Question

A conducting rod (length = 2.0 m) spins at a constant rate of 2.0 revolutions per second about an axis that is perpendicular to the rod and through its center. A uniform magnetic field (magnitude = 8.0 mT) is directed perpendicularly to the plane of rotation. What is the magnitude of the potential difference between the center of the rod and either of its ends?

Answer 1

Answer:

0.050V

Explanation:

To solve this problem it is necessary to use the concepts related to the potential between two objects that have a magnetic field, this concept is represented in the equation.

$\int dV = \int_{0}^{l/2} Bv (dl)$

Where,

v= tangencial velocity

B = Magnetic Field

We know for definition that,

$v= l\omega$

Where,

L = length

$\omega =$ Angular velocity

We can replace this values in our first equation then,

$\int dV = \int_{0}^{l/2} B (l\omega) (dl)$

Integrating we have,

$V = \frac{1}{8} Bl^2 \omega$

Replacing the values,

$V= \frac{1}{8} (8*10^{-3})(12.56)(4)$

$V = 0.050V$

Therefore the potential difference between the  center of the rod and the other rod is 0.050V