Q.

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.

Question Image
 List-I List-II
A(sinωtj^+cosωtk^)P)0
B(sinωti^+cosωtj^)Q)α4i^
C(sinωti^+cosωtk^)R)3α4i^
D(cosωtj^+sinωtk^)S)α4j^
  T)3α4i^

Which one of the following options is correct?

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a

AS,BT,CQ,DP

b

AQ,BP,CS,DR

c

AQ,BP,CS,DT

d

AT,BQ,CP,DR

answer is C.

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Detailed Solution

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=B2A2ω2Rsin2π6=α4i^

(I)Q

II. ϕ=0, hence no current and no torque

(II)P

(III)ϕ=BA2cos(ωt)i=BAω2Rsin(ωt)m=BA2ω2Rsin(ωt)k^τ=m×B=B2A2ω2×2Rsinωt(k^×(sinωti^+cosωtk^))τ=B2A2ω2Rsin2(ωt)j^τ=α4j^IIIS

 IV. ϕ=BA2sin(ωt)ϕ=BA2sin(ωt)i=BAω2Rcos(ωt)m=BA2ω2Rcosωt(k^)τ=m×B=B2A2ω2R(k^×j^)cos2(ωt)τ=+B2A2ω2R(i^)cos2π6τ=+34αi^

(IV)R.

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