4. Inside a very long cylinder, a uniform current density J0...

4. Inside a very long cylinder, a uniform current density $J_0$ A/m² is flowing parallel to the axis of the cylinder (i.e. the z-axis). a. Solve the Poisson's equation in cylindrical coordinates to obtain the vector magnetic potential $\vec{A}$ inside the cylinder. To fix the constant of integration, take $\vec{A} = 0$ at r = 0. Write the steps in details to get the final answer in the space below. b. From the result in part (a), obtain the expression for magnetic field $\vec{B}$ inside the cylinder. Write the steps in details to get the final answer in the space below. c. Use the integral form of Ampere's law to obtain the magnetic field $\vec{B}$ inside the cylinder. Write the steps in details to get the final answer in the space below.

Transcript

00:01 Here we consider a cylinder in a z direction with a radius a.

00:12 So let me write the radius with the red here.

00:15 That is a.

00:17 This is in the z direction.

00:20 Here is the x -axis.

00:22 So we can represent the z direction by a unit vector in k -kap.

00:26 And here is our y.

00:28 The current density is not constant across the cross -section of the wire.

00:33 So let's say the cylinder is a wire.

00:36 So the current is not uniform.

00:39 It's not constant.

00:40 We represent this current by, so for part a, j current density is b over r times exponential r minus a divided by delta in the k direction.

00:56 So k -k.

00:58 R is the distance from the center.

01:01 So for a d -i current, for element dr here, so dr, whose area will be da, which will be equal to 2 pi r times the dr.

01:21 Then we can write the di will be equal to j times da using the definition of a current density.

01:29 Then we can write the di so the i will be then j times 2 pi r d r then we bring down the j that we wrote here so we'll substitute this j and find di so di i will be then b over r times exponential r minus a a or a delta in a k -kap direction, so j times the 2 pi r times the d r.

02:03 Then the total current passing through the wire equals to the integration over the whole wire from r from r 0 to r to r is equal to a.

02:15 So i nod is equal to integral from 0 to a, di that is 2 pi b integral of 0 to a e exponential r minus a over delta d r then the integral evaluates into a 2 pi b delta er minus a over a delta from 0 to a this gives us 2 pi b delta b exponential minus a or delta then our i nod will be 2 pi b delta 1 minus e minus a or a delta the radius of the cylinder is a 5 centimeter which is a is equal to 5 centimeter b is a constant equals to 60 600 sorry ms per meter d is a constant which is equal to 2 .5 cm.

03:23 We substitute those values and find i .0 to be 2 pi times 600 times 0 .025 times 1 minus e exponential 0 .050 over 0 .025.

03:40 This gives us the current of current of i nod of 81 .5 empire...

Question

Please give Ace some feedback