Question

Ignoring air resistance and the little friction from the plastic tube, the magnet was a freely-falling object in each trial. If a freely-falling object is travelling twice as fast after it has fallen 40 m than after falling 10 m, what do you predict the maximum emf would be if you drop the magnet through the same coil from a height of 40 cm? Explain your answer.

Answer 1

Answer:

emf will also be 10 times less as compared to when it has fallen $40 \mathrm{m}$

Explanation:

We know, from faraday's law-

$e m f=-N \frac{\Delta \Phi}{\Delta T}$

and $\Phi=B . A$

So, as the height increases the velocity with which it will cross the ring will also increase. $(v=\sqrt{2 g h})$

Given

$\mathrm{V} 1(\text { Speed at } 40 \mathrm{m})=2 \mathrm{x} \mathrm{V} 2(\text { speed at } 10 \mathrm{m})$

$\sqrt{2 g h_{2}}=2 \times \sqrt{2 g h_{1}}=28.28 \mathrm{m} / \mathrm{s}$

Now, from $40 \mathrm{cm}$

$V_{3}=\sqrt{2 g h_{3}}=\sqrt{2 \times 10 \times 0.4}=2.82 \mathrm{m} / \mathrm{s}$

From equation a and b we see that velocity when dropped from $40 \mathrm{m}$ is 10 times greater when height is 40 $\mathrm{cm}$ so, emf will also be 10 times less as compared to when it has fallen $40 \mathrm{m}$