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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Magnetic Field due to a Current

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Magnetic Field due to a Current

Two large metal sheets carry surface currents as depicted in the...

Two large metal sheets carry surface currents as depicted in the figure. The current through a strip of width dldl is K dlK \, dl, where KK is a constant. Determine the magnetic field at the points PP, QQ, and RR.
Two large metal sheets carry surface currents as depicted in the figure. The current through a strip of width dl is K

Options

  1. Aμ0K\mu_0 K towards left at PP, 00 at QQ, μ0K\mu_0 K towards right at RR
  2. B12μ0K\dfrac{1}{2}\mu_0 K towards right at PP, μ0K\mu_0 K towards right at QQ, 12μ0K\dfrac{1}{2}\mu_0 K towards right at RR
  3. C00 at PP, μ0K\mu_0 K towards right at QQ, 00 at RR
  4. D00 at PP, μ0K\mu_0 K towards left at QQ, 00 at RR

Correct answer

C. 00 at PP, μ0K\mu_0 K towards right at QQ, 00 at RR

Step-by-step solution

The magnetic field due to a large infinite metal sheet carrying a surface current density KK is uniform and its magnitude is given by B=μ0K2B = \dfrac{\mu_0 K}{2}.

For the top sheet, the current is coming out of the plane of the paper. By the right-hand thumb rule, the magnetic field produced by it is directed towards the left above the sheet and towards the right below the sheet.

For the bottom sheet, the current is going into the plane of the paper. The magnetic field produced by it is directed towards the right above the sheet and towards the left below the sheet.

At point PP (above both sheets):
The field due to the top sheet is μ0K2\dfrac{\mu_0 K}{2} towards the left.
The field due to the bottom sheet is μ0K2\dfrac{\mu_0 K}{2} towards the right.
Net magnetic field at P=0P = 0.

At point QQ (between the sheets):
The field due to the top sheet is μ0K2\dfrac{\mu_0 K}{2} towards the right.
The field due to the bottom sheet is μ0K2\dfrac{\mu_0 K}{2} towards the right.
Net magnetic field at Q=μ0K2+μ0K2=μ0KQ = \dfrac{\mu_0 K}{2} + \dfrac{\mu_0 K}{2} = \mu_0 K towards the right.

At point RR (below both sheets):
The field due to the top sheet is μ0K2\dfrac{\mu_0 K}{2} towards the right.
The field due to the bottom sheet is μ0K2\dfrac{\mu_0 K}{2} towards the left.
Net magnetic field at R=0R = 0.
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