# Homework Help: Coaxial cable and magnetic field

1. Jun 21, 2008

### scholio

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

a coaxial cable carries current as follows: (see attachment)

at point P, 0.3 meters from the central axis, the magnetic field is 0.005 Teslas, what is current, I?

2. Relevant equations

magnetic field solenoid, B = mu_0(n)I where mu_) is constant = 4pi*10^-7, n is number of coils/turns, I is current

magnetic field, Biot-Savart, B = mu_0/4pi([integral(IdL/r^2)]) where dL is change in length, r is radius

3. The attempt at a solution

should i assume that the coaxial cable is a solenoid? i am doubtful because the equation does not involve radius, which i am supposed to involve to solve for I.

i'm sure i'm supposed to use the biot-savart equation, do i assum dL remains constant? how do the involve the two different currents going in opposite directions?

my initial plan was to set up three biot-savart equations, one using the center conductor, the other using the outer conductor. i would then equate the sum of them two, to the third which would use r = 0.3 meters.

i think it may work except, i do not know how to differentiate the different radius of the center/outer conductors.

help appreciated

#### Attached Files:

• ###### cox.JPG
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2. Jun 21, 2008

### dynamicsolo

Your attachment hasn't been cleared yet, but I'm going to hazard a guess that the distance 0.3 m puts point P outside the cable. You want to use Ampere's Law for this problem, since the current in the cable will be entirely "enclosed" by a circular "Amperian loop" of radius 0.3 m. (You could use the Biot-Savart Law, but it's sort of excessive. And, no, a cable does not act like a solenoid. The cable will form magnetic field loops around itself, while a solenoid is a sort of "stack of current rings" and produces a magnetic field running down the axis of the solenoid -- it is a magnetic analogue of the electric field within a parallel-plate capacitor.)

Since the cable has symmetry along its axis, you don't need to know the distribution of current as a function of radius within the components of the coaxial cable. There is a theorem (analogous to the one for a spherically symmetric distribution of charge) that the magnetic field outside an axially symmetric current flow is the same as if the entire current were flowing along the axis. So you can just sum the two currents (as vectors) in the "co-ax" and place them on the axis of the cable, a distance 0.3 meters from point P. (If the flow in each part of the cable were equal and opposite, the net current would be zero and the magnetic field outside the cable would also be zero.)

Last edited: Jun 21, 2008
3. Jun 21, 2008

### scholio

you are correct, point P is outside the coaxial cable.

i have started with ampere's law , [closed integral(B dr)] = mu_0(I_encircled) = 2pi(r)B, now this is where i get confused:
based on "(If the flow in each part of the cable were equal and opposite, the net current would be zero and the magnetic field outside the cable would also be zero.)" , is the net current equal to +I - 4I = -3I?

also i use B = 2pi(r)/mu_0(I_encirc) where B = 0.005 Teslas, r = 0.3, I_encirc = -3I, mu_ 0 = 4pi*10^-7, what am i solving for?

am i missing something, where does the I am solving for come into play, i don't see it in the ampere's law i tried?

cheers

4. Jun 21, 2008

### dynamicsolo

In Ampere's Law, the I_enc is the net current "enclosed" by the 0.3 meter circle (in your problem). So in your set-up, you will be solving for I_enc, which has a magnitude of 3I here (you can deal with the sign later).

Upon finding I, you would then assign its value in one direction to the component of the cable carrying +I and 4I pointing in the opposite direction to the other component.

5. Jun 21, 2008

### scholio

oh so i solve for the I in 3I, correct?

so:

B = 2pi(r)/mu_0(I_encirc)
0.005 = 2pi(0.3)/((4pi*10^-7)(3I))
I = 2pi(0.3)/((4pi*10^-7)(3)(0.005)) = 1.885/(1.885*10^-8) = 1*10^8 ampere

" Upon finding I, you would then assign its value in one direction to the component of the cable carrying +I and 4I pointing in the opposite direction to the other component. "
--->so +I - 4I = -3I = 3(1*10^8) = -3*10^8 ampere

is that what you meant, dynamicsolo?

6. Jun 21, 2008

### dynamicsolo

You had written Ampere's Law correctly [ mu_0(I_encircled) = 2pi(r)B ], so I didn't comment on your expression for B at the time. But I think you mean

B = mu_0·(I_encirc)/[2·(pi)·r]

or

I_enc = 3I = [ 2·(pi)·r·B ]/mu_0 .

You should get a pretty big current, but not quite as whopping as you found...

Last edited: Jun 21, 2008
7. Jun 21, 2008

### scholio

oh okay, i revised my calculations using the correct equations you supplied.

I_enc = 3I = [ 2·(pi)·r·B ]/mu_0

3I = (2pi)(0.3)(0.005)/(4pi*10^-7)

I = 0.0094/3.77*10^-6 = 2494.03 ampere

since +I - 4I = - 3I = -3(2494.03) = -7492.09 ampere

is that a "pretty big" yet not "whopping" current i should be getting?

8. Jun 22, 2008

### dynamicsolo

So you get a reference current of I close to 2500 A , making the current in the other cable component 4I or about 10,000 A in the opposite direction.

This gives you an idea of what a weak force magnetism is. Point P is only 0.3 meters (about a foot) away from a net current of 7500 A, and a field of only 5 millitelsas is observed there (or 50 gauss -- around a hundred times stronger than Earth's magnetic field, but not all that large compared to a lot of electromagnets...).

Last edited: Jun 22, 2008
9. Jun 22, 2008

### Redbelly98

Staff Emeritus
And it explains why people go to the trouble of making hundreds of windings in electromagnets and motors, in order to get the required current down to a reasonable level.