The magnetic field B at all points within the colored circle shown in Fig. E29.15 has an initial magnitude of 0.750 T.

(The circle could represent approximately the space inside a long, thin solenoid.) The magnetic field is directed into the plane of the diagram and is decreasing at the rate of -0.0350 T/s. (d) What is the emf between points a and b on the ring?

Video transcript

Hi, everyone. For this particular practice problem, we are asked to actually determine the value of induced E M F between points P and Q on the loop using the shortest line between the two points where we will have a circular loop with a diameter of centimeters inserted into a magnetic field directed into the page. And initially, the magnetic field will have a magnitude of 0.85 tesla. And the rate of decrease of the magnetic field is minus 0. tesla per seconds. And we were asked to actually determine the value of the induced E M F between points B and Q. So first, I'm gonna start us off with um writing down all the necessary information. So first, we have the diameter which is D which is 40 centimeters. But we want to use or convert this into the radius which is 20 centimeters which will equals in si to be 0.2 m. And it will have the magnetic field which is B of 0.85 tesla. And we will have the rate of decrease of the magnetic field which is DB over DT at a rate of time which is -0.04th per seconds. So next, what we wanna do is to use the sign convention for the law to get direction off the induced electric field. E, so we want to use our right hand rule where we will point our thumb to be uh for the magnetic field directed into the page and we will curl our fingers or the rest of our fingers to be the direction of the induced electric field which is going to be clockwise. So E is going to be clockwise just like. So, so we wanna then recall Fire Days law. So Fire Days Law using Fire Days Law, we can actually try to solve this problem and find the value for the indued E M F. So, uh according to far law, the integral of E multiplied by D L uh will equals to minus D del uh D of the magnetic flux or five B over D P which is the rate of change of the magnetic flux over time. So I'm going to actually break this down into the left side and the right side. So first, I'm gonna start us off with the left side which is the integral of E D L uh For this particular um problem, the E D L is going to be E multiplied by two pi er two pi R which is R is the radius of the circular loop. So the two pi R is actually the circumference of the loop itself. And then for the right side, we have minus D magnetic flux B over D T. And you wanna recall that magnetic flux can actually be calculated by multiplying B and A which is the magnetic field multiplied by the circum uh the area. And the A for this loop is going to be just by R squared. So we wanna substitute this formula of P five B into our derivative here. So this will be D multiplied by B pi R squared over D T and the pi R squared is going to be constant. So we can pull it out and this will then be minus uh pi R squared multiplied by DB over D T just like. So OK, so now that we break uh the two sides down, we can actually combine them together. So combining the two together, we can um I'm just gonna write down the mastered initial equation here just like. So, and then the left side is E multiplied by two pi R. The right side will be minus pi R squared multiplied by DB over D T. And we actually have all this information except for the electric field. So we can actually rearrange this so that we can find what the electric field is. I'm gonna cross out the pi and the R from both sides. And then the electric field can be calculated by this formula right here. Minus half, R multiplied by DB over D T just like. So, so this will be minus half multiplied by R which is 0.2 m given in the problem statement. And the DB over D T is minus 0.4 that per seconds just like. So and then from here, the minus sign will cancel out and then the elected field will be four times 10 to the power of minus three fold per meter. So that will be the electric field. And now what we wanna do next is to actually we wanna actually determine what the actual induced E M F is from a point of B to Q between the two points. So what we wanna know now as we can see that points B and Q are actually separated by an angle of 115 degrees from the figure uh given in the problem statement. And I'm just gonna write that down as theta here, which is 1 15 degrees. And we want to find what that ratio is to the ratio of the whole circle because that is essentially uh uh that is essentially what we need to calculate the in U C M F. So actually the ratio, I'm just gonna write that ratio is going to be theta over the overall um circle which is going to be 1 15 degrees over 3 60 degrees, which is going to be 0.32. So the points of P and Q are actually separated by a distance which is equal to 0.32 times the circumference or the whole C circumference of the loop. So because of this, the induced E M F can be calculated E M F can be calculated by E will equal to the overall um electric field multiplied by the ratio Multiplied by the circumference which is two by R. So this is essentially just normal induce E M F formula which comes from just the E multiplied by our epsilon equals E multiplied by uh the circumference which is essentially just the integral of E multiplied by D L just like. So the ratio here is to take an account of the distance or the location of the P and the Q. So now we can actually plug everything in the electric field is four times to the power of minus three fold per meter, which is uh determined previously, the ratio is 0.32 the two pi R is two pi multiplied by 0.2 m. And this will give us an induced EMF value of 1.6, 1 times 10 to the power of -3 fault. So 1.61 times 10 to the power of minus three fold is going to be the answer to this problem which is going to correspond to option D. So option D with an induced E M F between um the points B and Q of 1.61 times 10 to the power of minus three fold is going to be the answer to this practice problem. And if you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and that'll be all for this video. Thank you.