Question

A magnet that is moving towards a 20 turn coil of resistance 12♎ at .5 m/s results in a current of .25A produced in the coil.
What is the force exerted by the coil on the magnet?

Answer 1

Answer on Question #77020, Physics / Electromagnetism

Question. A magnet that is moving towards a $N=20$ turn coil of resistance $R=12\ \Omega$ at $v=0.5\ \mathrm{m/s}$ results in a current of $I=0.25\ \mathrm{A}$ produced in the coil. What is the force exerted by the coil on the magnet?

Solution.

Assume that $z \parallel r$ ($z$ is a distance between the magnet and the coil). We have

$$F = I B_r l = I B_r 2 \pi r,$$

where

$$B_r = \frac{r}{2} \frac{\Delta B_z}{\Delta z}$$

So,

$$E = I R = N \frac{\Delta \Phi}{\Delta t} = N \frac{\Delta (B_z S)}{\Delta t} = N S \frac{\Delta B_z}{\Delta t} = N S \frac{\Delta B_z}{\Delta z} \frac{\Delta z}{\Delta t} = N S \frac{\Delta B_r}{\Delta z} v \Rightarrow \frac{\Delta B_z}{\Delta z} = \frac{I R}{N S v}$$

We get

$$F = I B_r 2 \pi r = I \frac{r}{2} \frac{I R}{N S v} 2 \pi r = I \frac{r}{2} \frac{I R}{N \pi r^2 v} 2 \pi r = \frac{I^2 R}{N v}$$

Finally

$$F = \frac{I^2 R}{N v} = \frac{0.25^2 \cdot 12}{20 \cdot 0.5} = 0.075 N$$

Answer. $F = \frac{I^2 R}{N v} = 0.075 N$

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