- #1

latentcorpse

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(i)Find the current density, assumed uniform, within the mercury. Using Ampere's Law, find the magnetic field [itex]\mathbf{B}[/itex] at radius r from the axis.

I said that [itex]\mathbf{J}=\zeta (\mathbf{E} + \mathbf{v} \wedge \mathbf{B})=\zeta \mathbf{E}[/itex] as [itex]\mathbf{v}=0[/itex] as mercury is stationary.

Then I took an Amperian loop around the cylinder of radius r.

Ampere's Law tells us that [itex]\oint_C \mathbf{B} \cdot \mathbf{dr}=\mu_0 \int_S \mathbf{J} \cdot \mathbf{dA}[/itex]

Then,

[itex]\oint_C \mathbf{B} \cdot \mathbf{dr} = \mu_0 |\mathbf{J}| \int_S dA \Rightarrow \mu_0 \zeta |\mathbf{E}| \pi r^2 = \oint_C \mathbf{B} \cdot \mathbf{dr}[/itex]

Firstly, how does this look up to this point?

If ok, how do I find B from that last bit?

cheers!