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Problem 66 Hard Difficulty

An $N$ -turn square coil with side $\ell$ and resistance $R$ is pulled to the right at constant speed $v$ in the positive $x$ -direction in the presence of a uniform magnetic field $B$ acting perpendicular to the coil, as shown in Figure $\mathrm{P} 20.66$ . At $t=0$ , the right side of the coil is at the edge of the field. After a time $t$ has elapsed, the entire coil is in the region where $B=0 .$ In terms of the quantities $N, B, \ell, v,$ and $R,$ find symbolic expressions for $(a)$ the magnitude of the induced emf in the loop during the time interval $t,(\mathrm{b})$ the magnitude of the induced current in the coil, (c) the power delivered to the coil, and (d) the force required to remove the coil from the field. (e) What is the direction of the induced current in the loop? (f) What is the direction of the magnetic force on the loop while it is being pulled out of the field?

Answer

a. N B l v
b. \frac{N B l v}{R}
c. \frac{N^{2} B^{2} l^{2} v^{2}}{R}
d. \frac{N^{2} B^{2} l^{2} v}{R}
e. \text { clockwise }
f. \text { left }

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Video Transcript

for part a of our question were asked to find a symbolic expression for the magnitude of the induced E M f in the loop during the time interval T. Okay, so to do this, um, we're gonna use Faraday's law of induction, where the induced E m. F is defined as the rate of change of magnetic flux and closed in the circuit. So for a coil with in number of turns, the induced Ian meth is gonna be expressed as follows. Absalon is equal to minus the number of turns in the coil times the changing magnetic flux belt. If I over the change in time Delta T Okay, well, the magnetic flux fi is equal to the magnetic field be multiplied by the area. But the area here it's a square with the with sides of length. L. So it's gonna be magnetic field times, Elk Square and where the velocity here will call it V, of course, is just distance divided by time distance. Here it's traveled is l divided by Time T so we can go ahead and solve for tea because we want tea and our expression of the induced e meth where t is equal to l divided by velocity V. Therefore, the induced E M F absalon is equal to minus the number of turns in the coil multiplied by the magnetic field times l squared. And the reason here that it's not dealt ified is because we're told that the final statement initial magnetic field zero um, initial time is zero. So those values just go away. So we're here. We're left with the expression for Phi, which is be times l squared, divided by the expression for tea, which is l overby. So the l on the bottom cancels the squared l on top. So we end up with, ah, the number of turns, the magnetic field times be and this is Ah, there's actually a minus here, So there's minus signs will cancel eso number turns the moment Magnetic field Times B, which is the magnetic field lines. Make that expression look a little nicer here for B so you can read it. Here we go. And again there's a minus. Sign here, so let's rewrite that. OK, so time's a magnetic field times the length times of loss TV. We can go ahead and box that in as our answer for part? A. So then, for part B, we are asked to find an expression for the magnitude of the induced current in the coil. Okay, let's get a new page going. So this is part B the magnitude of the induced currents. So the current here I using owns law is equal to the induced E m f divided by the resistance. Well, we just found the induced ium f The inducing meth is the number of turns in the coil times the magnetic field times the length l times, the velocity V. And then, of course, this is just divided by our and that there is our expression. So it was simple enough we can box sudden is our solution for part B now part see, for part C were asked to find an expression for the power power here is equal to the current squared times the resistance So playing in the value for the current that we just found in part B and squaring it, we're left with number of turns in the coil times, the magnetic field times the length, times the velocity. All of that is squared. And then the R squared on the bottom. One of those about ours is gonna cancel with e r. That we're multiplying the power bi. So we're left with Just are on the bottom when we come box it in as their solution for part c. Okay, start a new page here, and this is part D for part d were asked thio to find an expression for, uh, let me see for the force required to remove the coil from the field. Okay, so we can do that by using the expression for work which says that it work is equal to the force times a displacement amount s so solving for force, we find that force is equal to work divided by displacement s where work can be expressed in terms of power. Since power is equal to work divided by time, we find that work is equal to power times time. So therefore, our force is equal to power. Times time divided by displacement s. Okay, well, velocity is distance divided by time, so time divided by distance is equal to one over the velocity. So this is equal to power divided by velocity. So if we go back to this page, we found our expression for power. Let's plug in the expression for power from parts. See, into this, we find that this is equal to course. One of those velocity components is gonna cancel. So now we have a number of turns in the coil times the magnetic field times the distance. L All those values air still squared from the power term. One of the velocities is gonna cancel, so that velocity is no longer squared, so we can pull it out of that. And then this is still divided by resistance are that's our expression for the force that could be boxed in as their solution for part D. Okay, now for part e another page going for part E were asked, What is the direction of the induced current in the loop? Okay, so here the magnetic field is in the clockwise direction and decreasing as the loop is moving out of the field. Therefore, the direction should be in a clockwise direction, opposing the inward flux. Right? Because the direction is always to impose, oppose the flux so we can say clockwise to oppose the inward flux. We'll box. I didn't is our solution for e and then for the very last part, part f were asked what is the direction of the magnetic force on the loop while it is being pulled out of the field? Okay, so as the coil is going towards the right, the forces acting towards the right hence the magnetic field is going to act towards the left to oppose the motion of the coil because the oil is going towards the right. So I think the best way to answer this would be, say to say, the magnetic field, or I like to write it as the field acts towards the left to oppose the coil, the coils motion weaken box that in as a solution for part F and that's the last part of our problem.

University of Kansas
Top Physics 102 Electricity and Magnetism Educators
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