FIGURE P30.47 shows a 1.0-cm-diameter loop with R = 0.50 Ω ins...

Question

FIGURE P30.47 shows a 1.0-cm-diameter loop with R = 0.50 Ω inside a 2.0-cm-diameter solenoid. The solenoid is 8.0 cm long, has 120 turns, and carries the current shown in the graph. A positive current is cw when seen from the left. Determine the current in the loop at t = 0.010 s.

Accepted Answer

Welcome back. Everyone. In this problem. A 0.75 centimeter radius copper loop with a resistance of 0.75 ohms is positioned inside a 1.25 centimeter radius solenoid that is 100 centimeters long and consists of 800 turns. The solenoid carries a current that changes with time as shown in the graph. And in this system, a positive current flows clockwise. We want to find the current in the copper loop at T equals 15 milliseconds for our answer choices A is negative 17 micro amperes B is negative 35 micro amperes C is negative 39 micro amperes and D is negative 51 micro amperes. Now, let's think about what we're trying to find here. We're trying to find the current in the copper loop at a particular time. Now, there are a few assumptions we make here. First, we're assuming that the magnetic field produced by the solenoid is uniform across the area of the copper loop. We're also assuming that the solenoid is long enough so that we can ignore edge effects and hence treat the magnetic field as being parallel to the axis of the solenoid. Finally, we're also assuming that the copper loop is small enough compared to the solenoid diameter. So we can treat the entire loop as experiencing the same magnetic field. Now, before we try to figure out the current in the copper loop, let's make a note of all the things we know. First off, we're concerned with the current when the time is 15 milliseconds on our left, we have a graph of the current of the solenoid versus the time. And at 15 milliseconds, our current in the solenoid is positive, we can notice that here. And since it's positive, like our assumption says here in this system, a positive current flows clockwise. So at 15 milliseconds, our current is flowing clockwise. So it's flowing in this direction. Let me show the clockwise direction here on our diagram. So if that's the case, what do you think that means for our induced current in the loop? Well, the magnetic field produced by the solenoid runs parallel to the axis of the loop and it establishes a magnetic flux through the loop as the current in the solenoid varies. This leads to a change in the magnetic flux. And the change in flux induces an EMF in the opposite direction, thereby generating a corresponding induced current in the opposite direction within the loop as shown here. So that means our current induced in the loop is going to be going anti clockwise. So if we can imagine here, if it's represented by these red arrows that would represent the current inside the loop. And because it is going in the opposite direction because it is flowing anti clockwise, then we can assume our current would be negative. So already we know that the current in our loop has to be negative. What else do we know? Well, we also know that the resistance of our loop is 0.75 ohms. OK. We also know that the radius of our loop is 0.75 centimeters. So that's 0.75 times 10 to the negative 2 m. And we know that the length of our solenoid is 100 centimeters. So that's 100 times 10 to the negative 2 m. Now, what do we know about the current in a cyanide? What do we know? Well, for any current we know, so we can recall that the current would be equal to the EF divided by the resistance of our loop. We know the resistance, but we don't know the EMF however, recall that the EMF was induced because of because of the change in flux. As a matter of fact, that's the definition of emfemf equals the change in flux over time. So we can replace our AM F in our diagram here in our formula here rather as are changing flus over time. But again, here in our problem, we don't know what our flux is. So what else do we know Well, we know that our flux, we know the definition for flux. And let's again, let's write our formula here. The definition of flux or flux is the product of the magnetic field and the cross sectional area of arsenide. In this case, our cross sectional area because it's a circle would be pi R squared. And in this situation, if we're talking about the change in flux, as we've said, our radius stays the same, but it is our magnetic field that is changing because that change in flux is because of a change in magnet change in magnetic field. So we can write that then as one divided by R multiplied by the area of our loop, which is pi R squared multiplied by our change in magnetic field over time. But here we run into another roadblock because we don't know the value for our magnetic field and we don't know how it's changing over time. However, we do know that the magnitude or the strength of our magnetic field in a solenoid equals mu knot multiplied by the number of turns multiplied by the current. In this case, the current running through the solenoid divided by its length. OK. So that's what we know about the strength of our magnetic field. Now again, here we know that mu knot is a constant. The number of turns here is a constant that's not changing, neither is the length of the solenoid, but the current for our solenoid is changing. So that tells us, then that our change in magnetic field would be equal to mu knot multiplied by and multiplied by the change in our solenoid. The change in our current, the change in the current sorry of our solenoid and all of that would have been divided by the length of the solenoid. So if we substitute that value here into our formula for the current of the loop that tells us, then that the current in the loop equals pi R squared, that the radius of the loop divided by the resistance of the loop multiplied by mu and divided by L multiplied by or a change in the current of the solenoid divided by our change over time. Now, let's take a break. Let's think about what we have so far. So if we look on our formula, we know what the radius of the loop is. We know what its resistance is. We know what Muno is. That's the permeability of free space. We know how many turns are in the solenoid. And we know the length of the solenoid, but we don't know the change in current over time for the solenoid. Is there a way that we can figure that out? Well, if we go back to our information that we have notice that we have a graph of the solenoid versus the time are the current in the salon rather versus the time. So since we have this information here what does that mean? The slope of our graph represents the change in the current in the solenoid over time. So if we can find the value for the slope of our graph, that slope. So let's write that here, our slope would be equal to the change in current over the change in time. So what's the slope of our graph? Well, we can use any two points but I think I like to use these points here. Notice that going from zero milliseconds and negative 0.7 empires, our current goes up to or at 10 milliseconds. Our current is no zero and PS so in other words, our current increases by 0.7 pires over 10 seconds. So that means our slope would be equal to our change in current, which is 0.7 S. So let's make a note of it down here. So our change in current over time would be 0.7 S divided by our change in time, which is 10 milliseconds, 10 times 10 to the negative three seconds. And if we work that out, 0.7 divided by 10 times 10 to the negative three equals 70. So a change in current over time is 70 s per second. No, that would represent our change in current over time in our formula. So we now have a value for the change in current of the solenoid over time. And since we have all known values, we can substitute to solve for the current in the loop. So therefore, that means the current in the loop would be equal to pi multiplied by R squared. That's the radius of our loop. 0.75 times 10 to the negative 2 m R squared multiplied by the permeability of free space which is four pi times 10 to the negative seven tesla meters per pi multiplied by the number of turns which is 800. And all of that is being divided by our, the resistance of our loop, which is 0.75 ohms multiplied by the length of our solenoid, which is 100 times 10 to the negative 2 m. And all of that is being multiplied by our chains in the current in the solenoid over time, which is 70 pairs per second. So this would give us the current in the loop. All we need to do now is to substitute these values into our calculator. Now, when you put this into your calculator to where the salt, you should find that the current in the loop is 1.658 times 10 to the negative five amperes. Notice that all of our answers are listed in micro amperes. So if we write it in micro amperes, you get the current in the loop to be equal to 17 micro amperes. There's one last thing we have to put on our final answer. Remember that our current is flowing in the counter clockwise direction. So that means the current in our loop would be negative. So our answer should actually be, let me put that in a different color here. Negative 17 micro amperes. In other words, the induced current is negative 17 micro amperes. When we look back on our answer choices that would have been answer choice. A thanks a lot for watching everyone. I hope this video helped.

FIGURE P30.47 shows a 1.0-cm-diameter loop with R = 0.50 Ω inside a 2.0-cm-diameter solenoid. The solenoid is 8.0 cm long, has 120 turns, and carries the current shown in the graph. A positive current is cw when seen from the left. Determine the current in the loop at t = 0.010 s.

Key Concepts

Electromagnetic Induction

Ohm's Law

Magnetic Flux

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