A 10 A current is charging a 1.0-cm-diameter parallel-plate capac...

Question

A 10 A current is charging a 1.0-cm-diameter parallel-plate capacitor.b. What is the magnetic field strength at a point 2.0 mm radially from the center of the capacitor?

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with the Empire Maxwell law. In this problem, we have a hybrid car that's equipped with a high voltage battery that supplies a current of 50 amps to a 1.5 centimeter diameter parallel plate capacitor. If a sensor is placed three millimeters radially from the center of the capacitor, what would it register as the magnetic field strength? We're given four possible choices as our answers. Choice A is 1.3 multiplied by 10 to the negative four teslas. Choice B is 2.2 multiplied by 10 to the negative four teslas. Choice C is 3.3 multiplied by 10 to the negative four teslas. And choice D is 5.3 multiplied by 10 to the negative four teslas. We're gonna start off by drawing a diagram of what we're given the problem. So I'm gonna draw my two parallel plates, there's circular in nature since we were given a dia diameter and we have the current coming in, we'll have each plate connected to a wire. And why of the current coming in from the left side and it will be exiting the right side of the parallel plate. Por I'm also going to draw a box in the middle that's off the center line between the two plates and that's gonna be our sensor. So that's gonna be three millimeters off of the center line connecting the two plates. Now, I'm going to um, be charging this plate since I have a current coming in, it's going to put charge on that plate. And so it's going to generate an electric field between the two plates that's going horizontally one plate to the next. And that electric field is going to be changing because the charge, the amount of charge on the plate is gonna be changing. And so the strength of the electric field is also going to be changing. That change in electric field is actually going to create magnetic loops in between the two capacitors. So go and create magnetic field loops. This will be labeled as B and that's what the sensor is going to pick up. And that's ultimately what the question is asking us to find. Since we have both magnetic and electric fields and electric fields changing, we need to use the um Empire Maxwell Law. So we call the Empire Maxwell Law that is going to be the line integral of E dot DL. And that's gonna be equal to you not multiplying the quantity of I plus epsilon not EDT of by and that is being enclosed parts. So we have the mu knot, which is constant, multiplying the enclosed current plus epsilon knot, multiplying the time derivative of the electric block. And that is the electric block that's enclosed by our amerian loop. Let's look at our imper loop, we've actually drawn the magnetic field. And so we'll just have a loop that follows that magnetic field line. That means that the magnetic field is going to be parallel to the path that we take to RDL. And it also means that the magnetic field strength is going to remain constant. So since the magnetic field and path are in the same direction that gets rid of the dot product magnetic field strength being constant, I can pull it outside the interval. So I'll just have B and the integral of DL is just length of that loop, which is the circumference of the circle to pi R. So our here is gonna be the radius of our loop that's gonna be measured by the sensor. Now that has to equal a Muno. And if I look at my diagram, I don't actually have any current flowing between the two plates, there's no charge is actually leaving the plate on the left and moving to the plate on the right. It's only generating an electric field. That means that the current that's been enclosed by our loop is actually zero. But I still have the second term. So I have the new knot multiplied by the epsilon knot. And then we'll have the DDT. And here I need to recall our definition for the electric flux. Recall that our electric flux ye is equal to the integral of E dot D A. So here the electric field is going to the right, the area vector is also going to be going right because it's perpendicular to the area that's closed. So that gets rid of the dot product, the electric field strength is constant 22 parallel plate capacitors. So that could be pulled outside the integral and just left with the interval of D A which is just gonna be the area. That means that my by E gonna be equal to E multiplied by A. Now, the area here is the area that is being enclosed by the parian loop. So in this case, pie is going to be equal to e multiplied by the area that imperial loop, which is just I are squared. So I can plug that into our formula for the Phi. And so I'll have the DDT of E multiplying Ir squared. So now I need to actually write the electric field in terms of the quantities that we are given. So I'll rewrite this equation that I have. So I'll have B multiplied by two pi R that's equal to, you're not, absolutely not, the PI R squared is a constant. So I can pull it outside the derivative, I'll have PI R squared, then I'll have the DDT and for the electric field, we need to recall our formula for the electric field of a parallel plate capacitor. So in that case, E is going to be equal to the charge on plate Q by, by epsilon knot multiplied by the area of the plate. So the area in this case is the area of the plate. So we'll have to write that formula out. We'll have E is equal to Q by by not. And the area here is that of a circle. It's gonna be pi R squared, but we weren't given the radius, we were given the diameter. So I have pi multiplying D divided by two quantity squared. There D is the diameter of that we were given. So I can plug that into our formula for the Maxwell Empire Law. And so I have the DDT of you divided by epsilon knot, I by the quantity of D over two quantity squared. So I can actually simplify the right hand side there a bit. So I have B multiplied by two pi R equal to Muno, I have an epsilon knot in the numerator and an epsilon knot in the dom denominator inside the derivative, but it's constant. So it can move through the derivative. So that will cancel out same thing happens with the pies. And so what I'm left with is R squared divided by, I can pull out the D divided by two quantity squared since it's a constant can be pulled outside the derivative, I'll have D divided by two quantity squared and then I'll just have the time derivative of the charge DQDT. So at this point I need to solve for the magnetic field B I have B equal to you not divided by two pi the R that I would divide over cancels out with one of the RS. So I'm left with just R divided by D divided by two widely squared. And the QDT is by definition, our current that is just equal to I. So I have values for everything on the right hand side of the equation. So we'll go ahead and plug in values. They will have B equal to you. Not that is four pi multiplied by 10 to the negative seven tesla meters or amp that gets divided by P. The R here is going to be the three millimeters. We need to convert that into meters. So that'll be three multiplied by 10 to the negative three eaters. We'll divide that by the D over two. So RD in this case, the diameter was 1.5 centimeters. We'll convert that 2 m by dividing by 100. That'll be 0.015 eaters. The diameter gets divided by two, that fraction gets squared. And then I need to more play that by the current which is 50 amps. So everything here on the right hand side is just number and plug it into my calculator. That means my magnetic field is equal to 5.3 multiplied by 10 to the negative four teslas. And that corresponds to answer B. So just a quick recap, we started off by using the Empire Maxwell Law equation. But in order to use that, we've had to also recall our definition for electric flux and the electric bill due to a parallel flight capacitor. Once we were able to combine all those, we were, we were able to uh calculate our magnetic field inside the capacity. So I hope that this has been useful and I'll see you in the next video.

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A 10 A current is charging a 1.0-cm-diameter parallel-plate capacitor.b. What is the magnetic field strength at a point 2.0 mm radially from the center of the capacitor?

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