A metal ring 4.50 cm in diameter is placed between the north and ...

Question

A metal ring 4.50 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.250 T/s. (a) What is the magnitude of the electric field induced in the ring?

Accepted Answer

Hi everyone today, we are going to calculate the magnitude of the induced electric field in the conductor where we have a circular conductor with a radius of 2. centimeters placed in a uniform magnetic field of 1.23 Tesla decreasing gradually at a rate of 0.45 Tesla per second. So we are going to just start with identifying all the known information here. We have the radius in centimeter which is going to equal to 2. times 10 to the power of minus two m. And we have the magnetic field itself which is 1.23 Tesla. And last we have the actual decreasing rate of the magnetic field itself. And I'm going to represent that with the B over DT which is basically just a change in magnetic field over time, it is going to be minus 0.45 Tesla per second just because it is decreasing. Okay, so now that we've identified all our necessary information, we want to recall that a changing magnetic flux to a coil will actually induce E. M. F. In that coil, which means that an electric field is actually induced in the material of the coil itself. And according to Faraday's law, the induced electric field will actually go base this particular equation here. So, I'm just gonna write it down the according to Faraday's Law to induce electric field, E multiplied by D. L. Which is the length will equals to minus d of the change in the flux magnetic flux over the over time. Okay, so for the magnitude of the induced electric field using this formula at a circular conductor, the Faraday's Law will actually give us, I am just going to write it down real quick. So the integral of E. D. L. Equals minus D. Five B. Which is the flux over DT. And this solving for this for a circular conductor, this will be E multiplied by the length. For circular conductor is going to be the perimeter of a circle. So it is going to be two pi R. Which will equals two minus D. Recall that the magnetic flux is going to be calculated using B multiplied by A. Which will equals to be multiplied by pi R. Square. So this is going to be B. This is going to be the magnetic flux here. This is going to be multiplied by pi R. Square. Just like so over DT. Okay, so we can actually kind of manipulate this. Now recall that the pi R squared or the dimension of the conductor is going to be constant. So we can put it outside of our derivative here. So this will then be E. Multiplied by two pi R equals minus minus um pi R squared, multiplied by D. B. Over D. T. And then we can kind of cross out the pie and one of the are here And from there we can rearrange to find what E. S. So E is then going to be minus R. Over two, multiplied by D. B. Over D. T. Right okay, now that we have this, we can actually start plugging in all the information that we have here into this formula right here. So we know that the R. Is going to be 2.5 times 10 to the power of minus two m. I'm just gonna actually put it down here to make it easier for you guys to see. So we know that the R. Is going to be 2.5 times 10 to the power of - m. And then we have over to here and then recall that the D. B over D. T. Is minus 0.45. So this is going to be minus 0.45. Tesla per second. This is going to be e. And this will actually give us an e value of 5.6, 2 times 10 to the power of minus tree, full per meter just like so okay, so this will be the answer to this particular problem, which is 5.62 times 10 to the power of minus three, which is going to correspond to option B right here. So option B is going to be the answer to this particular problem. Um, let me know if you guys have any confusion but make sure to check out our other lesson videos regarding this particular topic because there will be a lot of other videos similar to this that will help you guys to gain better knowledge and understanding of this example. That will be all for this problem and thank you.

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