# Homework Help: Magnetic field inside cavity

1. May 18, 2013

### Saitama

1. The problem statement, all variables and given/known data
A constant direct current of uniform density $\vec{j}$ is flowing in an infinitely long cylindrical conductor. The conductor contains an infinitely long cylindrical cavity whose axis is parallel to that of the conductor and is at a distance $\vec{\ell}$ from it. What will be the magnetic induction $\vec{B}$ at a point inside the cavity at a distance $\vec{r}$ from the centre of cavity?

A)$\frac{\mu_0(\vec{j} \times \vec{r})}{2}$
B)$\frac{\mu_0(\vec{j} \times \vec{l})}{2}$
C)$\frac{\mu_0(\vec{j} \times \vec{r})+\mu_0(\vec{j} \times \vec{\ell})}{2}$
D)$\frac{\mu_0(\vec{j} \times \vec{\ell})+\mu_0(\vec{j} \times \vec{r})}{2}$

2. Relevant equations

3. The attempt at a solution
I have done similar questions in electrostatics so I think the same procedure can be applied. The procedure in electrostatics involves considering the cavity as oppositely charged cylinder and find the resultant field (I hope I explained it correctly). So here in this case, I can consider the cylindrical cavity with a uniform current density of $\vec{-j}$. I can easily calculate the magnetic field inside a cylinder which comes out to $\frac{\mu_0Iz}{2\pi R^2}$ where z is the distance from axis but the problem is how would I write it in the form of cross products?

Any help is appreciated. Thanks!

2. May 18, 2013

### rude man

You know the B (or H) field for the cylinder in the absence of the cavity in the coordinate system xyz of the cylinder. Use the cartesian system thruout.

You also know the field for the opposite current "flowing in the cavity" with zero current in the rest of the cylinder, but in the coordinate system of the cavity which is z' = cavity axis and (x' = 0, y' = 0) is the center of the cavity cross-section.

What you need to do now is to translate the coordinates of the cavity B field in terms of the coordinate system of the cylinder. Put the center of the cavity in the xyz system at (x = l,y = 0). So x' = x - l, y' = y, z' = z.

You also need to invoke the cartesian expression for the B field around an amperian circulation surrounding a uniform current density. You'll do this twice, once for each independent current density. Hint: it's a cross-product easily derived from ampere's law.

Last edited: May 18, 2013
3. May 18, 2013

### TSny

First consider the cylinder without a cavity. Let $\vec{s}$ be a vector perpendicular to the axis of the cylinder that subtends from the axis to a point inside the cylinder where you want an expression for $\vec{B}$. (The magnitude $s$ would take the place of your $z$.) Can you find a way to express the magnetic field vector $\vec{B}$ at that point in terms of the two vectors $\vec{s}$ and $\vec{j}$ using a cross product? You might start by taking your expression for the magnitude of $\vec{B}$, $\frac{\mu_0Iz}{2\pi R^2}$, and expressing it in terms of $j$ and s. Then think of a way to construct a vector expression that would have that magnitude and also give the direction of $\vec{B}$.

4. May 18, 2013

### rude man

PS your teach has a sense of humor - C and D are the same answer!

But I wouldn't necessarily pick one of those just because you have an a priori chance of 50% of being right!

5. May 20, 2013

### Saitama

Thank you both of you!

The formula for magnetic field inside a current carrying cylinder in vector form is $\frac{\mu_o(\vec{j}\times \vec{z})}{2}$.

Using the above formula, I get B.

Thank you!

6. May 20, 2013

### rude man

Good work!
What intersting thing can you say about the B field inside the hole?

Last edited: May 20, 2013
7. May 20, 2013

### Saitama

The interesting thing is that it is same everywhere depending only on the distance between the axis of cavity and cylinder. :tongue2:

8. May 21, 2013

### rude man

Exactly! The same both in magnitude and direction! I found that quite amazing.

9. May 21, 2013

### TSny

Yes, it is amazing. There's a similar situation with E fields. If you take a sphere with uniform volume charge density and cut out a spherical cavity anywhere inside, then you find $\vec{E}$ is uniform inside the cavity.

10. May 22, 2013

### rude man

@TSny: Awesome! (to employ a current expletive).
And I imagine the proof proceeds analogously?

11. May 22, 2013

### TSny

Yes, very similar.