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Ampére's Law Cylindrical Conductor with Hole 
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#1
Mar109, 09:24 AM

P: 8

1. The problem statement, all variables and given/known data
A long circular rod of radius R, made of conducting material, has a cylindrical hole of radius a bored parallel to its axis and displaced from the centre of the rod by a distance d. The rod carries a current I distributed uniformly over its crosssection. 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 density j and one with current density j (the latter representing the hole) 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 that r is 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 


#2
Mar109, 10:46 AM

HW Helper
PF Gold
P: 1,197

When you apply Ampere's law, you'll need to draw a loop so that you can evaluate the line integral, as well as calculating the current which is enclosed by that loop. You'll be able to evaluate the magnitutde of B this way for the rod with +I and the rod with I. Since you've used [itex]\oint_ L \mathbf{B} \cdot d\mathbf{L}[/itex] as [itex]2 \pi r B[/itex] in your calculation, you've already figured out the direction of B. Note that the two wires of current +I and I will have different current densities as their areas are different. Finally, find the vector sum of the magnetic field due to the contribution of the two rods to get your final magnetic field in the empty region. 


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