Potential difference in rod rotating in magnetic field

[kbar1]

Homework Statement

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 anyone 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}}

Homework 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

The Attempt at a Solution

The answer provided is (D).

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?

 

[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

 

[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!