# Current carrying wire and induced electric field within the wire

#### thepatient

So I just had an exam and I'm so unsure about this problem. It seems that I got it right, but I could be completely wrong. XD

1. Homework Statement

A long, straight current-carrying wire carries an increasing current I that is uniformly distributed over the cross-section of the wire (the current density j is uniform). The magnitude of the induced electric field inside the wire is (r is the distance to the center of the wire):
a) zero at all points inside wire
b) proportional to r
c) proportional to r^2
d) proportional to 1/r
e) proportional to 1/r^2

2. Homework Equations
closed-line integral ∫B*dl = µ0 *I (enclosed)
j = I/A
closed line integral ∫E*dl = -d/dt (magnetic flux)
magnetic flux = ∫B*dA

3. The Attempt at a Solution

So basically I thought, since there is a current changing in time, there is a magnetic field being induced. This magnetic field, since it is proportional to the current, also changes in time, so it creates an electric field. So I began with:

∫B*dl = µ0 *I (enclosed)
The line integral was a distance r from the center where r is less than the radius of the exterior of the wire.

The I enclosed wasn't the entire I, but the ratio between the enclosed current and net current.

j = I(enclosed) /A = I(enclosed)/pir^2

I(net)/piR^2 = I(enclosed)/pir^2
r^2/R^2 * I(net) = I(enclosed)

Then substituting back into the line integral:
∫B*dl = µ0 *r^2/R^2 *I(net)
B*2pir = µ0 *r^2/R^2 *I(net)
B =1/2pi * ( µ0 *r/R^2 *I(net))

Then calculated magnetic flux:
∫B*dA =
And here is where it gets kind of fuzzy. XD I think I might have assumed that the flux was in the area of pir^2 within the wire and the B field going in the same direction as the wire. But I think the B field is actually circling that area of pir^2, so I think I made a mistake there. :\

I ended up with magnitude of E = rµ0* I/(4piR^2) and chose B. Was I completely wrong? XD

Related Introductory Physics Homework Help News on Phys.org

#### Norcalli

Firstly, you're in Freund's 4B and he's omitting the question. The first question I'll ask do you know where the B field lines are? Before you start doing any equations you have to know where the field lines will be.
If you use the equation for a magnetic field from a point charge (and apply it a lot), you can figure out that the B field should look like concentric circles with the center of the wire, and at the center it is going to have a magnitude of 0. Now if you take a surface which is a circle that is a cross section of the wire, you can figure out that the flux through this surface is going to be 0 because the field lines are going to be on the same plane as the surface (i.e. angle = pi/2 => B.dA=0). This tells you that there is no induced E field parallel to the current. Now you can take this surface and rotate it and show that net flux will be 0. So, the answer that is most likely to be correct is zero at all points (and he graded mine and said it was right).

#### thepatient

Ahh thats hilarious. I never would have thought I would find someone in my class here. XD nice to know.

I got the problem confused with a solenoid for some reason while doing this. XD everything seemed to make sense until I got home. I looked at the problem again and noticed that this wasn't a solenoid but a straight current carrying wire. That changes the magnetic field to a circulating field clockwise. I was just hoping that because there is a flux within a window of Length l times dr, then maybe there was a line integral of E*dl, but then at home i noticed at all points cos(theta) =90 so E*dl=0, also you mentioned that net magnetic flux =0.

I asked anyway, since sometimes I would think I'm completely wrong, but I'm right, or vice-versa. Thanks for the input. :]

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving