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il63 [147K]
3 years ago
6

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.
Physics
1 answer:
horsena [70]3 years ago
4 0

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}

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