# Time Dependent Current in a Wire

1. Mar 23, 2014

### Flop880

1. The problem statement, all variables and given/known data

Problem is attached

2. Relevant equations

A formula sheet is also attached

3. The attempt at a solution

flux=$\int$B dA from .31m to .82
B=u I(enclosed)/2(pi)(d)
d=x
dA=dx L
so ∫ (u)(I)(L)dx / 2(pi)(x) from .31m to .82m remember x=d in the pic. My answer is 3.9687e-7 and its wrong

integral came out to be (u)(I)(L)/2pi (ln(.82) - ln(.31))
u=4(pi)e-7
I=4A

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2. Mar 23, 2014

### Simon Bridge

That "dA" indicates an area - so why is this not a double integral?

I dont think you've used the formula correctly or you got confused between two different uses.
$$B=\frac{\mu_0 I}{2\pi d}$$... would be the magnetic field strength a distance d from a long straight wire.

The magnetic flux through area dA at position (x,y) would be $d\Phi = B(x,y,t)\;dA$
You'd have to integrate over the whole LxW area to find the total flux.

3. Mar 23, 2014

### Flop880

I just got the right answer by multiplying by the width, not length. so dA= dx w, since dx is the length thats changing times the width which gives area. What do you mean by double integral? Doesn't that give volume?

4. Mar 23, 2014

### Simon Bridge

Well done.

note:
Triple integrals give volume. dV=dx.dy.dz,

dA=W.dx is only true when the thing you are integrating does not vary with y
- which is what you have.

Last edited: Mar 23, 2014