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JEE Main 20205 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
5 September
Shift
Morning Shift
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Main 2020 · 5 September · Morning Shift · Physics · Magnetic Effects of Current

A square loop of side 2a, and carrying current I is kept in XZ...

A square loop of side 2a,2a, and carrying current II is kept in XZXZ plane with its centre at origin. A long wire carrying the same current II is placed parallel to the zz-axis and passing through the point (0,b,0),(b>>a)( 0, b, 0 ) , ( b > > a ). The magnitude of the torque on the loop about zz-axis is given by.A square loop of side 2a, and carrying current I is kept in XZ plane with its centre at origin. A long wire carryingA square loop of side 2a, and carrying current I is kept in XZ plane with its centre at origin. A long wire carrying

Options

  1. Aμ0I2a22\pib\dfrac{{\mu}_{0}{I}^{2}{a}^{2}}{2\pib}
  2. Bμ0I2a32πb2\dfrac{{\mu}_{0}{I}^{2}{a}^{3}}{2\pi{b}^{2}}
  3. C2μ0Ia22\pib\dfrac{2{\mu}_{0}I a 2 2}{\pib}
  4. D2μ0Ia32πb2\dfrac{2{\mu}_{0}I a 3 2}{\pi{b}^{2}}

Correct answer

C. 2μ0Ia22\pib\dfrac{2{\mu}_{0}I a 2 2}{\pib}

Step-by-step solution

A square loop of side 2a, and carrying current I is kept in XZ plane with its centre at origin. A long wire carrying

Net force acting on wire ABAB & CDCD are zero

A square loop of side 2a, and carrying current I is kept in XZ plane with its centre at origin. A long wire carrying

F1=ILB=I×(2a)×(μ0I2πa2+b2)=μ0I2aπa2+b2{F}_{1} = ILB = I \times \left(2a\right) \times \left(\dfrac{{\mu}_{0}I}{2\pi\sqrt{{a}^{2} + {b}^{2}}}\right) = \dfrac{{\mu}_{0}{I}^{2}a}{\pi\sqrt{{a}^{2} + {b}^{2}}}

τ=F1cosθ×2a=μ0I2aπa2+b2×ba2+b2×2aτ = {F}_{1}cos \theta \times 2a = \dfrac{{\mu}_{0}{I}^{2}a}{\pi\sqrt{{a}^{2} + {b}^{2}}} \times \dfrac{b}{\sqrt{{a}^{2} + {b}^{2}}} \times 2a

τ=2μ0I2a2bπ(a2+b2)≈2μ0I2a2\pib,(a≪b)τ = \dfrac{2{\mu}_{0}{I}^{2}{a}^{2}b}{\pi\left({a}^{2} + {b}^{2}\right)} \approx \dfrac{2{\mu}_{0}{I}^{2}{a}^{2}}{\pib}, \left(a ≪ b\right)

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