8) If the current density in a wire or radius R is given by J = kr, 0 < r < R, what is the curre

Step 1: To find the total current in the wire, we need to integrate the current density over the

8) If the current density in a wire or radius R is given by...

Transcript

00:01 We have a wire with a radius of 3 .4 millimeters, and the current density in both cases features this constant, which is 5 .5 times 10 to the 4th amper per square meter.

00:12 And so we want to evaluate the total current in this wire in the case where we have this current density and then this current density.

00:20 And so the total current, i'll write this on the bottom.

00:23 Of course, there's just going to be the integral of j of r, d ,a.

00:28 And since there is no angular dependence in the current density here, we can do the angular integral, the d -theta integral, and this will just be the integral from zero to r of j of r times r, dr, so let's do that integral for both cases.

00:44 So here we can fact out the constant j -0 in front of the integral, so we have 2 pi times j -not, times the integral from 0 to r of little r over big r times r, d -r, so we can write this as a, like r squared d r and so that integral is a one -third r cubed one -third little r cubed and when we evaluate it and everything um and plug in the answers we'll get something like two pi over three times j -not times r squared um and so with the numbers we're given this is about 1 .33 amps next up similar thing going on here we can factor out this constants once again and we have the integral from zero to capital r of one minus little r over big r times r d r and so the first term in the integral is going to produce a one -half r squared the second term in the integral is going to produce you know a negative one -third r cubed neglecting the constants and so we'll have like two pi times j -not times one -half capital r squared minus one -third third capital r squared so 1 half minus the third is a sixth.

02:00 So this is just a sixth of the previous, or sorry, it's half of the previous current density, basically...

Question

8) If the current density in a wire or radius R is given by J = kr, 0 < r < R, what is the current in the wire? A) $kR^2/2$ B) $kR^3/3$ C) $kR^2$ D) $2\pi kR^3/3$ E) $3\pi kR^3/2$

Question

8) If the current density in a wire or radius R is given by J = kr, 0 < r < R, what is the current in the wire? A) $kR^2/2$ B) $kR^3/3$ C) $kR^2$ D) $2\pi kR^3/3$ E) $3\pi kR^3/2$

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