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 counter clockwise 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=π6ω. Let α=A2μ02m2I2ω2R. List-I List-IIA(sinωtj^+cosωt k^)P)0B(sinωti^+cosωt j^)Q)−α4i^C(sinωti^+cosωt k^)R)3α4i^D(cosωtj^+sinωt k^)S)α4j^ T)−3α4i^Which one of the following options is correct?

ϕ=BAk^⋅n^ϕ=BA2cos⁡(ωt)ε=BAω2sin⁡(ωt)i=BAω2Rsin⁡(ωt)m→=iAk^=BA2ω2Rsin⁡(ωt)k^T→=m→×B→=B2A2ω2Rsin2⁡(ωt)(k^×j^)=−B2A2ω2R[i^]sin2⁡(ωt)T=−B2A2ω2Rsin2⁡π6=−α4i^(I)→QII. ϕ=0, hence no current and no torque(II)→P(III)ϕ=BA2cos⁡(ωt)⇒i=BAω2Rsin⁡(ωt)⇒m→=BA2ω2Rsin⁡(ωt)k^τ→=m→×B→=B2A2ω2×2Rsin⁡ωt(k^×(sin⁡ωti^+cos⁡ωtk^))τ→=B2A2ω2Rsin2⁡(ωt)j^⇒τ→=α4j^⇒III→S IV. ϕ=BA2sin⁡(ωt)⇒ϕ=BA2sin⁡(ωt)⇒i=−BAω2Rcos⁡(ωt)m→=−BA2ω2Rcos⁡ωt(k^)τ→=m→×B→=−B2A2ω2R(k^×j^)cos2⁡(ωt)⇒τ=+B2A2ω2R(i^)⋅cos2⁡π6⇒τ→=+34αi^(IV)→R.

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 counter clockwise 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=π6ω. Let α=A2μ02m2I2ω2R. List-I List-IIA(sinωtj^+cosωt k^)P)0B(sinωti^+cosωt j^)Q)−α4i^C(sinωti^+cosωt k^)R)3α4i^D(cosωtj^+sinωt k^)S)α4j^ T)−3α4i^Which one of the following options is correct?

ϕ=BAk^⋅n^ϕ=BA2cos⁡(ωt)ε=BAω2sin⁡(ωt)i=BAω2Rsin⁡(ωt)m→=iAk^=BA2ω2Rsin⁡(ωt)k^T→=m→×B→=B2A2ω2Rsin2⁡(ωt)(k^×j^)=−B2A2ω2R[i^]sin2⁡(ωt)T=−B2A2ω2Rsin2⁡π6=−α4i^(I)→QII. ϕ=0, hence no current and no torque(II)→P(III)ϕ=BA2cos⁡(ωt)⇒i=BAω2Rsin⁡(ωt)⇒m→=BA2ω2Rsin⁡(ωt)k^τ→=m→×B→=B2A2ω2×2Rsin⁡ωt(k^×(sin⁡ωti^+cos⁡ωtk^))τ→=B2A2ω2Rsin2⁡(ωt)j^⇒τ→=α4j^⇒III→S IV. ϕ=BA2sin⁡(ωt)⇒ϕ=BA2sin⁡(ωt)⇒i=−BAω2Rcos⁡(ωt)m→=−BA2ω2Rcos⁡ωt(k^)τ→=m→×B→=−B2A2ω2R(k^×j^)cos2⁡(ωt)⇒τ=+B2A2ω2R(i^)⋅cos2⁡π6⇒τ→=+34αi^(IV)→R.

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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 $\hat{n}$ of a long solenoid. The solenoid has m turns per unit length and carries current I counter clockwise as shown in the figure. The magnetic field due to the solenoid is in $\hat{n}$ direction. List-I gives time dependences of $\hat{n}$ in terms of a constant angular frequency $ω$ . List-II gives the torques experienced by the circular loop at time $t = \frac{π}{6 ω}$ . Let $α = \frac{A^{2} μ_{0}^{2} m^{2} I^{2} ω}{2 R}$ .
List-I List-II
A $(\sin ω t \hat{j} + \cos ω t \hat{k})$ P) 0
B $(\sin ω t \hat{i} + \cos ω t \hat{j})$ Q) $- \frac{α}{4} \hat{i}$
C $(\sin ω t \hat{i} + \cos ω t \hat{k})$ R) $\frac{3 α}{4} \hat{i}$
D $(\cos ω t \hat{j} + \sin ω t \hat{k})$ S) $\frac{α}{4} \hat{j}$
T) $- \frac{3 α}{4} \hat{i}$
Which one of the following options is correct?

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$A \to S, B \to T, C \to Q, D \to P$

$A \to Q, B \to P, C \to S, D \to R$

$A \to Q, B \to P, C \to S, D \to T$

$A \to T, B \to Q, C \to P, D \to R$

answer is C.

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Detailed Solution

$\begin{array}{l} ϕ = B A \hat{k} \cdot \hat{n} \\ ϕ = \frac{B A}{\sqrt{2}} \cos (ω t) \\ ε = \frac{B A ω}{\sqrt{2}} \sin (ω t) \\ i = \frac{B A ω}{\sqrt{2} R} \sin (ω t) \\ \vec{m} = i A \hat{k} = \frac{B A^{2} ω}{\sqrt{2} R} \sin (ω t) \hat{k} \\ \vec{T} = \vec{m} \times \vec{B} = \frac{B^{2} A^{2} ω}{\sqrt{2} R} \sin^{2} (ω t) (\hat{k} \times \hat{j}) \end{array}$

$\begin{array}{l} = - \frac{B^{2} A^{2} ω}{2 R} [\hat{i}] \sin^{2} (ω t) \\ T = - \frac{B^{2} A^{2} ω}{2 R} [\sin^{2} (\frac{π}{6})] = \frac{- α}{4} \hat{i} \end{array}$

$(I) \to Q$

$I I . ϕ = 0, hence no current and no torque$

$(I I) \to P$

$\begin{array}{l} (I I I) ϕ = \frac{B A}{\sqrt{2}} \cos (ω t) \Rightarrow i = \frac{B A ω}{\sqrt{2} R} \sin (ω t) \Rightarrow \vec{m} = \frac{B A^{2} ω}{\sqrt{2} R} \sin (ω t) \hat{k} \\ \vec{τ} = \vec{m} \times \vec{B} = \frac{B^{2} A^{2} ω}{\sqrt{2} \times \sqrt{2} R} \sin ω t (\hat{k} \times (\sin ω t \hat{i} + \cos ω t \hat{k})) \\ \vec{τ} = \frac{B^{2} A^{2} ω}{2 R} \sin^{2} (ω t) \hat{j} \Rightarrow \vec{τ} = \frac{α}{4} \hat{j} \Rightarrow I I I \to S \end{array}$

$\begin{array}{l} IV. ϕ = \frac{B A}{\sqrt{2}} \sin (ω t) \Rightarrow ϕ = \frac{B A}{\sqrt{2}} \sin (ω t) \Rightarrow i = - \frac{B A ω}{\sqrt{2} R} \cos (ω t) \\ \vec{m} = - \frac{B A^{2} ω}{\sqrt{2} R} \cos ω t (\hat{k}) \\ \vec{τ} = \vec{m} \times \vec{B} = - \frac{B^{2} A^{2} ω}{2 R} (\hat{k} \times \hat{j}) \cos^{2} (ω t) \Rightarrow τ = + \frac{B^{2} A^{2} ω}{2 R} (\hat{i}) \cdot \cos^{2} (\frac{π}{6}) \Rightarrow \vec{τ} = + \frac{3}{4} α \hat{i} \end{array}$

$(I V) \to R .$

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