# Homework Help: Time Dependent Current in a wire

1. Mar 7, 2012

### kjlchem

1. The problem statement, all variables and given/known data
An infinite straight wire carries a current I that varies with time as shown above. It increases from 0 at t = 0 to a maximum value I1 = 2.1 A at t = t1 = 14 s, remains constant at this value until t = t2 when it decreases linearly to a value I4 = -2.1 A at t = t4 = 24 s, passing through zero at t = t3 = 21.5 s. A conducting loop with sides W = 20 cm and L = 57 cm is fixed in the x-y plane at a distance d = 49 cm from the wire as shown.

What is ε1, the induced emf in the loop at time t = 7 s? Define the emf to be positive if the induced current in the loop is clockwise and negative if the current is counter-clockwise.

2. Relevant equations

B = μI/2∏r

Flux = B*A

-dflux/dt = ε

3. The attempt at a solution

I don't understand what I'm doing wrong with this problem.

This is what I have so far...

(dB*A)/dt= ε, A = L(W)

μ(dI)(L)W/(2∏rdt) = ε

μ=12.566*10^-7
dI = 2.1 A
L = .57 m
W = .2 m
dt=14 s.

On the left side of the box, r = .49 m and the current is negative, so the emf is positive.
On the right side of the box, r = 1.06 m and the current is positive, so the emf is negative.
Putting the 2 emf's together by subtracting the right side from the left side, I get an emf of -3.753*10^-9V.

What am I doing wrong?

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2. Mar 7, 2012

### Redbelly98

Staff Emeritus
I'm not sure about the method you are using. For calculating EMF for a straight section of wire, I am only familiar with doing that for the wire moving through a magnetic field.

Since the wire loop is not moving, I think you have to use ε=-dflux/dt instead. So first I would first calculate the flux through the loop -- as a function of time, during the time interval that contains 7 s.

3. Mar 7, 2012

### kjlchem

Yeah, I used ε = -dflux/dt. The flux as a function of time = μ(dI)(L)W/(2∏rdt)

4. Mar 8, 2012

### Redbelly98

Staff Emeritus
That won't work here; for one thing, there is no r given in this problem. I think I see your problem though:
That only works if B is uniform over the whole area. It isn't; B is stronger at the side of the rectangle closest to the wire, and weaker at the far side.

Instead, you'll need to do an integral to calculate the flux:
Flux = $\int B \cdot dA$​

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