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.

 

[HalfLostAndSearching]

L=1mThe time dependent current in a wire can be calculated using the formula B=uI(enclosed)/2(pi)(d), where B is the magnetic field, u is the permeability of the material, I is the current, and d is the distance from the wire. In this case, the current is time dependent, which means that it is changing over time. To calculate the total magnetic flux, we need to integrate the formula over the entire length of the wire.

The integral of B=uI(enclosed)/2(pi)(d) over the length of the wire can be written as ∫ (u)(I)(L)dx / 2(pi)(x) from .31m to .82m, where u is the permeability, I is the current, L is the length of the wire, and dx is the infinitesimal distance along the wire. Remember that x is equal to d in the picture.

Substituting the given values of u=4(pi)e-7, I=4A, and L=1m, we get ∫ (4(pi)e-7)(4A)(1m)dx / 2(pi)(x) from .31m to .82m. Evaluating this integral, we get (4(pi)e-7)(4A)(1m)/2(pi) (ln(.82) - ln(.31)). Plugging in the values, we get a final answer of 3.9687e-7.

It is important to note that the units of the final answer are Tesla-meters squared (Tm^2), which is the unit of magnetic flux. Also, make sure to double check the units of all the given values to ensure consistency in the final answer.