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## Induced current in conductor moving circularly in constant B-field

1. The problem statement, all variables and given/known data

A light bulb with resistance R is attached on a metal rod which is rotating around the point O on the figure. The metal rod is in contact with an electrical conductor which is a part of a circle with radius d. The metal rod and the circular electrical conductor is a closed circuit. The rod now rotates with angular velocity $\omega$ through the constant magnetic field pointing out from the paper.

a)

Find an expression for the induced current through the light bulb, expressed in terms of $\omega$, d, B and R.

2. Relevant equations

IR=vBr

where v is the tangential speed of the rod perpendicular to the B-field (every speed is perpendicular to the B-field, since we are looking at a plane) and r is the length of the rod moving at this speed.

I=$\frac{\omega Br^{2}}{R}$

v substituted for $\omega r$

3. The attempt at a solution

Since the every part of the rod is moving with different linear speeds, we should integrate the RHS from the 0 to d with respect to r and that should be it right?

i get:

I=$\int^{d}_{0}\frac{\omega Br^{2}}{R}$

I=$\frac{d^{3}B\omega}{3R}$

But when i look up the solution it says:

I=$\frac{Bd^{2}\omega}{2R}$

so who's right?

Edit: Problem solved!
Attached Thumbnails

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Maybe i should add that RI=vBr is derived from Faradays law of induction, stating that the induced EMF is equal to the closed path integral of E+v X B with respect to l (path of the circuit), and Ohm's law stating that the EMF is equal to RI when looking at the entire circuit. I only integrate over the rod since this is the only thing moving relative to the B-field. The cross product in faradays law reduces to the magnitudes of v and B multiplied, since they are always perpendicular to each other in this problem and since i only need to find the magnitude of the EMF.

Nevermind i solved it!  After reading my last post over, i realized that i should use Faradays law of induction as the more general law, rather than IR=vBl which is a solution to Faradays law in a particular situation. I then obtained the same answer as in the solutions sheet.

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