How Is Induced Current Calculated in a Moving Bar Near a Wire?

[Lizziecupcake]

Problem: A long vertical wire carries a steady 17 A current. A pair of rails are horizontal and are 0.2 m apart. A 7.6 Ω resistor connects points a and b, at the end of the rails. A bar is in contact with the rails, and is moved by an external force with a constant velocity of 0.5 m/s as shown. The bar and the rails have negligible resistance. At a given instant t1, the bar is 0.18 m from the wire, as shown. In Figure, at time t1, the induced current in µA is:
Express the answer in three decimal places.

My attempt at the question:
Current in the wire, i = 17 A
Distance between the rails, d = 0.20 m
Resistance, R = 7.6 Ω
Velocity , v = 0.50 m/s
At time t1, the bar is 0.18 m from the wire

B = μ0 i / 2 π d
= (4 π *10-7 T.m/A) (17 A) / 2π (0.20 m)
B = 1.7 *10-5 T
------------------------------------------------------
Induced emf , ε = dφ / dt
= B L v
= (1.7 *10-5 T) (0.18 m) (0.50 m/s)
= 1.53 *10-6 V
-----------------------------------------------------------
Induced current, I = ε / R
= (1.53 *10-6 V) / (7.6 Ω)
= 2.01 *10-7 A
= 0.201 μ A

It seems to be wrong so I don't exactly know what I'm doing wrong so could someone help me out?

 

[BruceW]

I'm still trying to get my head round the set-up. It sounds as though the bar, rails and resistor make a rectangle which is perpendicular to the vertical wire with steady current. Is this right? But in this case, the vertical wire would create a magnetic field which is not perpendicular to the rectangle. So the magnetic flux would be zero, at all times... But it seems unlikely that the answer would just be zero, so I think I have misunderstood the set-up somehow.

 

[issacnewton]

cupcake, you said some figure is there. figure will help