- #1
Lizziecupcake
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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?
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?