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Question

A small circular loop of area A and resistance R is fixed on a horizontal xyplane 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=π6ω Let α=A2μ20m2I2ω2R.


List-I List-II
(I) 12(sinωt ^j+cos ωt ^k) (P) 0
(II) 12(sin ωt^i+cos ωt ^j) (Q) α4^i
(III) 12(sin ωt ^i+cos ωt ^k) (R) 3α4^i
(IV) 12(cos ωt ^j+sin ωt ^k) (S) α4^j
(T) 3α4^i

Which one of the following options is correct

A
IQ,IIP,IIIS,IVR
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B
IQ,IIP,IIIS,IVT
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C
IS,IIT,IIIQ,IVP
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D
IT,IIQ,IIIP,IVR
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Solution

The correct option is A IQ,IIP,IIIS,IVR
I.

ϕ=B A ^k^n

ϕ=BA2cos(ωt)

ε=BAω2sin(ωt)

i=BAω2Rsin(ωt)

m=iA ^k=BA2ω2Rsin(ωt)^k

τ=m×B=B2A2ω2Rsin(ωt)(^k×^n)

=B2A2ω2R[^i]sin2(ωt)

τ=B2A2ω2R[sin2(π6)]=α4^i

(I) Q

II.

ϕ=O

(II) P

III.

ϕ=BA2cos(ωt)

i=BAω2Rsin(ωt)

m=BA2ω2Rsin(ωt)^k

τ=m×B=B2A2ω2×2Rsin(ωt)(^k×(sinωt ^i+cosωt ^k))

τ=B2A2ωsin(ωt)2Rsin(ωt)^j

=B2A2ω2Rsin2(ωt)^j

=α4^j

(III) S

IV.

ϕ=BA2sin(ωt)

i=BAω2Rcos(ωt)

m=BA2ω2Rcosωt(^k)

τ=m×B=B2A2ω2R(^k×^j)cos2(ωt)

τ=+B2A2ω2R(^i).cos2(π6)

=+34α^i

(IV)R.
Hence, option (C) is correct.

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