Question

A very thin circular disk of radius R = 17.00 cm has charge Q = 50.00 mC uniformly distributed on its surface. The disk rotates at a constant angular velocity ? = 5.00 rad/s around the z-axis through its center. Calculate the magnitude of the magnetic field strength on the z axis at a distance d = 1.700 × 10^3 cm from the center. Note that d >> R.

Answer 1

Break up the disk into a series of loops of infinitesimal width. A given loop of radius r and thickness dr will have a total area dA=2?r dr, and thus contain a total charge dQ =2??r dr. This loop is rotating at an angular velocity ?, which means that the charge dQ on our loop makes a circuit around the axis every period of rotation, T = 2?/? seconds. Since a charge dQ makes a circuit every T seconds, our infinitesimally thin ring is a current loop:

For this single loop, equivalent to a current dI, we can easily calculate the field a distance z above the axis. We did this in class, it is also an example problem in your textbook. Applied to the present case, the solution is:

In order to find the total field, we have to integrate over all possible infinitesimal rings, from r ? 0 to r=R:

The surface charge density is ? = Q/(?R²)

Putting corresponding values in above obtained equation we get:

B(z) = 7.35*10^-14 T