# Induced current due to a Solenoid

## Homework Statement

Two metallic rings A and B, identical in shape and size but of different resistivities Pa and Pb are kept of the top of two identical vertical solenoids. When current I is switched on in both the solenoids in an identical manner, the rings A and B jump to heights ha and hb resp, with ha > hb. What are the possible relations between their resistivities and their masses ma and mb?

## The Attempt at a Solution

Let us assume that the current in the two solenoids is towards right (as seen from a side) and anticlockwise as seen from the top.
The anticlockwise flow of current at the top creates a north pole there. The magnetic field through the solenoid is directed outwards which induces a clockwise current in the ring. As seen from the bottom, the flow of current in the ring creates a south pole.

The north pole of the solenoid and the south pole of the ring attract each other. Why will the ring jump?

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Lenz law Hint: B-field is weaker when staying farther from the solenoid.
Besides, your analysis is not correct It's actually the north poles of both face each other. This concurs with Lenz law.

See the figure.

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The direction of induced current is not correct Tell me the reason why you think it should be in your way.

The direction of induced electric field is given by right hand thumb rule. So current flow is in that direction.

The direction of induced electric field is given by right hand thumb rule.
See picture.

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Ah, I never knew that there will be a difference between the increasing and decreasing currents.

In case 1, the induced current is in the same direction of I
In case 2, it is opposite to that of I.

Now apply back to your problem Lesser resistivity ring will get greater impulse.
So Pa<Pb and ma><=mb (can take any values - such that ha is always greater than hb). Is it fine?

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Not really. We have to do some math here.
_ The solenoids will give rise to the same induced emf e = IR.
_ We have the magnetic force on each ring F = kI, where k depends on the B-field of the solenoid and the geometry of the ring and thus is the same for the 2 rings.
_ Therefore: F = ke/R. That means, the impulse given to the ring X ~ 1/R.
_ We have: X = mv, where v is the initial speed. Now assume that the ring will go farther from the solenoid such that it will no longer be affected by B-field of solenoid. Then we have: h = v2/2g.
_ Therefore: 1/mR ~ sqrt(h).
That means: ma*Pa < mb*Pb. Think about it. Even if Pa>Pb, if ma <<<<<<<< mb (ma is very small), then you can lift ring ma more easily than ring mb _ The solenoids will give rise to the same induced emf e = IR.
How did you get that expression?

Agree that induced emf inside each ring must be the same?
Apply Ohm's law: e = IR Induced e.m.f using Ohm's law?? You took 'I' (given in question) as the current flowing in the solenoid or the induced current?

No, I used Ohm's law to deduce the relation between I and e.
Induced emf is deduced by B-field, geometries of the solenoid and the ring, blah blah blah, but I don't care about that arduous process. The only thing I care is that induced emf is the same. And after I obtain induced emf, I turn my attention to the induced current inside the ring, by Ohm's law: e = IR You denoted 'I' as the induced current. I was confused there. In the question 'I' is the maximum current passing through the solenoid (which builds up from 0)

Oops, sorry, I didn't notice that :uhh:

Then I understood everything. Thanks!!!