JEE Challenger
More from Electromagnetic Induction
PhyHardAdv 2022 • Paper 1

Torque on Circular Loop in Varying Solenoid Field

A small circular loop of area AA and resistance RR is fixed on a horizontal xyxy-plane with the center of the loop always on the axis n^\hat{n} of a long solenoid. The solenoid has mm turns per unit length and carries current II counterclockwise as shown in the figure. The magnetic field due to the solenoid is in n^\hat{n} direction. List-I gives time dependences of n^\hat{n} in terms of a constant angular frequency ω\omega. List-II gives the torques experienced by the circular loop at time t=π6ωt = \frac{\pi}{6\omega}.

Let α=A2μ02m2I2ω2R\alpha = \frac{A^2 \mu_0^2 m^2 I^2 \omega}{2R}.

List-IList-II(I) 12(sin⁡ωt j^+cos⁡ωt k^)(P) 0(II) 12(sin⁡ωt i^+cos⁡ωt j^)(Q) −α4 i^(III) 12(sin⁡ωt i^+cos⁡ωt k^)(R) 3α4 i^(IV) 12(cos⁡ωt j^+sin⁡ωt k^)(S) α4 j^(T) −3α4 i^\begin{array}{ll} \text{List-I} & \text{List-II} \\[4pt] \text{(I) } \frac{1}{\sqrt{2}}(\sin \omega t \, \hat{j} + \cos \omega t \, \hat{k}) & \text{(P) } 0 \\[4pt] \text{(II) } \frac{1}{\sqrt{2}}(\sin \omega t \, \hat{i} + \cos \omega t \, \hat{j}) & \text{(Q) } -\frac{\alpha}{4} \, \hat{i} \\[4pt] \text{(III) } \frac{1}{\sqrt{2}}(\sin \omega t \, \hat{i} + \cos \omega t \, \hat{k}) & \text{(R) } \frac{3\alpha}{4} \, \hat{i} \\[4pt] \text{(IV) } \frac{1}{\sqrt{2}}(\cos \omega t \, \hat{j} + \sin \omega t \, \hat{k}) & \text{(S) } \frac{\alpha}{4} \, \hat{j} \\[4pt] & \text{(T) } -\frac{3\alpha}{4} \, \hat{i} \end{array}

Which one of the following options is correct?

Question Diagram 1

Options

A

I →\rightarrow Q, II →\rightarrow P, III →\rightarrow S, IV →\rightarrow T

B

I →\rightarrow S, II →\rightarrow T, III →\rightarrow Q, IV →\rightarrow P

C

I →\rightarrow Q, II →\rightarrow P, III →\rightarrow S, IV →\rightarrow R

Correct
D

I →\rightarrow T, II →\rightarrow Q, III →\rightarrow P, IV →\rightarrow R

Topics & Concepts

Electromagnetic InductionFaraday's LawTorqueMagnetic Force

Step-by-Step Solution

To find the torque experienced by the circular loop, we analyze the electromagnetic induction in the loop due to the changing magnetic field of the solenoid.

1. Magnetic Field and Flux

The magnetic field produced by the long solenoid along its axis n^\hat{n} is given by: B⃗=μ0mIn^\vec{B} = \mu_0 m I \hat{n}

The circular loop of area AA lies in the horizontal xyxy-plane, so its area vector is: A⃗loop=Ak^\vec{A}_{\text{loop}} = A \hat{k}

The magnetic flux Φ\Phi passing through the loop is: Φ=B⃗⋅A⃗loop=μ0mIA(n^⋅k^)\Phi = \vec{B} \cdot \vec{A}_{\text{loop}} = \mu_0 m I A (\hat{n} \cdot \hat{k})

2. Induced EMF and Current

By Faraday's Law of Electromagnetic Induction, the induced EMF in the loop is: E=−dΦdt=−μ0mIAddt(n^⋅k^)\mathcal{E} = -\frac{d\Phi}{dt} = -\mu_0 m I A \frac{d}{dt} (\hat{n} \cdot \hat{k})

The induced current flowing through the loop of resistance RR is: iind=ER=−μ0mIARddt(n^⋅k^)i_{\text{ind}} = \frac{\mathcal{E}}{R} = -\frac{\mu_0 m I A}{R} \frac{d}{dt} (\hat{n} \cdot \hat{k})

3. Magnetic Dipole Moment and Torque

The magnetic dipole moment of the induced current loop is: M⃗=iindA⃗loop=−μ0mIA2R(ddt(n^⋅k^))k^\vec{M} = i_{\text{ind}} \vec{A}_{\text{loop}} = -\frac{\mu_0 m I A^2}{R} \left( \frac{d}{dt}(\hat{n} \cdot \hat{k}) \right) \hat{k}

The torque experienced by the circular loop is: τ⃗=M⃗×B⃗=[−μ0mIA2R(ddt(n^⋅k^))k^]×(μ0mIn^)=−μ02m2I2A2R(ddt(n^⋅k^))(k^×n^)\vec{\tau} = \vec{M} \times \vec{B} = \left[ -\frac{\mu_0 m I A^2}{R} \left( \frac{d}{dt}(\hat{n} \cdot \hat{k}) \right) \hat{k} \right] \times (\mu_0 m I \hat{n}) = -\frac{\mu_0^2 m^2 I^2 A^2}{R} \left( \frac{d}{dt}(\hat{n} \cdot \hat{k}) \right) (\hat{k} \times \hat{n})

Given α=A2μ02m2I2ω2R\alpha = \frac{A^2 \mu_0^2 m^2 I^2 \omega}{2R}, we can write: τ⃗=−2αω(ddt(n^⋅k^))(k^×n^)\vec{\tau} = -\frac{2\alpha}{\omega} \left( \frac{d}{dt}(\hat{n} \cdot \hat{k}) \right) (\hat{k} \times \hat{n})

At time t=π6ωt = \frac{\pi}{6\omega}, we have ωt=π6\omega t = \frac{\pi}{6}, which yields sin⁡ωt=12\sin \omega t = \frac{1}{2} and cos⁡ωt=32\cos \omega t = \frac{\sqrt{3}}{2}.

Evaluation for Each Case in List-I:

Case (I): n^=12(sin⁡ωt j^+cos⁡ωt k^)\hat{n} = \frac{1}{\sqrt{2}}(\sin \omega t \, \hat{j} + \cos \omega t \, \hat{k})

  • n^⋅k^=12cos⁡ωt  ⟹  ddt(n^⋅k^)=−ω2sin⁡ωt\hat{n} \cdot \hat{k} = \frac{1}{\sqrt{2}} \cos \omega t \implies \frac{d}{dt}(\hat{n} \cdot \hat{k}) = -\frac{\omega}{\sqrt{2}} \sin \omega t
  • At t=π6ωt = \frac{\pi}{6\omega}: ddt(n^⋅k^)=−ω22\frac{d}{dt}(\hat{n} \cdot \hat{k}) = -\frac{\omega}{2\sqrt{2}}
  • k^×n^=12sin⁡ωt(k^×j^)=−122i^\hat{k} \times \hat{n} = \frac{1}{\sqrt{2}} \sin \omega t (\hat{k} \times \hat{j}) = -\frac{1}{2\sqrt{2}} \hat{i}
  • Torque: τ⃗=−2αω(−ω22)(−122i^)=−α4i^\vec{\tau} = -\frac{2\alpha}{\omega} \left( -\frac{\omega}{2\sqrt{2}} \right) \left( -\frac{1}{2\sqrt{2}} \hat{i} \right) = -\frac{\alpha}{4} \hat{i}
  • Match: I→Q\text{I} \rightarrow \text{Q}

Case (II): n^=12(sin⁡ωt i^+cos⁡ωt j^)\hat{n} = \frac{1}{\sqrt{2}}(\sin \omega t \, \hat{i} + \cos \omega t \, \hat{j})

  • n^⋅k^=0  ⟹  ddt(n^⋅k^)=0\hat{n} \cdot \hat{k} = 0 \implies \frac{d}{dt}(\hat{n} \cdot \hat{k}) = 0
  • Torque: τ⃗=0⃗\vec{\tau} = \vec{0}
  • Match: II→P\text{II} \rightarrow \text{P}

Case (III): n^=12(sin⁡ωt i^+cos⁡ωt k^)\hat{n} = \frac{1}{\sqrt{2}}(\sin \omega t \, \hat{i} + \cos \omega t \, \hat{k})

  • n^⋅k^=12cos⁡ωt  ⟹  ddt(n^⋅k^)=−ω2sin⁡ωt\hat{n} \cdot \hat{k} = \frac{1}{\sqrt{2}} \cos \omega t \implies \frac{d}{dt}(\hat{n} \cdot \hat{k}) = -\frac{\omega}{\sqrt{2}} \sin \omega t
  • At t=π6ωt = \frac{\pi}{6\omega}: ddt(n^⋅k^)=−ω22\frac{d}{dt}(\hat{n} \cdot \hat{k}) = -\frac{\omega}{2\sqrt{2}}
  • k^×n^=12sin⁡ωt(k^×i^)=122j^\hat{k} \times \hat{n} = \frac{1}{\sqrt{2}} \sin \omega t (\hat{k} \times \hat{i}) = \frac{1}{2\sqrt{2}} \hat{j}
  • Torque: τ⃗=−2αω(−ω22)(122j^)=α4j^\vec{\tau} = -\frac{2\alpha}{\omega} \left( -\frac{\omega}{2\sqrt{2}} \right) \left( \frac{1}{2\sqrt{2}} \hat{j} \right) = \frac{\alpha}{4} \hat{j}
  • Match: III→S\text{III} \rightarrow \text{S}

Case (IV): n^=12(cos⁡ωt j^+sin⁡ωt k^)\hat{n} = \frac{1}{\sqrt{2}}(\cos \omega t \, \hat{j} + \sin \omega t \, \hat{k})

  • n^⋅k^=12sin⁡ωt  ⟹  ddt(n^⋅k^)=ω2cos⁡ωt\hat{n} \cdot \hat{k} = \frac{1}{\sqrt{2}} \sin \omega t \implies \frac{d}{dt}(\hat{n} \cdot \hat{k}) = \frac{\omega}{\sqrt{2}} \cos \omega t
  • At t=π6ωt = \frac{\pi}{6\omega}: ddt(n^⋅k^)=3ω22\frac{d}{dt}(\hat{n} \cdot \hat{k}) = \frac{\sqrt{3}\omega}{2\sqrt{2}}
  • k^×n^=12cos⁡ωt(k^×j^)=−322i^\hat{k} \times \hat{n} = \frac{1}{\sqrt{2}} \cos \omega t (\hat{k} \times \hat{j}) = -\frac{\sqrt{3}}{2\sqrt{2}} \hat{i}
  • Torque: τ⃗=−2αω(3ω22)(−322i^)=3α4i^\vec{\tau} = -\frac{2\alpha}{\omega} \left( \frac{\sqrt{3}\omega}{2\sqrt{2}} \right) \left( -\frac{\sqrt{3}}{2\sqrt{2}} \hat{i} \right) = \frac{3\alpha}{4} \hat{i}
  • Match: IV→R\text{IV} \rightarrow \text{R}

Conclusion

The correct matching is: I→Q,II→P,III→S,IV→R\text{I} \rightarrow \text{Q}, \quad \text{II} \rightarrow \text{P}, \quad \text{III} \rightarrow \text{S}, \quad \text{IV} \rightarrow \text{R}

This corresponds to Option C.

PreviousNext