Question

A solid metal disk of radius \(R\) is rotating around its center axis at a constant angular speed of \(\omega .\) The disk is in a uniform magnetic field of magnitude \(B\) that is oriented normal to the surface of the disk. Calculate the magnitude of the potential difference between the center of the disk and the outside edge.

Short answer

Answer: The magnitude of the potential difference between the center of the disk and the outside edge of the rotating solid metal disk is given by: $$ V = \frac{1}{2}\omega BR^2 $$

Step-by-step solution

Establishing the formula for calculating the magnetic force

Firstly, we need to determine the expression for the magnetic force acting on a small radial element on the disk. The magnetic force acting on a charged particle of charge q and velocity v in a magnetic field B is given by: $$ F_m = qvB \sin{\theta} $$ Where \(\theta\) is the angle between the velocity vector of the charged particle and the vector direction of the magnetic field. Here, the magnetic field B is perpendicular to the surface of the disk (normal), so \(\theta = 90^{\circ}\). Thus: $$ F_m = qvB $$

Relating linear velocity to angular velocity

As the disk is rotating with a constant angular velocity, \(\omega\), we can relate linear velocity(v) of the small radial element at distance r from the center to the angular velocity(\(\omega\)) using the formula: $$ v = r\omega $$

Computing magnetic force on the radial element

Now we can find the magnetic force experienced by the small radial element on the disk at a distance r from the center using the formula for F_m and the relation between linear and angular velocity: $$ F_m = q(r\omega)B $$

Calculating EMF induced in the radial element

Now, let's find the EMF induced in the small radial element of width dr. The potential difference (dV) associated with the magnetic force acting on the small radial element will be: $$ dV = F_m \,\frac{dr}{q} $$ Substituting the expression for F_m from step 3: $$ dV = (qr\omega B)\,\frac{dr}{q} $$ The q's will cancel out: $$ dV = r\omega B\,dr $$

Calculating the total EMF induced in the disk

Now, we need to integrate dV along the radius from 0 to R in order to find the total EMF induced in the entire disk. $$ V = \int_{0}^{R} r\omega B\,dr $$ The constants can be moved out of the integral: $$ V = \omega B \int_{0}^{R} r\,dr $$ Now, we will integrate: $$ V = \omega B\left[\frac{1}{2}r^2\right]_0^R $$ After plugging in the limits, we will get: $$ V = \omega B\left(\frac{1}{2}R^2\right) $$

Getting the potential difference between the center and the outside edge

Finally, we have the magnitude of the potential difference between the center of the disk and the outside edge: $$ V = \frac{1}{2}\omega BR^2 $$ In conclusion, the magnitude of the potential difference between the center of the disk and the outside edge of the rotating solid metal disk with radius R and constant angular speed \(\omega\) in a uniform magnetic field B perpendicular to the disk's surface is given by: $$ V = \frac{1}{2}\omega BR^2 $$