Time Dependent Current in a Wire

[Flop880]

Homework Statement

Problem is attached

Homework Equations

A formula sheet is also attached

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

 

[Simon Bridge]

Homework Statement

Problem is attached

Homework Equations

A formula sheet is also attached

The Attempt at a Solution

flux=$\int$B dA from .31m to .82

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

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

I don't 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.

 

[Flop880]

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

 

[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.