# Wire with current and field problem

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1. Oct 30, 2015

### Callix

1. The problem statement, all variables and given/known data
An electric current is uniformly distributed throughout a long, straight wire that has a diameter of 0.05m. If the current through the wire is 6.0A, calculate the magnitude of the magnetic field:

a). 0.02m radially away from the wire center
b). 0.05m radially away from the wire center
c). What must the current be to create a magnetic field of magnitude 1.0 T and 0.05m radially away from the wire center.

2. Relevant equations
In solution

3. The attempt at a solution

I was hoping that someone could check my work and make sure that what I have is correct! Any help would be appreciated.

a). This one I wasn't too sure about. I looked up some information but I wasn't sure if it applied here.
I found that:
$$J= \frac{I}{\pi R^2}$$$$I_{enc}=J \pi r^2= \frac{Ir^2}{R^2}$$$$\int B\cdot dl=\mu_0 I_{enc}=2 \pi r B$$$$2\pi rB= \frac{\mu_0 I}{2 \pi R^2}$$$$B= \frac{\mu_0 Ir}{2 \pi R^2}$$
Is this the correct approach? And if so, could someone explain the steps they took and why? It seems they used two equations and then did some algebra to combine both, but I can't seem to understand what they did in the 4th step to get there...

b). $$\int B\cdot dl=\mu_0 I \rightarrow B=\frac{\mu_0 I}{2 \pi r}$$
Where I = 6.0 A and r = 0.05m

c). $$I= \frac{2\pi r B}{\mu_0}$$
Where B = 1.0 T amd r = 0.05m

2. Oct 31, 2015

### rude man

The third step in (a) is wrong. It should read 2πrB = μ0 Ir2/R2. (The last step is right).
Your (b) and (c) look right.

3. Oct 31, 2015

### Callix

Ah okay, thanks! Would you be able to explain to me how they got to that point, too? For some reason I can't seem to understand how they manipulated the two equations to get the result that they did. It would be greatly appreciated! Thanks again

4. Oct 31, 2015

### rude man

Which of the 5 steps don't you understand? Seems pretty clearly written up to me.

5. Oct 31, 2015

### Callix

I don't understand how they combined them to form the final equation. I understand the equation for current density and the equation for Ampere's Law. I can't seem to understand what was substituted into the other to get the final equation.

6. Oct 31, 2015

### rude man

If current density is I/πR^2 everywhere across a cross-section and the radius of the cross-section I'm interested in is r, what is the current inside r?

7. Oct 31, 2015

### Callix

I'm not sure... This is why I was asking if you could explain the steps to me