A straight cylindrical wire lying along the x axis has a...

A straight cylindrical wire lying along the $x$ axis has a length of $0.500 \mathrm{m}$ and a diameter of $0.200 \mathrm{mm} .$ It is made of a material that obeys Ohm's law with a resistivity of $\rho=$ $4.00 \times 10^{-8} \Omega \cdot \mathrm{m} .$ Assume that a potential of $4.00 \mathrm{V}$ is maintained at $x=0,$ and that $V=0$ at $x=0.500 \mathrm{m} .$ Find (a) the electric ficld $\mathbf{E}$ in the wire, (b) the resistance of the wire, (c) the electric current in the wire, and (d) the current density $J$ in the wire. Express vectors in vector notation. (e) Show that $\mathbf{E}=\rho \mathbf{J}$.

Transcript

00:01 All right, so we have a wire that is half a meter long.

00:04 And this is not going to be drawn to scale by any stretch of imagination.

00:07 It's half a meter long, and it has a radius of 0 .2 millimeters.

00:16 And so we're also told that the voltage difference between the ends of the wire is 4 volts.

00:23 And the wire has a resistivity of 4 times 10 to the negative 8 oom meters.

00:30 And so we want to find out several things about this wire.

00:32 First, we want to find the electric field in the wire.

00:35 So this is easy.

00:36 The electric field is just the change and voltage across the wire over the change and distance.

00:42 So our voltage difference between the ends is four volts, and our distance is half a meter.

00:47 So our electric field is just going to be eight volts per meter.

00:52 Next up, we want to find what is the resistance of the wire.

00:58 So the resistance of a wire.

01:00 It's going to be the resistivity.

01:01 Times the length of the wire divided by the cross -sectional area.

01:05 So this will be 4 times 10 to the negative 8 oom meters times the length, which is half a meter, divided by pi times 0 .0002 meters squared.

01:20 So if you plug everything in, correctly, this should come out to about 0 .159 oms.

01:28 So that's our resistance.

01:29 Then part c we want to find the electric current in the wire so that we get this just from ohms law says the current is like the voltage difference voltage drop across the wire divided by the resistance so eight or sorry four volts over 0 .159 oms gives us a current of about 25 .13 amps which is a very large current and then part d asks us, what is the current density in the wire? so this is just going to be the current divided by the cross -sectional area of the wire.

02:12 So 25 .13 amps divided by pi times 0 .0002 meters squared...

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