(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A long circular rod of radiusR, made of conducting material, has a cylindrical hole of radiusabored parallel to its axis and displaced from the centre of the rod by a distanced. The rod carries a currentIdistributed uniformly over its cross-section.

Consider the superposition of two currents flowing in opposite directions, and hence show that the magnetic field in the hole is uniform and equal to:

[tex]B = \frac{ \mu_{0} dI }{2 \pi (R^{2} - a^{2})}[/tex]

Hint:

Consider the rod with a hole to be a superposition of two rods without holes: one with current densityand one with current densityj(the latter representing the hole)-j

Deduce from Ampére's law that:

[tex]\mathbf{B}(\mathbf{r})=\frac{1}{2} \mu_{0}(\mathbf{j} \times \mathbf{r})[/tex]

for each current, where j is the current density, r is the vector from centre of the rod.

2. Relevant equations

Ampéres law (integral form):

[tex]\oint_ L \mathbf{B} \cdot d\mathbf{L} = \mu_{0} I[/tex]

(differential form)

[tex]\mathbf{\nabla} \times \mathbf{B} = \mu_{0} \mathbf{j}[/tex]

3. The attempt at a solution

Using cylindrical polars, I worked through for the integral form of Ampere's law for the first rod (j, radius R) and got:

[tex]B = \frac{\mu_{0}}{2} j[/tex]

I think that's right, and it seems to be along the lines of what I should get. Except, it's a scalar, but what I'm given in the hints is vector.

Also, given thatris the vector from the centre of the rod to axis is going to be zero for the first rod, doesnt j x 0 give zero?

But if I try the differential form I end up with zero after the first step, I think this isn't the way to do it?

So I'm a bit stuck, any hints would be appreciated :)

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

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Ampére's Law Cylindrical Conductor with Hole

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