A long vertical w ire carries a steady 70 A current As shown in the figure, a pa

Question

A long vertical w ire carries a steady 70 A current As shown in the figure, a pair of horizontal nils are 0.20 m apart A 20-Ohm 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 horizontal velocity of 0.90 m/s to the right, as shown The bar and the rails have negligible resistance. At the instant that the bar is 0.20 m from the wire, what are the induced current in the resistor (4 pis) and its direction through the resistor?(mu_0=4 pi times 10^-7 T.m/A) 0.63 muA, from a to b 0.63 mu A, from b to a 0.32 mu A, from a to b 0.32 mu A, from b to a 1.9 mu A, from b to a

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)) * 0.20 m * 70 A * 0.90m/s / ( 2 * 3.1415 * 20 Ohm * 0.20 m)

= 0.63 uA, from a to b

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