- #1
Meow12
- 45
- 20
- Homework Statement
- Analyze the direction of electric current in a stationary copper ring as a bar magnet moves through it at constant velocity north-pole first.
- Relevant Equations
- ##\Phi_B=\vec{B}\cdot\vec{A}##
##\displaystyle\epsilon=-\frac{d\Phi_B}{dt}##
Suppose the ring is held stationary such that its area vector points upward. The magnet moves downward toward the ring at constant velocity north-pole first. (The starting position is shown in the figure.)
Since ##\vec{B}## is downward while ##\vec{A}## is upward, ##\Phi_B=\vec{B}\cdot\vec{A}## is negative and decreasing (becoming more negative). So, ##\displaystyle\frac{d\Phi_B}{dt}## is negative. Thus, ##\displaystyle\epsilon=-\frac{d\Phi_B}{dt}## is positive. By the right-hand rule, the induced current in the ring flows counterclockwise when viewed from above.
##\Phi_B## keeps decreasing until it reaches its minimum (##\displaystyle\frac{d\Phi_B}{dt}=0##) when the magnet is midway through the ring. Then, ##\Phi_B## starts increasing (becoming less negative). So, ##\displaystyle\frac{d\Phi_B}{dt}## is positive. ##\displaystyle\epsilon=-\frac{d\Phi_B}{dt}## becomes zero (at the minimum of ##\Phi_B##), then becomes negative, and eventually becomes zero when the magnet is far away from the ring. By the right-hand rule, the induced current stops momentarily, reverses direction (flows clockwise when viewed from above), and eventually dies out.
Does this look okay? Thanks in advance.
Last edited: