A long, thin solenoid has 900 turns per meter and radius 2.50 ...

Question

A long, thin solenoid has 900 turns per meter and radius 2.50 cm. The current in the solenoid is increasing at a uniform rate of 36.0 A/s. What is the magnitude of the induced electric field at a point near the center of the solenoid and (a) 0.500 cm from the axis of the solenoid; (b) 1.00 cm from the axis of the solenoid?

Accepted Answer

Hi everyone. In this particular example, we are asked to calculate the electric field induced measure near the midpoint of the solenoid. And the radial distance of 0.75 centimeters and 1.25 centimeters from the soil needs access where the soil itself will have a current increasing uniformly at a rate of 38 per seconds through the solenoid with a radius of three centimeters. And also I will have 800 turns per meter. So let's start with identifying all the different things that is known in this particular example. First, we have the rate at which the current is increasing, which is going to be the D I over DT, which is going to be 38 M per second. And we also have the radius of the solenoid which is going to be three centimeters Which will equal to three times 10 to the power of -2 m. And unless we have the end or the number of turns, which is going to be 800 turns. So recall that in order for us to actually solve this problem, we want to apply Faraday's law. So Faraday's law itself is essentially just this E multiplied by D L equals minus D magnetic flux five B over D T. And because we will only look at the magnitude of E or we will only need to calculate the magnitude of E. We can just take a magnitude an absolute value of this and kind of neglect the minus here. So we want to apply this to an integration path that is a circle of Radius R where R is less than the solenoid radius. Because we know that at a radio distance of 0.75 cm and 1.25 cm, it will be less than the actual radios off the solenoid itself. Okay. So in order to solve this, we want to start with creating a uh formula or equation that we will be able to use. So let's just start solving like each step here, I'm gonna start with the left side. So the left side, we will have e multiplied by D L that will equals two multiplied by, in a solenoid, we will be having the actual circumference of a circle. So this will equals two multiplied by two pi R. So next, we can recall that the magnetic flux can actually be calculated by multiplying B and A together. And for a solenoid, the A is just going to be the area of a circle. So it is going to be pi R square just like this where this are, is our soul. Just like that. Okay. Next, we want to input this into the right side equation here. So we have the D five B over D T in the absolute value, I'm just going to remove the minus sign immediately. So this will essentially be D being multiplied by pi R sol squared over DT just like. So, and then from here, we know that our of or the area of the soil is constant so we can pull it out of the derivative. So this will just be by R squared saw more supplied by D B over D T just like so okay, next, we can actually combine this two equation which will give us I'm just going to rewrite this whole equation here, I guess to make it easier to understand the L uh this is going to just be defi b over D T just like. So and the left side is going to be E multiplied by two by our. So, and the right side is going to be pi R sol squared DP over DT. And in this case, we can kind of simplify this, the arsenal can actually be crossed out. So does the pie. So in order to find E, which is the one that we wanted, this is just going to be 1/2 multiplied by our sol multiplied by D B over D C just like. So okay, now that we get this equation right here, we know that the be here can actually be calculated by multiplying you not multiplied by N multiplied by I. So we want to recall this equation here for a solenoid. And we want to input this value this B into this equation here so that this equation will then become multiplied by D. And I, and we want to recall that the mu naught and the end is going to be constant in the solenoid. So we can pull it out. And the only thing in the derivative is then going to just be the eye. So this will then be equals to half our saw new not and D I over DT just like so okay. So this is essentially the final equation that we will use to calculate part one and part two. So we will just actually start by solving our first part. Um So for the first part actually ask us To calculate it where the radial distance is 0.75 cm. Okay. So for red distance, 0.75 cm we have E equals half our saw, you're not and D I. So um in this case, what we can do is to Ashley, start plugging every single thing into the equation. Uh In this case, our our soul here can actually be substituted with the radio distance here because that is the distance that we want to find the E at. Okay. So this will just be half time, 0.75 times 10 to the power of minus two m Times four pi times stand to the power of - Tesla multiplied by meter over emp Multiplied by N which is going to be 800 m To the power of -1 and the D over DT is going to be 38 amp per seconds just like so and this will actually come out to the value of E equals 1.43 times 10 to the power of minus four full per meter at red, this of 0.75 centimeters just like so and then for the second part, four red dist one point 25 cm here, 25 cm, we essentially have the same thing. The only difference is just the are here, it's going to be substituted With a value of 1.25 cm. So this is going to be half 1.25 then stand to the power of minus two m times the mu not which is just a constant value 800 for the end and the eye is 38 A M per second. And this will come out to be 2.39 times 10 to the power of minus four fold per meter at rat dist of 1.25 centimeters just like so okay. So that will pretty much be the answer to our problem with an E value of 1.43 times 10 to the power of minus four fall per meter at Reagan distance of 0.75 centimeters, an E value of 2.39 times 10 to the power of minus per meter at red distance of 1.25 centimeters. And that will correspond to option B here and that will pretty much be all of it for this particular practice problem. If you guys have any sort of confusion on this, please make sure to check out our other lesson videos on similar topics and that will be all for this practice problem. Thank you.

Key Concepts

Faraday's Law of Induction

Magnetic Field Inside a Solenoid

Induced Electric Field

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