# Potential difference in rod rotating in magnetic field

1. Feb 22, 2013

### kbar1

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

Q: A rod of length l rotates with a uniform angular velocity ω about the axis passing through its center and perpendicular to its length. A uniform magnetic field B exists with its direction normal to the plane of rotation. The e.m.f. induced between the center and any one end of the rod is:

(A)$Bl^2ω$ (B)${\displaystyle\frac{Bl^2ω}{2}}$ (C) ${\displaystyle\frac{Bl^2ω}{4}}$ (D)${\displaystyle\frac{Bl^2ω}{8}}$

2. Relevant equations

$ΔV = {\displaystyle\frac{\delta\phi}{\delta t}}$, where $\phi$ = magnetic flux = B∙Area

$ΔV = Blv$, where B is magnetic field, l is length of rod, v is velocity of rod perpendicular to length and magnetic field.

$ω = {\displaystyle\frac{2\pi}{T}}$

$v = rω$

3. The attempt at a solution

The given solution goes something like this:

Area swept in one rotation = $\pi{\displaystyle\frac{l^2}{4}}$

$ΔV = B{\displaystyle\frac{ΔA}{ΔT}} = B{\displaystyle\frac{\pi{\displaystyle\frac{l^2}{4}}}{{\displaystyle\frac{2\pi}{ω}}}} = {\displaystyle\frac{Bl^2ω}{8}}$

However, the method I tried was:
$ΔV = Bvl = B(rω)l = B{\displaystyle\frac{lω}{2}}{\displaystyle\frac{l}{2}} = {\displaystyle\frac{Bl^2ω}{4}}$.

Where did I go wrong?

2. Feb 22, 2013

### tiny-tim

hi kbar1!

Bvl only works if every point on the line of length l has speed v

but this line is stationary at one end, so its average speed is halved

3. Feb 22, 2013

### kbar1

Hello! The rod's center is stationary. The two ends are rotating. But I think I got your point. That has to be one of the quickest problem resolutions ever! Thanks a lot!