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The velocity can be written as the angular velocity times the radius and the differential length written as dr . Therefore,
which is the same solution as before.
From Faraday’s law, the emf induced in the coil is
The constant angular velocity is $\omega =d\theta \text{/}dt.$ The angle $\theta $ represents the time evolution of the angular velocity or $\omega t$ . This is changes the function to time space rather than $\theta $ . The induced emf therefore varies sinusoidally with time according to
where ${\epsilon}_{0}=NBA\omega .$
Check Your Understanding Shown below is a rod of length l that is rotated counterclockwise around the axis through O by the torque due to $m\overrightarrow{g}.$ Assuming that the rod is in a uniform magnetic field $\overrightarrow{B}$ , what is the emf induced between the ends of the rod when its angular velocity is $\omega $ ? Which end of the rod is at a higher potential?
$\epsilon =B{l}^{2}\omega \text{/}2,$ with O at a higher potential than S
Check Your Understanding A rod of length 10 cm moves at a speed of 10 m/s perpendicularly through a 1.5-T magnetic field. What is the potential difference between the ends of the rod?
1.5 V
A bar magnet falls under the influence of gravity along the axis of a long copper tube. If air resistance is negligible, will there be a force to oppose the descent of the magnet? If so, will the magnet reach a terminal velocity?
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