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Alternating Current question

2007 · Shift 2 · Q20
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Alternating Current question

2007 · Shift 2 · Q20

JEE AdvancedPhysicsAlternating CurrentMCQ+3 / −1
STATEMENT 1 A vertical iron rod has a coil of wire wound over it at the bottom end. An alternating current flows in the coil. The rod goes through a conducting ring as shown in the figure. The ring can float at a certain height above the coil. IIT-JEE 2007 Paper 2 Offline Physics - Alternating Current Question 5 English Because STATEMENT 2 In the above situation, a current is induced in the ring which interacts with the horizontal component of the magnetic field to produce an average force in the upward direction.
  1. A
    Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
  2. B
    Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
  3. C
    Statement-1 is True, Statement-2 is False.
  4. D
    Statement-1 is False, Statement-2 is True.
View written solutionFree

Correct answer: A

Analysis of Statement 1

  1. The coil at the bottom end of the iron rod carries an alternating current (AC). This setup acts as an electromagnet with a time-varying magnetic field.
  2. The iron rod serves to concentrate the magnetic field lines, increasing the magnetic flux.
  3. This time-varying magnetic field passes through the conducting ring placed above the coil, causing a change in magnetic flux through the ring.
  4. According to Faraday's law of electromagnetic induction, a changing magnetic flux induces an electromotive force (EMF) and hence a current in the conducting ring.
  5. According to Lenz's law, the direction of the induced current is such that it opposes the change in magnetic flux that produced it. This results in a net repulsive force between the coil and the ring.
  6. Let the current in the coil be I1=I01sin⁡(ωt)I_1 = I_{01} \sin(\omega t)I1​=I01​sin(ωt). The magnetic flux through the ring is Φ2∝I1\Phi_2 \propto I_1Φ2​∝I1​. The induced EMF in the ring is E2=−dΦ2/dtE_2 = -d\Phi_2/dtE2​=−dΦ2​/dt, so E2E_2E2​ is proportional to −cos⁡(ωt)-\cos(\omega t)−cos(ωt).
  7. The ring has some resistance (R) and self-inductance (L). The induced current I2I_2I2​ in the ring will lag the induced EMF E2E_2E2​ by a phase angle ϕ=arctan⁡(ωL/R)\phi = \arctan(\omega L/R)ϕ=arctan(ωL/R). Because of this phase lag, the primary current I1I_1I1​ and the induced current I2I_2I2​ are, on average, out of phase.
  8. The force between the coil and the ring is proportional to the product of the currents, F(t)∝I1(t)I2(t)F(t) \propto I_1(t) I_2(t)F(t)∝I1​(t)I2​(t). Due to the phase difference, the repulsive part of the force cycle is stronger and lasts longer than the attractive part. This results in a net average repulsive force (upward force) over a complete cycle.
  9. If this average upward force is strong enough to balance the gravitational force (weight) acting on the ring (mgmgmg), the ring will levitate or float at a certain height above the coil. This is a well-known phenomenon, often demonstrated as Thomson's jumping ring experiment.
  10. Therefore, Statement 1 is True.

Analysis of Statement 2

  1. The statement claims that a current is induced in the ring. As explained above, this is correct due to Faraday's law.
  2. The statement then claims this induced current interacts with the horizontal component of the magnetic field to produce the upward force.
  3. Let's analyze the forces. The magnetic field produced by the coil has both a vertical component (BzB_zBz​) and a radial (horizontal) component (BrB_rBr​) at the location of the ring. The field lines loop out from the top of the coil.
  4. The induced current (I2I_2I2​) in the ring flows azimuthally (in a circle). Let's consider a small element of the ring wire, dl⃗d\vec{l}dl. The direction of dl⃗d\vec{l}dl is tangential to the ring.
  5. The Lorentz force on this element is given by dF⃗=I2(dl⃗×B⃗)d\vec{F} = I_2 (d\vec{l} \times \vec{B})dF=I2​(dl×B).
  6. The magnetic field is B⃗=Brr^+Bzz^\vec{B} = B_r \hat{r} + B_z \hat{z}B=Br​r^+Bz​z^. The current element is dl⃗=dlθ^d\vec{l} = dl \hat{\theta}dl=dlθ^ (in cylindrical coordinates).
  7. The force is dF⃗=I2dl(θ^×(Brr^+Bzz^))=I2dl(Br(θ^×r^)+Bz(θ^×z^))d\vec{F} = I_2 dl (\hat{\theta} \times (B_r \hat{r} + B_z \hat{z})) = I_2 dl (B_r (\hat{\theta} \times \hat{r}) + B_z (\hat{\theta} \times \hat{z}))dF=I2​dl(θ^×(Br​r^+Bz​z^))=I2​dl(Br​(θ^×r^)+Bz​(θ^×z^)).
  8. Using the cross products for cylindrical unit vectors: θ^×r^=−z^\hat{\theta} \times \hat{r} = -\hat{z}θ^×r^=−z^ and θ^×z^=r^\hat{\theta} \times \hat{z} = \hat{r}θ^×z^=r^.
  9. So, dF⃗=I2dl(−Brz^+Bzr^)d\vec{F} = I_2 dl (-B_r \hat{z} + B_z \hat{r})dF=I2​dl(−Br​z^+Bz​r^).
  10. The vertical component of the force is dFz=−I2BrdldF_z = -I_2 B_r dldFz​=−I2​Br​dl. This component is produced by the interaction of the azimuthal current I2I_2I2​ with the radial (horizontal) component of the magnetic field, BrB_rBr​. Integrating over the ring gives the total vertical force FzF_zFz​.
  11. The radial component of the force, dFr=I2BzdldF_r = I_2 B_z dldFr​=I2​Bz​dl, points radially outward or inward, and its net effect over the entire ring is zero (it only tends to stretch or compress the ring).
  12. As established in the analysis of Statement 1, the time average of the vertical force FzF_zFz​ is non-zero and directed upwards.
  13. Thus, Statement 2 accurately describes the physical mechanism: an induced current interacts with the horizontal component of the magnetic field to produce an average upward force.
  14. Therefore, Statement 2 is True.

Conclusion

Statement 1 is true, and Statement 2 provides the correct and detailed physical explanation for the phenomenon described in Statement 1. The levitation is due to the average upward force, which arises from the interaction between the induced current and the horizontal component of the AC magnetic field. Hence, Statement 2 is the correct explanation for Statement 1.

This corresponds to option A.

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