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 given time dependence 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=A2μ02m2i2ωR List-I List-III(sinωt j^+cosωt k^)P0II(sinωt i^+cosωt j^)Q−α4i^III(sinωt i^+cosωt k^)R3α4i^IV(cosωt j^+sinωt k^)Sα4j^ T−3α4i^Which one of the following options is correct?

(I)B→=μ0mI2(sinωt j^+cosωt k^) ϕ=B→.A→=μ0mI2cos(ωt)A i=εR=μ0mIωA2Rsin(ωt)So, M→=iA→=μ0mIωA22Rsin(ωt)k^ τ→=M→×B→=μ02m2I2ωA22Rsin2(ωt)(−i^)For t=π6ω τ→=μ02m2I2ωA22R×14(−i^) τ=A2μ02m2I2ω2R×14(−i^)=α4(−i^)so(I)→QSo (I)→Q(II) solving as (I), ϕ=0,ε=0,i=0,M¯=0. So(II)→PSimilarly, (III)→S (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 counterclockwise as shown in the figure. The magnetic field due to the solenoid is in n^ direction. List-I given time dependence 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=A2μ02m2i2ωR List-I List-III(sinωt j^+cosωt k^)P0II(sinωt i^+cosωt j^)Q−α4i^III(sinωt i^+cosωt k^)R3α4i^IV(cosωt j^+sinωt k^)Sα4j^ T−3α4i^Which one of the following options is correct?

(I)B→=μ0mI2(sinωt j^+cosωt k^) ϕ=B→.A→=μ0mI2cos(ωt)A i=εR=μ0mIωA2Rsin(ωt)So, M→=iA→=μ0mIωA22Rsin(ωt)k^ τ→=M→×B→=μ02m2I2ωA22Rsin2(ωt)(−i^)For t=π6ω τ→=μ02m2I2ωA22R×14(−i^) τ=A2μ02m2I2ω2R×14(−i^)=α4(−i^)so(I)→QSo (I)→Q(II) solving as (I), ϕ=0,ε=0,i=0,M¯=0. So(II)→PSimilarly, (III)→S (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 counterclockwise as shown in the figure. The magnetic field due to the solenoid is in $\hat{n}$ direction. List-I given time dependence 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 $α = \frac{A^{2} μ_{0}^{2} m^{2} i^{2} ω}{R}$

List-I List-II
I $(\sin ω t \hat{j} + \cos ω t \hat{k})$ P $0$
II $(\sin ω t \hat{i} + \cos ω t \hat{j})$ Q $- \frac{α}{4} \hat{i}$
III $(\sin ω t \hat{i} + \cos ω t \hat{k})$ R $\frac{3 α}{4} \hat{i}$
IV $(\cos ω t \hat{j} + \sin ω t \hat{k})$ 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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$(I) \to (S); (I I) \to (T); (I I I) \to (Q); (I V) \to (P)$

$(I) \to (Q); (I I) \to (P); (I I I) \to (S); (I V) \to (R)$

$(I) \to (Q); (I I) \to (P); (I I I) \to (S); (I V) \to (T)$

$(I) \to (T); (I I) \to (Q); (I I I) \to (P); (I V) \to (R)$

answer is C.

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Detailed Solution

$(I) \vec{B} = \frac{μ_{0} m I}{\sqrt{2}} (\sin ω t \hat{j} + \cos ω t \hat{k}) ϕ = \vec{B} . \vec{A} = \frac{μ_{0} m I}{\sqrt{2}} \cos (ω t) A i = \frac{ε}{R} = \frac{μ_{0 m I ω A}}{\sqrt{2} R} \sin (ω t)$
So, $\vec{M} = i \vec{A} = \frac{μ_{0} m I ω A^{2}}{\sqrt{2} R} \sin (ω t) \hat{k}$
$\vec{τ} = \vec{M} \times \vec{B} = \frac{μ_{0}^{2} m^{2} I^{2} ω A^{2}}{2 R} \sin^{2} (ω t) (- \hat{i})$
For $t = \frac{π}{6 ω}$
$\vec{τ} = \frac{μ_{0}^{2} m^{2} I^{2} ω A^{2}}{2 R} \times \frac{1}{4} (- \hat{i})$
$τ = \frac{A^{2} μ_{0}^{2} m^{2} I^{2} ω}{2 R} \times \frac{1}{4} (- \hat{i}) = \frac{α}{4} (- \hat{i}) s o (I) \to Q$
So $(I) \to Q$
(II) solving as (I), $ϕ = 0, ε = 0, i = 0, \bar{M} = 0. S o (I I) \to P$
Similarly, $(I I I) \to S$
$(I V) \to R$

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