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here is an AC current in a wire loop, e is a changing B field within the loop. this changing B field ce an EMF (voltage) of own in the loop of wire? 1) Yes, it must 2) No, it does not. B field AC current produces a changing B field. • A changing B field induces an EMF in a loop, even the loop that produces the changing B field • This is called self-inductance

If there is an AC current in a wire loop, there is a changing B field within the loop

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B field. If there is an AC current in a wire loop, there is a changing B field within the loop. Does this changing B field induce an EMF (voltage) of its own in the loop of wire ? 1) Yes, it must 2) No, it does not. • AC current produces a changing B field. - PowerPoint PPT Presentation

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Page 1: If there is an  AC  current in a wire loop,  there is a  changing  B field  within the loop

If there is an AC current in a wire loop, there is a changing B field within the loop.

Does this changing B field induce an EMF (voltage) of its own in the loop of wire?

1) Yes, it must

2) No, it does not.B field

• AC current produces a changing B field.• A changing B field induces an EMF in a loop, even the

loop that produces the changing B field• This is called self-inductance

Page 2: If there is an  AC  current in a wire loop,  there is a  changing  B field  within the loop

Self-Induction in Daily Life When you turn off your toaster by unplugging it

current changes rapidly to zero rapidly changing B field produced very large voltage induced in wires of toaster spark

voltage difference large enoughfor electrons to flow through air

If you unplug your computer while it is still on, large voltages may be generated, possibly destroying the CPU chip.

Page 3: If there is an  AC  current in a wire loop,  there is a  changing  B field  within the loop

A solenoid has an inductance of 87.5mH and a resistance of 0.250. Find:

The time constant for this circuit:

and how long it takes for the current to reach ½ its maximum once connected to a 3volt battery.

= L/R = 87.510-3H/0.250 = 0.350 seconds

Maximum current is I = 3v/0.250 = 12 A

6A = 12A(1-e-t/) ½ = e-t/

ln(½) = -t/0.350sec

t = 0.2426 sec