What EMF is induced in the loop at t=0?

[says]

Homework Statement

A current of 10A is switched off at time t = 0 in a long cable (picture in attachments). It powers a device, with the current return wire having 5 mm separation. A wire loop of dimension 5 by 5 cm is located at a distance 3 cm from the cable.

1) What Emf is induced in the loop at t = 0 if it is in the same plane as the wires in the cable?

2) What direction would the induced current flow in the loop?

3) What Emf would be induced if the cable were rotated 90 degrees but otherwise not moved?

Homework Equations

emf = - N ΔΦ_B/Δt
B = μ₀ I / 2πd
Φ_B = B*Area

The Attempt at a Solution

B[/B] = (4*π*10^-7)*(10A) / (2*π*(0.03))
B = 6.67 * 10^-5 T

Φ_B = 6.67 * 10^-5 * (0.05²)
Φ_B = 1.67 * 10^-7 Wb

This is where I'm a bit confused. The question asks for the emf at t=0. I assumed that because the current is 10 A, that at t=1 second, the current will be 10A.

emf = (-1) (1.67 * 10^-7) / (1)
emf = -1.67 * 10^-7 V

 

[scottdave]

They tell you that the power is turned off at t=0, so you know that whatever flux was there, is going to start collapsing. The long conductor and its return wire will act like a 1 Turn inductor.

 

[says]

I think it might be better to use Ampere's Law: ∫B dl to solve this problem.

Won't the cable that is further away from the loop cancel out some of the magnetic field because the current is going in the opposite direction?

 

[Dadface]

Is there enough information in the question?

 

[says]

B = μ₀ I / 2πs

I've changed the d to an s from my original equation.

∫B ds = ∫ (μ₀ I / 2πs) ds

Then I could just compute 2 integrals with different boundary conditions.

 

[berkeman]

Won't the cable that is further away from the loop cancel out some of the magnetic field because the current is going in the opposite direction?

Yes. When computing the initial flux through the loop at time t=0-, you subtract the fields generated by the two currents.

They tell you that the power is turned off at t=0, so you know that whatever flux was there, is going to start collapsing. The long conductor and its return wire will act like a 1 Turn inductor.

But the inductance and parasitic capacitance of that long wire loop will determine the amplitude and frequency of the flyback ringout waveform. They do not seem to have supplied that, unless the OP is leaving some information out of his post.

Is there enough information in the question?

I don't think so either. We need the physical dimensions of the long wire loop and some way to approximate its L and parasitic capacitance in order to calculate what the loop current does after the switch is opened at t=0, IMO.

 

[scottdave]

I agree that we do not have enough information. For instance, we know that it powers a device, so it could still draw a load for a moment, after the circuit is deenergized.