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=A2μ02m2I2ωR . List-I List-2I)sinωt j^ + cosωt k^P)0II)sinωt i^ + cosωt j^Q)−α4 i^III)sinωt i^ + cosωt k^R)3α4 i^IV) cosωt j^ +sinωt k^S)α4 j^ T)−3α4 i^Which one of the following options is correct?

I ϕ=BA k^.n^ ϕ=BA2cosωt ε=BAω2Rsinωt m→=iA k^ =BA2ω2Rsinωtk^ τ→=m→×B→=B2A2ω2Rsinωtk^× n^ =−B2A2ω2Rsin2ωt τ→=−B2A2ω2Rsin2π6=−α4i^ (I) →Q.(II). φ=O(II) → P(III). ϕ=BA2cosωt i=BAω2Rsinωt m→=BA2ω2Rsinωtk^ τ→=m→ × B→=B2A2ω2×2Rsinωtk^ ×sinωt i^ τ=B2A2ωsinωt2Rsinωt j^ =B2A2ω2Rsin2ωtj^ =α4j^ III → SIV. ϕ=BA2sinωti=−BAω2Rcosωt vecm=−BA2ω2Rcosωtk^ τ→=m→ ×B→=−B2A2ω2Rk^ × j^cos2ωt τ=+B2A2ω2Ri^.cos2π6 =+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=A2μ02m2I2ωR . List-I List-2I)sinωt j^ + cosωt k^P)0II)sinωt i^ + cosωt j^Q)−α4 i^III)sinωt i^ + cosωt k^R)3α4 i^IV) cosωt j^ +sinωt k^S)α4 j^ T)−3α4 i^Which one of the following options is correct?

I ϕ=BA k^.n^ ϕ=BA2cosωt ε=BAω2Rsinωt m→=iA k^ =BA2ω2Rsinωtk^ τ→=m→×B→=B2A2ω2Rsinωtk^× n^ =−B2A2ω2Rsin2ωt τ→=−B2A2ω2Rsin2π6=−α4i^ (I) →Q.(II). φ=O(II) → P(III). ϕ=BA2cosωt i=BAω2Rsinωt m→=BA2ω2Rsinωtk^ τ→=m→ × B→=B2A2ω2×2Rsinωtk^ ×sinωt i^ τ=B2A2ωsinωt2Rsinωt j^ =B2A2ω2Rsin2ωtj^ =α4j^ III → SIV. ϕ=BA2sinωti=−BAω2Rcosωt vecm=−BA2ω2Rcosωtk^ τ→=m→ ×B→=−B2A2ω2Rk^ × j^cos2ωt τ=+B2A2ω2Ri^.cos2π6 =+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 $α = \frac{A^{2} μ_{0}^{2} m^{2} I^{2} ω}{R}$ .
List-I List-2
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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$I \to S, II \to T, III \to Q, IV \to P$

$I \to Q, II \to P, III \to S, IV \to T$

$I \to Q, II \to P, III \to S, IV \to R$

$I \to T, II \to Q, III \to P, IV \to R$

answer is C.

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Detailed Solution

I
$ϕ = B A \hat{k} . \hat{n} ϕ = \frac{B A}{\sqrt{2} \cos} (ω t) ε = \frac{B A ω}{\sqrt{2} R} \sin (ω t) \vec{m} = i A \hat{k} = \frac{B A^{2} ω}{\sqrt{2} R} \sin (ω t) \hat{k} \vec{τ} = \vec{m} \times \vec{B} = \frac{B^{2} A^{2} ω}{2 R} \sin (ω t) (\hat{k} \times \hat{n}) = - \frac{B^{2} A^{2} ω}{2 R} [\sin^{2} (ω t)] \vec{τ} = - \frac{B^{2} A^{2} ω}{2 R} [\sin^{2} (\frac{π}{6})] = \frac{- α}{4} \hat{i}$
(I) →Q.
(II). $φ = O$
(II) → P
(III). $ϕ = \frac{B A}{\sqrt{2}} \cos (ω t)$
$i = \frac{B A ω}{\sqrt{2} R} \sin (ω t) \vec{m} = \frac{B A^{2} ω}{\sqrt{2} R} \sin (ω t) \hat{k} \vec{τ} = \vec{m} \times \vec{B} = \frac{B^{2} A^{2} ω}{\sqrt{2} \times \sqrt{2} R} \sin ω t (\hat{k} \times (\sin ω t \hat{i})) τ = \frac{B^{2} A^{2} ω \sin ω t}{2 R} \sin (ω t) \hat{j} = \frac{B^{2} A^{2} ω}{2 R} \sin^{2} (ω t) \hat{j} = \frac{α}{4} \hat{j}$
III → S

IV. $ϕ = \frac{B A}{\sqrt{2} \sin} (ω t)$

$i = - \frac{B A ω}{\sqrt{2} R} \cos (ω t) v e c m = - \frac{B A^{2} ω}{\sqrt{2} R} \cos ω t (\hat{k}) \vec{τ} = \vec{m} \times \vec{B} = - \frac{B^{2} A^{2} ω}{2 R} (\hat{k} \times \hat{j}) \cos^{2} (ω t) τ = + \frac{B^{2} A^{2} ω}{2 R} (\hat{i}) . \cos^{2} (\frac{π}{6}) = + \frac{3}{4} α \hat{i}$

(IV) → R

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