(II) A 2.44-m-long coil containing 265 loops is wound on an ir...

Question

(II) A 2.44-m-long coil containing 265 loops is wound on an iron core (average μ = 1850μ₀) along with a second coil of 115 loops. The loops of each coil have a radius of 2.00 cm. (a) Determine the mutual inductance M. (b) If the current in the first coil drops uniformly from 12.0 A to zero in 98.0 ms, determine the emf induced in the second coil.

Accepted Answer

Welcome back. Everyone in this problem, an iron rod characterized by an average permeability mu equal to 1850 times the permeability of free space. Mu knot is the core around which two coils are wound. The first coil with a length of 5 m consists of 400 turns. While the second coil has 180 turns, each turn has a radius of three centimeters A us to determine the mutual inductance between these two coils. And B says if the current in the first coil decreases uniformly from 20 to 0 over a period of 110 milliseconds, what will be the induced EMF in the second coil A says the mutual inductance is 0.087 henries and the induced EMF is 16.72 volts. B says they are 0.092 Henries and 16.72 volts respectively. C 0.095 Henries and 18.62 volts and D 0.095 Henries and 18.62 volts respectively. Now for us to solve, let's try to first make a note of all the information that we have, OK. From our problem. So in our problem, we were told that the length of the coil, we can call that L is 5 m. We also know that the number of turns in the first coil and one is 400. While the number of turns in the second coil and two is 180 the radius of each turn is three centimeters. So we can write that as 0.03 m. And we were also told that the permeability of free space mu is equal to sorry. The permeability mu is 1850 times the permeability of free space mu not OK. So if we substitute that value, that's 1850 multiplied by four pi multiplied by 10 to the negative seventh Henry per meter, which is going to be equal to 0.0023 Henrys per meter. Ok? We know as well that the initial current in the first coil I one is 20 S and all of this happens over 0.11 2nd. OK. So a change in time is 0.11 2nd. Now we're told that our iron rod is the core around which two coils are wound. So since we're talking about a core here, we know that it's cylindrical and the cross sectional area is a circle. So we can go ahead and find the cross sectional area of our core. It's going to come in handy a bit later. So that cross sectional area A is going to be equal to pi R squared, which is pi multiplied by 0.03 m squared. Which when we, we, we get it to be equal to 0.0028 square meters. So now that we have all this information, how can we use it to help us find the mutual inductance A and then the induced EMF B let's start with our mutual inductance. What do we know? Well, recall. Ok. Recall that the mutual inductance M is equal to the permeability multiplied by the number of turns in co one and two multiplied by the area divided by the length. No, we already went ahead and we have all this information. The permeability is 0.0023 Henry per meter. The number of turns in the first coil is 400 in the second. It's 180 our area is 0.0028 square meters. We're dividing all of that by our length of 5 m. Ok. Now, when we go ahead and do that, you should find that you get the mutual inductance to be 0.092 Henry. So we've solved the first part of our problem. Next, let's go to the other part of our problem. Remember, it wants us to calculate the EF in the second coil. If the current in the first coil decreases uniformly from 20 amps to zero over a period of 110 milliseconds. So how can we figure out what the induced EMF is? Well, at this point, the mutual inductance helps because recall that the induced is equal to the negative value of the mutual inductance multiplied by the change in current divided by the change in time. We already know if we go ahead and substitute here, our mutual inductance is 0.092 Henry. OK. Our change in current is the difference between zero and 20 pi because remember its initial current is 20 then it goes down to zero pi and we know this happens over 110 milliseconds or as we said, 0.11 seconds. OK. So now to solve for the induced EMF we can go ahead and substitute these values. Now this value is going to be negative. And when we multiply a negative by a negative, we get a positive. So the induced F is going to be 16.72 votes, know that we have our mutual inductance and our induced ef when we look at both of those values that tells us, then that B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Mutual Inductance

Electromotive Force (emf)

Faraday's Law of Induction

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