00:01
In this problem on the topic of the magnetic field, we are told that a sphere of radius r has a constant volume charge density, which we'll call row.
00:10
We want to find the magnetic field at the center of the sphere when it rotates as a rigid body, and the angular velocity of rotations is omega about an axis through its center.
00:21
Now we'll consider the sphere as we made up of little rings of radius r centered on the rotation axis.
00:26
The contribution to the field from each ring is db, and db is equal to mu not r squared di over 2 into x squared plus r squared to the power 3 over 2, where the charge on the current each ring, di, is equal to dq over t, which is omega dq over 2 pi.
01:16
Now the charge element on each ring dq is equal to the volume charge density row times the volume element dv, which we can write as row into 2 pi r d r d x.
01:39
So we therefore can see that d b is equal to mu not times row times omega r cubed the r d x over 2 into x squared plus r squared to the power 3 over 2 now again the volume charge density row is equal to the total charge q over the volume of a sphere which is four thirds pi r cubed and so if we integrate this we can find the magnetic field so the magnetic field b is equal to the integral from minus r to r times the integral from zero to the square root of r squared minus x squared of mu not times row times omega over 2 times r cubed the r d x over x squared plus r squared to the power 3 over 2.
03:11
Now we let v be r squared plus x squared, which means that dv is equal to 2r d r and r squared be v minus x squared...