  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=2√3 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)2⇒x=√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  Suggest corrections   