Magnetic Fields!
A nonconducting disk of Radius 5 m has charge density (units C/m^2) of p(r)=2pi r (at the center there is essentially no charge, and each ring of radius r around the center has charge density per area of 2pi r). The disk spins at angular velocity 10 rad/s around an axis perpendicular to the plane that the disk is on and through the center of the disk. What is the magnitude of the magnetic field at the center of the disk? Provide your answer to three significant figures (and do not use scientific notation).
The answer is 0.000493.
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1 solution
By definition, p = dq/dA since p is the area density at any location. (dq is the amount of charge at that location, and dA is the area of that location) Let X be a constant equal to 1 C/(m^3)
So dq/dA = X 2pi*r (where r is the distance from the center of the disk)
And dq= 2Xpi*r * dA
since A=pi*r^2
dA=2pi r dr
Then dq=4X (pi)^2 r^2 dr.
For clarification, this expression says that in this disk, the amount of charge present in a ring centered at the center of the disk with radius r is r^2*dr.
When the disk spins, the charge in each ring spins, creating a current.
To find this current for a ring of radius r, we should find how much charge goes through a location on the ring in 1 second (note we will count charge that goes through that location twice by doubling that charge). If the period of rotation is T, it takes the entire ring of charge T seconds to go through that location on the ring. From rotational motion, we know that T=2 pi/w where w is the angular velocity of the disk's spinning. To find how many revolutions in 1 second, we take 1/T = w/(2pi). So if di is the current produced in that ring when the disk spins, we have di=w/(2pi) *dq, or di=2 X pi w*r^2.
Now we shall find the magnetic field. Each ring's current contributes to the magnetic field in the center. In fact, if a ring has current I, the magnetic field in the center of the ring is B=u0 I/(2R) where u0 is the permeability constant of free space, and R is the radius of the ring. So each ring in the disk contributes dB= u0 di/(2r) where r is the radius of the ring in the disk. Substituting with out earlier equation for di gives
dB=u0 X w pi r*dr.
(Units Check: u0 has units of T m s/C. w has units of 1/sec. pi is unitless. rdr has units of m^2. X has units of C/m^3 So the units multiply to Teslas). To find the total magnetic field, we can just integrate that expression from R=0 to R=5m.
B=0.000493 Teslas from integrating.