The magnetic field within a long, straight solenoid with a circul...

Question

The magnetic field within a long, straight solenoid with a circular cross section and radius R is increasing at a rate of dB/dt. (a) What is the rate of change of flux through a circle with radius r_1 inside the solenoid, normal to the axis of the solenoid, and with center on the solenoid axis? (b) Find the magnitude of the induced electric field inside the solenoid, at a distance r_1 from its axis. Show the direction of this field in a diagram. (c) What is the magnitude of the induced electric field outside the solenoid, at a distance r_2 from the axis?

Accepted Answer

Hi everyone in this particular problem, we are asked to first find the rate of change of flux to the circle where in this case, we will have a Illinois with a diameter of 20 centimeters creating a magnetic field that increases at a rate of 0.0 5000 per second. And for part one, we will have uh circular cross section with a diameter of centimeters centered on the solenoid axis perpendicular to the axis as well. And we're asked to find the rate of change of flux, which is this one is going to be D five B over D T. And then for the second part, we are asked to calculate the magnitude of the induced electric field at a radio distance of six centimeters from the soil it axis. And the third one is at a radial distance of 12 centimeters from the solenoid axis, which is very similar to the second part. Okay. So first, what we want to do is to just identify all the things that we know. So we have a diameter of the solenoid of 20 centimeters. I'm just gonna write down the are immediately the our soul is going to be set centimeters which equals to 0.1 m. And then we will have the rate of change at which the magnetic field increases, which is the D B over D T. And this is going to be the 0.5 tesla per second. Okay. So the way we want to solve this is to Ashley just by tackling things one by one. So first I'm just going to start up, start us off with part one. This is going to just be the D five B over D T. So we want to recall that FI B can actually be calculated by multiplying B and A together. And for solenoid, the A is going to just be the area of a circle which is going to just be be multiplied by pi R squared. Okay. So in this case, you want to substitute this equation into this D five B over D T. So the D five B over D T can then be D multiplied by B pi R squared over D T. And we know that the pi R squared is going to be constant in this case. So this will just be by R squared multiplied by D B over D T just like. So, okay. So now we can actually solve this problem. So we know that we're asked to find the D five D over DT, which is just this equation here by R squared multiplied by D Phi D B over D T Sorry. And we are asked to evaluate it at a diameter of 12 centimeter. So that will just be an R value of divided by two times 10 to the power of minus two m squared. And D B over D T is known which is are given here 0.5 Tesla per second just like so and this will actually come out to be A D five B over D T value 0.56, five times 10 to the power of -3. That's love per second, just like. So. Okay. So we can kind of neglect option B and option D because we know that the defi B over D T is going to be 0. times standard of our gratitude that slot for seconds. Okay. So now moving on to part two. So for part two, we're asked to find the magnitude of induced electric field at a radial distance of six centimeters from the syringe access. So what we wanted to first identify is to actually identify that the six centimeters is going to be less than the solenoid access here. So, right, just six centimeters less than our soul just like so. Okay. So we want to evaluate the integral for a path which is a circle of radius R and concentric with the solenoid, the magnetic field of the solenoid is going to be confined to the region inside of the cell itself. So B R will equals to zero for our a bigger than our sol just like. So. Okay. So um the way we want to solve this is to actually by solving Faraday's law for a solenoid. So the Faraday's law itself is essentially just this integral of E times D L equals to the D five B or the magnetic flux over D T. And this is just to calculate the magnitude. So I'm neglecting the minus sign here. But we have essentially this formula right here. So the D5B over DT is essentially pretty much calculated using this here that we have decomposed earlier. But the left side of things is where things get different. So we have two integral E times D L. And for solenoid, this will be solved to be E multiplied by the circumference of a circle two pi R Just like. So, and substituting this here, we will then get e multiplied by two pi R Equals D five B for D C. And the D five B over D T is the one that we have find earlier, which is essentially this right here, which is going to be pi R squared times D B over D T. So this is going to be pi R squared times D B over D T and we can kind of simplify this a little bit I'm just gonna find for E because that is what we're being asked. So E will then become pi R squared over two pi R time's D B over D T just like. So, and the pipe can be crossed out. And one of the art can also be crossed out because we're looking at a radial distance less than the art of the solenoid. That's why we can actually cross this out. Next. This can actually be simplified to be one health times R times D B over D T. And this R S the are of where we are evaluating the radio distance. So this can actually be solved. Now, the E is then going to be health multiplied by 0.06 m over two 0 0. m because we have the over to here multiply by 0.5 Tesla per seconds, which is the D B over D T. So the radial distance for the second part is going to be six centimeters here. That's why we have 0.6 m and D B over D T is the same, which is going to be 0.5 tesla per seconds here. This will actually come out to a value of 1.5 times 10 to the power of minus three fold just like so and that will be the answer for the second part. Now, for the third part, we have pretty similar things actually. So for the third part, all the flux is going to be within are less than the R sol. So in this case, the red distance Of 12 cm, it's going to be less, more than our soul. So we are going to kind of go through pretty similar step by step here. But in this case, from this stage here, we will not be able to actually simplify this because the art or the radial distance is bigger than our saw. So I'm going to write down this equation here. So this is going to be E multiplied by two by our equals pi R squared DP over D T. The difference is that in this case, our, our square here is our soul and this R S the radio distance that we have. So in this case, we can simplify E to then be um why are sol squared D B over D T divided by two pi R, we can kind of neglect the pi there. And this can be used to actually calculate R E. So the E Will then Ashley just be our soul, which is 0.1 m squared over to our, which is two times the radio distance is 12 centimeters. So this will be 10.12 m just like. So and the D B over D T is going to be the same which is 0. tesla per seconds. And this will then be Equals to 2. times 10 to the power of -3 fold. And that will be the answer for the 3rd part. So that will actually all corresponds to option C right here where we have the D five B over D T of 0.565 times 10 to the power of minus three tesla per second. We have the um induced electric field magnitude of 1.5 times 10 to the power of minus three fall for a radial distance of six centimeters from the standard access and 2.8 times to the power of minus three fold for a radial distance of centimeters from the solenoid axis. And that will be all for this particular example. If you guys have any sort of confusion, please make sure to check out our other lesson videos on similar topics and that will be all for this particular problem. Thank you very much.

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