A long vertical wire carries a steady 9 A current. A pair of rails are horizonta

Question

A long vertical wire carries a steady 9 A current. A pair of rails are horizontal and are 0.2 m apart. A 13.8 ? 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 0.13 m from the wire, as shown. In Figure, at time t1, the induced current in µA 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)) * 0.2 m * 9 A * 0.7m/s / ( 2 * 3.1415 * 13.8 Ohm * 0.13 m)
= 1.405 * 10^-7 A
= 0.141 µA

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