Animated Solution for Physics - Electromagnetic Induction: A small circular loop of area A and resistance R is fixed on a horizontal xy-plane with the center of the loop always on the axis n^ of a long solenoid. The solenoid has m turns per unit length and carries current I counterclockwise as shown in the figure. The magnetic field due to the solenoid is in n^ direction. List-I gives time dependences of n^ in terms of a constant angular frequency ω. List-II gives the torques experienced by the circular loop at time t=6ωπ, Let α=2RA2μ02m2I2ω.
List-I
(P)
21(sinωtj^+cosωtk^)
(Q)
21(sinωti^+cosωtj^)
(R)
21(sinωti^+cosωtk^)
(S)
21(cosωti^+sinωtk^)
List-II
(1)
0
(2)
−4αi^
(3)
43αi^
(4)
4αj^
(5)
−43αi^
Select Matching Pairs:
* Multiple Allowed
PMatches
QMatches
RMatches
SMatches
Visualized Solution
\text{Physical Setup}
A=Ak^
B=μ0mIn^
\text{Magnetic Flux and Torque}
ϕ=B⋅A=μ0mIA(n^⋅k^)
ε=−dtdϕ,i=Rε
τ=M×B=(iAk^)×(μ0mIn^)
\text{Master Equation}
τ=−Rμ02m2I2A2dtd(n^⋅k^)(k^×n^)
α=2RA2μ02m2I2ω
τ=−ω2αdtd(n^⋅k^)(k^×n^)
\text{Case I: Torque Expression}
n^=21(sinωtj^+cosωtk^)
dtd(n^⋅k^)=−2ωsinωt
k^×n^=−21sinωti^
τ=−αsin2ωti^
\text{Case I: Evaluation}
t=6ωπ⟹ωt=6π
sin(6π)=21
τ=−α(21)2i^=−4αi^→(Q)
\text{Case II}
n^=21(sinωti^+cosωtj^)
n^⋅k^=0⟹ϕ=0
τ=0→(P)
\text{Case III}
n^=21(sinωti^+cosωtk^)
dtd(n^⋅k^)=−2ωsinωt
k^×n^=21sinωtj^
τ=αsin2ωtj^=4αj^→(S)
\text{Case IV: The Calculation}
n^=21(cosωti^+sinωtk^)
dtd(n^⋅k^)=2ωcosωt
k^×n^=21cosωtj^
τ=−αcos2ωtj^=−43αj^(Not in options!)
\text{Case IV: Resolving the Typo}
Assume typo: i^→j^
n^corrected=21(cosωtj^+sinωtk^)
k^×n^corrected=−21cosωti^
τ=αcos2ωti^=43αi^→(R)
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The Sigma Insight: Faraday's Laws of Electromagnetic Induction
Solution Diagram
This problem is a beautiful exercise in vector calculus and electromagnetic induction, but it also comes with a fascinating twist—a typo that caused it to be officially dropped in JEE Advanced! Let's break down the physics step-by-step and uncover exactly what went wrong.
Analyzing the Setup
Imagine a circular loop of area A resting perfectly flat on the xy-plane. Because it lies in this plane, its area vector points straight up along the z-axis. We can write this mathematically as:
A=Ak^
Now, a long solenoid is generating a magnetic field B. The direction of this field is given by a unit vector n^, which changes over time. The magnetic field is:
B=μ0mIn^
To find the torque on the loop, we first need to understand how the magnetic flux through the loop is changing. The magnetic flux ϕ is the dot product of the magnetic field and the area vector:
ϕ=B⋅A=μ0mIA(n^⋅k^)
Notice that because the area vector is strictly along k^, only the z-component of the magnetic field contributes to the flux. Any magnetic field lines parallel to the xy-plane simply skim over the loop without piercing it.
The Master Equation for Torque
According to Faraday's Law, a changing magnetic flux induces an electromotive force (EMF) in the loop:
ε=−dtdϕ
This EMF drives an induced current i through the loop's resistance R:
i=Rε=−Rμ0mIAdtd(n^⋅k^)
Once a current flows, the loop behaves like a magnetic dipole with a magnetic moment M=iA=iAk^. This magnetic dipole then interacts with the external magnetic field to experience a torque:
τ=M×B=(iAk^)×(μ0mIn^)
Substituting our expression for the induced current i, we get a comprehensive master equation for the torque:
τ=−Rμ02m2I2A2dtd(n^⋅k^)(k^×n^)
The problem kindly provides us with a constant α=2RA2μ02m2I2ω. By factoring out ω/2 from our master equation, we can rewrite it in a beautifully compact form:
τ=−ω2αdtd(n^⋅k^)(k^×n^)
Now, we possess a powerful tool. Instead of recalculating the physics from scratch for every option in List-I, we simply need to plug the given n^ vector into this master equation!
Evaluating the Cases
Case I:
Given n^=21(sinωtj^+cosωtk^).
First, we extract the z-component: n^⋅k^=21cosωt.
Differentiating this with respect to time gives −2ωsinωt.
Next, we compute the cross product: k^×n^=21sinωt(k^×j^)=−21sinωti^.
Plugging these into our master equation yields:
τ=−αsin2ωti^
Evaluating this at t=6ωπ, we know sin(π/6)=1/2, so sin2(π/6)=1/4. The torque is −4αi^, which perfectly matches option (Q).
Case II:
Given n^=21(sinωti^+cosωtj^).
Look closely—there is no k^ component! This means n^⋅k^=0. The magnetic field is entirely parallel to the loop. Consequently, the flux is zero, the induced current is zero, and the torque is exactly 0. This matches option (P).
Case III:
Given n^=21(sinωti^+cosωtk^).
The z-component is identical to Case I, so the derivative is again −2ωsinωt.
However, the cross product changes: k^×n^=21sinωt(k^×i^)=21sinωtj^.
The torque becomes αsin2ωtj^. At t=6ωπ, this evaluates to 4αj^, matching option (S).
The Typo in Case IV
Case IV:
Given n^=21(cosωti^+sinωtk^).
The z-component is 21sinωt, and its derivative is 2ωcosωt.
The cross product is k^×n^=21cosωt(k^×i^)=21cosωtj^.
Plugging these into the master equation gives:
τ=−αcos2ωtj^
At t=6ωπ, cos(π/6)=3/2, so cos2(π/6)=3/4. The torque is −43αj^.
But wait! If you look at List-II, this option does not exist!
This is where the examiners made a mistake. If we assume there was a typo in the question paper, and the i^ in Case IV was actually meant to be a j^, let's see what happens.
If n^corrected=21(cosωtj^+sinωtk^), the cross product becomes k^×n^=−21cosωti^.
The torque equation then yields αcos2ωti^. Evaluating this at t=6ωπ gives exactly 43αi^, which perfectly matches option (R).
Because of this typographical error, the question was officially dropped from the JEE Advanced 2022 grading scheme. However, the physics behind it remains a fantastic learning opportunity!
