Assume there is no friction. In one trial, you hold the magnet so the front is 2

Question

Assume there is no friction. In one trial, you hold the magnet so the front is 24.4 cm above the top of a coil which has 6.14 turns/cm, and release it from rest. An EMF of 8.06 V is induced in the coil.

a) In procedure 1: now you hold the magnet 35.9 cm above the edge of the coil, and release it from rest. What EMF will be induced in the coil now?
EMF = V

b) In procedure 2: you hold the magnet 24.4 cm above the edge of a coil with 9.70 turns/cm, and release it from rest. What EMF will be induced in the coil now?
EMF = V

c) Now, you hold the magnet 77.6 cm above the edge of a coil with 15.7 turns/cm, and release it from rest.
- What EMF will be induced in the coil now? EMF = V
- If the effective resistance of the coil is 64.8 ?, what current is induced in the coil? I = A

Explanation / Answer

induced emf e = lvB

also v = sqrt (2gh)

and magnetic field B = u0nI

so putting all to ghter emf e = L(sqrt(2gh) * u0ni

so induced emf is directly sqrt(h) and directly proportional to no. of turns

a. emf e2/e1 = sqrt(35.5/24.4)

e2 = 1.206 * 8.06

e2 = 9.721 volts

b. apply e2/e1 = n2/n1

e2 = 9.7* 8.06/6.14

e2 = 12.733 volts

c. apply e2/e1 = sqrt(h2/h1) * n2/n1

e2 = 8.06* sqrt(77.6/24.4) * 15.7/6.14

e2 = 36.706 volts

d. induced current I = emf/R

I in case 1 = 9.721/64.8 = 0.15 A

I in case 2 = 12.733/64.8 = 0.196 A

I in case 3 = 36.706/64.8 = 0.566 A

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