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Question

Figure shows a uniform circular loop of radius a having resistance per unit length ρ placed in a unifrom magnetic field B perpendicular to plane of figure. A uniform rod of length 2a & resistance R moves with a velocity v as shown.
Find the current in the rod when it has moved a distance a2 from the centre of circular loop.
Given B=3,a=2,v=3,ρ=94π,R=23 all in SI units.


Solution



Using Pythogaras theorem, we can calculate the length rod on the circular loop. 
The half of the length of the rod on the loop is given by,
x2=a2(a2)2x=3a2
The total length of the rod on the circular loop is 2x=3a

The motional emf of the rod is given by ϵ=Blv=B×3a×v
Substituting the values we get, ϵ=18 V 

Let θ be the angle subtented by one end of the rod and the center of the rod at the center of the circular loop.
cosθ=12
θ=60
The total angle made by the rod at the center of the circular loop is 120
one third of the circular loop is to the right of the rod and two thirds is on the left of the rod.
their resistances are also in the ratio 1:2

The circumference of the circular loop is 2πa=4π
Given resistance per unit length of the loop 94π
the resistance of the whole loop is 94π×4π=9 Ω

Now dividing this resistance in the ratio 1:2 (for the right and the left parts of the loop w.r.t the rod) we have 3 Ω for the right part and 6 Ω for the left part.

And the resistance of the length of the rod that is on the circular loop is 1 Ω

Now the whole system can be written as shown in the figure.


6 Ω is parallel to 3 Ω and the this combination is parallel to 1 Ω.

Their effective resistance is 3 Ω.
The current through the rod is i=183=6 A

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