π)\dfrac{{\mu}_{0}i}{2r}\left(\dfrac{1}{2} + \dfrac{1}{2\pi}\right)
  4. Dμ0i2r(12+1π)\dfrac{{\mu}_{0}i}{2r}\left(\dfrac{1}{2} + \dfrac{1}{\pi}\right)

Correct answer

C. μ0i2r(12+12π)\dfrac{{\mu}_{0}i}{2r}\left(\dfrac{1}{2} + \dfrac{1}{2\pi}\right)

Step-by-step solution

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

Magnitude of magnetic field due to semi-infinite portion of wire AB at point C is

B1=μ0i4\pir{B}_{1} = \dfrac{{\mu}_{0}i}{4\pir}

Magnitude of magnetic field due to semicircular arc BD at point C is

B2=μ0i4r{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 B1→\overset{\to}{{B}_{1}} and B2→\overset{\to}{{B}_{2}} are both outwards from the plane.

Therefore, the magnitude of net magnetic field at point C is

B=B1+B2B = {B}_{1} + {B}_{2}

=μ0i2r[12+12π]= \dfrac{{\mu}_{0}i}{2r}\left[\dfrac{1}{2} + \dfrac{1}{2\pi}\right]

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