A metal blade spins at a constant rate of 5.0 revolutions per second about a pivot through one end of the blade.This rotation occurs in a region where the component of the Earth's magnetic field perpendicular to the blade is 30 $\mu$T.If the blade is 60 cm in length, what is the magnitude of the potential difference between its ends?

Question

Quizplus · Accepted Answer

The potential difference induced across the length of the blade is given by:
$\Delta V = B\ell v$, where $B$ is the magnetic field, $\ell$ is the length of the blade, and $v$ is its velocity.

Substituting the given values, we get:
$\Delta V = (30\ \mu T)(0.60\ m)(5.0\ rev/s)(2\pi\ rad/rev) = 0.566\ \mu V$

Since 1 V = 10^6 µV, the answer is:
$\Delta V = 0.566\ \mu V = \boxed{0.17\ mV}$ (to two significant figures)

A Metal Blade Spins at a Constant Rate of 5 μ\muμ

A Metal Blade Spins at a Constant Rate of 5 $\mu$