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JEE Main 202430 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
30 January
Shift
Morning Shift
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Main 2024 · 30 January · Morning Shift · Physics · Magnetic Effects of Current

Two insulated circular loop A and B radius a carrying a current of...

Two insulated circular loop AA and BB radius aa carrying a current of II in the anti clockwise direction as shown in figure. The magnitude of the magnetic induction at the centre will be:

Two insulated circular loop A and B radius a carrying a current of I in the anti clockwise direction as shown in

Options

  1. A2μ0Ia\dfrac{\sqrt{2}{\mu}_{0}I}{a}
  2. Bμ0I2a\dfrac{{\mu}_{0}I}{2a}
  3. Cμ0I2a\dfrac{{\mu}_{0}I}{\sqrt{2}a}
  4. D2μ0Ia\dfrac{2{\mu}_{0}I}{a}

Correct answer

C. μ0I2a\dfrac{{\mu}_{0}I}{\sqrt{2}a}

Step-by-step solution

Two insulated circular loop A and B radius a carrying a current of I in the anti clockwise direction as shown in

Right hand thumb rule states that, “If you imagine holding a current carrying wire in your right-hand with your thumb pointing towards the direction of electric current flow then the direction in which your fingers curl, gives the direction of lines of force of the magnetic field”. In this case, direciton of magnetic field at the centre for both wires is as shown below.

Two insulated circular loop A and B radius a carrying a current of I in the anti clockwise direction as shown in

Magnetic field due to a circular wire is given by, BA=BB=μ0I2a{B}_{A} = {B}_{B} = \dfrac{{\mu}_{0}I}{2a}

As both magnetic field lines at centre are perpendicular, therefore

∵Bnet=(BA)2+(BB)2=2μ0I2a=μ0I2a∵ {B}_{net} = \sqrt{{\left({B}_{A}\right)}^{2} + {\left({B}_{B}\right)}^{2}} = \dfrac{\sqrt{2}{\mu}_{0}I}{2a} = \dfrac{{\mu}_{0}I}{\sqrt{2}a}

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