A long vertical wire carries a steady 10 A current. A pair of rails are horizont
ID: 2258825 • Letter: A
Question
A long vertical wire carries a steady 10 A current. A pair of rails are horizontal and are 0.1 m apart. A 10.4 ? 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 0.7 m/s as shown. The bar and the rails have negligible resistance. At a given instant t1, the bar is 0.14 m from the wire, as shown. In Figure, at time t1, the induced current in
A long vertical wire carries a steady 10 A current. A pair of rails are horizontal and are 0.1 m apart. A 10.4 ? 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 0.7 m/s as shown. The bar and the rails have negligible resistance. At a given instant t1, the bar is 0.14 m from the wire, as shown. In Figure, at time t1, the induced current in mu A is: Express the answer in three decimal places.Explanation / Answer
The magnetic field strength at distance r of a long current carrying wire is
B = mu0 I / (2 pi r)
The magnetic flux when the bar is at distance x from the current carrying wire, is the integral of B over the surface enclosed by the circuit (the field is concentric to the current carrying wire and therefore everywhere perpendicular to the area of the circuit):
Phi = Integral B. dS = Integral(from x to 2.0m) mu0 I / (2 pi r) * L dr
= mu0 L I / (2 pi) ln( 2.0 / x)
The time derivative of this flux (which we need to calculate the EMF from Lenz' law) is
dPhi/dt = dPhi/dx * dx/dt
= -mu0 L I / (2 pi x) * dx/dt
= - mu0 L I /(2 pi ) * v
The induced V = - dPhi/dt, and using Ohm's law, I_induced = V_ind / R
SO
I_induced = mu0 L I v / (2 pi R x)
= 4 *3.1415*10^-7 (V s/(A m)) * 0.1 m * 10 A * 0.7m/s / ( 2 * 3.1415 * 10.4 Ohm * 0.14 m)
= 0.96 * 10^-7 A
= 0.096