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On a question from my book:

A long straight wire with a circular cross section of radius $R$ carries a current $I$. Assume the current density is not constant over the cross section of the wire, but rather varies as $J=\alpha r$ where $\alpha$ is a constant. Given $I, R$

Find $\alpha$

Find the magnetic field as a function of r both inside and outside the wire

I think it's just the calculus parts confuses me. My attempt:

$$J=\alpha r' = \frac{dI}{dA}$$ $$dI = 2 \pi r'^2 dr' \alpha$$

$$I = 2 \pi \alpha \int_0^R r'^2 dr'$$ $$I = 2 \pi R^3 \alpha /3$$ $$\alpha = \frac{3I}{2 \pi R^3}$$

from here you just use ampere's and I believe there's no variance issues?

$$\oint \vec{B} \cdot \vec{dl} = \mu_0 I_{in}$$

apply J=I/A

$$B 2 \pi r = \mu_0 \alpha r A$$

$$B = \frac{\mu_0 \alpha A}{2 \pi}$$]

$$B = \frac{3 \mu_0 I r^2}{2 R^3}$$

Is this right? The units seem to line up so I'm hopeful.

Outside the wire is treated as the general uniform wire case I assume and am not too worried about that.

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closed as off-topic by tpg2114, centralcharge, John Rennie, Qmechanic Nov 9 '13 at 16:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

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This is basically a question about how to check an answer.

To check the first part you can check the units and see if they work out. Then you can pretty much only be off by a prefactor. To check the prefactor I would just recalculate $I$ in terms of $\alpha$ and $R$. That is, I would evaulate $\int \alpha r 2 \pi r dr$, plugging in your claimed expression for $\alpha$, and see if this does indeed give me $I$.

To test the value for $B$, you can check dimensions. Your expression has the same dimensions of $\mu_0 I /R$ so that is good. Another thing you can do is test the limits $r \to 0$ and $r \to R$. Do you get what you expect in these cases?

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Dear friend pay attention to ampere's law which has in it the term I(enclosed) in it.So when you find field inside the wire current enclosed within a circle of radius r is something that you will obtain by integrating the current density and is not what you wrote.

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