Question

A thin metal rod of length L is rotating counterclockwise with an angular velocity omega on the plane of the paper about an axis through one of its ends, as shown in the Figure. A uniform magnetic field B_0 points into the plane in the region where the rod is rotating. Calculate the magnitude and direction of the magnetic force F_B that acts on the rod's charge carriers, resulting in the accumulation of opposite charges at the two ends of the rod. Sketch the rod and show the positive and negative ends. This separation of charges sets up an E in the rod. What is the direction of E? The E grows until F_E = - F_B. Equating the two forces, calculate the magnitude of the electric field. (E) Finally, calculate the potential difference between the two ends of the rod in terms of omega, L, and B_0 by recognizing that delta V = EMF = EL.

Answer 1

A)

Using right hand rule of cross product and Lorentz force we can find that charge carriers that electrons will have force F= q(v x B) = -e(v x B) towards the center of the circle as shown in the below figure.

Thus the magnetic force F_B on electron will be towards the center of the circle.

B)

C)

Electric field direction is from +ve charges to negative charges. Thus E is shown in the above figure away from the center.

D)

F_E = - F_B = - q(v x B) = -[-e(v x B)] = evB

E = F_E/e = vB

Since v=wL

E = wBL

E)

?V = EMF = EL = wB₀L ^2