A long, thin solenoid has 400 turns per meter and radius 1.10 cm. The current in the solenoid is increasing at a uniform rate di/dt. The induced electric field at a point near the center of the solenoid and 3.50 cm from its axis is 8.00*10^-6 V/m. Calculate di/dt.

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Hi everyone in this problem, we are asked to determine the numerical value of the current change rate where we have an experiment with a solenoid of radius 1. cm and 500 turns per meter. And we have a magnetic field director to the solenoid inducing a uniformly increasing Corinne and Amanda took off the magnets. The electric field induced is six times stands to the power of minus six fold per meter measure around the midpoint of the solenoid at the radial distance of 4.5 centimeters from the solenoid access. So we will just start with creating a list of everything that is known. First, we have the radius of the solenoid itself which is going to be 1.3 centimeter or 1.3 times 10 to the power of minus two m. And we also have the end or the number of turns for the solenoid which is we have the magnitude of the electric field induced which is going to be the E which is six times 10 to the power of minus six fold per meter. And we have the radial distance from the solenoid access at which this is measured. So this is going to be our, This, I guess, which is going to be 4.5 cm, which is going to be uh 4.5 times 10 to the power of -2 m. So from all of this, we know that we can actually solve this problem by using Faraday's law recall, that Faraday's law is essentially saying that the India CMF will actually equal to the E multiplied by D L which will equal to uh just the minus or essentially because we're looking at the magnitude, it's going to be an absolute value minus the derivative of the magnetic flux over time just like. So, okay. So now uh for a solenoid, we know that this side right here is going to actually be just e multiplied by the circumference over the perimeter of a circle which is two pi R and the right side, we're just going to neglect the minus sign here. The right side, I'm just going to actually leave it as D magnetic flux over DT because we want to recall before actually substituting this formula later on inside that the magnetic flux can actually be calculated by multiplying B times A which can be decomposed again to, to be mu naught times N times I times A where in this case, the A is still going to be the area, the eye is going to be the current. The end is going to be the number of turns and the mu not is going to be the permeability of free space. So also the mu not is essentially just going to be a constant of four pi times 10 to the part of minus seven and over a squared just like. So, okay, so now we can actually continue on this, so we can substitute this magnetic plus formula here into disparities law that we have decomposed. Kind of this will be equals to e multiplied by two pi air R equals D new, not N I A over DT just like. So okay. So now we know that the mu not the N N D A is pretty much all going to be constant, so we can pull it out of the derivative and leave the I N D I over DT is going to be the one that is being asked, which is the current change rate. So this is going to be E two pi R equals mu naught and multiplied by a multiplied by D I over D T just like so okay. So now we can actually rearrange this so that we can actually find what the eye over DT is. So rearranging this, the eye over D T, essentially, it's just going to be multiplied by two pi R divided by mu not. And so now we want to identify what the area is. So for a solenoid recall that the area is just going to actually be pi R square. So I'm just gonna write that down. Um Unite and uh bye R squared. Okay. So remember that earlier, we identified that we have two different ours. So we know that the, our for this first one here is going to be measured from the radio distance where the electric field, the magnitude of the electric field induced is measured on. So this R is going to be the artist. Well, the area is still going to just be the area of the solenoid. So it's going to be the radius of the solenoid. So this is just our sol okay, just like. So, so now that we figured that out, we can actually just start plugging everything in to find our D I over DT. So the E we know it's going to be six times 10 to the power of minus six fall per meter just like so and now we have pi and then we have the artist which is 4.5 times 10 to the power of minus two m. And then all of that is going to be over four pi times 10 to the power of minus seven and over A squared, which is the Um you're not. And then we have the end value which is m report of -1. And then we also have lastly are sol which is 0. m or essentially I'm just gonna write it down for consistency purposes, this is going to just be 1.3 times 10 to the part of minus two m square just like. So, so after solving all of this, we know that our D I over D T is then going to be five point oh eight M per second, Which is going to be the current change rate of 5.08 and per second, which will actually Chris Fund to option A that we have here. So option A is going to be the answer to this breakfast problem. So that will be all for this breakfast problem. If you just have any sort of question on anything, please make sure to check out our other lesson videos on similar topics and that will be all for this particular problem. Thank you.

A long, thin solenoid has 400 turns per meter and radius 1.10 cm. The current in the solenoid is increasing at a uniform rate di/dt. The induced electric field at a point near the center of the solenoid and 3.50 cm from its axis is 8.00*10^-6 V/m. Calculate di/dt.

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Key Concepts

Induced Electric Field

Solenoid

Faraday's Law of Induction