# Potential difference in rod rotating in magnetic field

## 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 any one end of the rod is:

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

## 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?

## Answers and Replies

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