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Question

Figure below shows a conducting rod of negligible resistance that can slide on a smooth U-shaped rail made of a wire of resistance 1 Ω m1. Position of the conducting rod at t=0 is shown. A time dependent magnetic field B=2T(tesla) is switched ON at t=0.
Following the situation of the previous question, the magnitude of force required to move the conducting rod at a constant speed of 5 cms1 at the same instant t=2s is equal to

A
0.16 N
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B
0.12 N
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C
0.08 N
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D
0.06 N
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Solution

The correct option is C 0.08 N
When a current carrying loop is placed in magnetic field it experiences a force given by,
F=I(l×B)
In terms of magnitude,
F=BIl(1)
Current through the loop after 2 sec when the rod is moving with velocity 5 cm/s,
I=ER
From Faraday's law induced emf, E=dϕdt
Since flux, ϕ=B.A
After t sec if wire is moving with velocity v area, A=0.2×(0.4v×t)
Also magnetic field, B=2t
Therefore flux, ϕ=B.A=2t×0.2×(0.4v×t)
ϕ=0.16t0.4vt2
dϕdt=0.160.8vt
At t=2 sec and v=5 cm/s = 0.05 m/s
dϕdt=0.160.8×0.05×2=0.08
E=dϕdt=0.08 V
In terms of magnitude, E=0.08 V
Length of the loop after 2 sec, L=0.2+2×(0.4v×t)
L=0.2+2×(0.40.05×2)=0.8 m
Therefore resistance, R=1Ω/m×0.8=0.8 Ω
Thus current, I=ER=0.080.8=0.1 A
Length of the rod, l=0.2 m
Since magnetic field, B=2t
Afetr 2 sec magnetic field, B=2×2=4 T
Substituting the values of B,I and l in (1),
Force required to move conducting rod, F=BIl=4×0.1×0.2=0.08 N

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