💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

# A 0.650-m-long metal bar is pulled to the right at a steady 5.0 m/s perpendicular to a uniform, 0.750 T magnetic field. The bar rides on parallel metal rails connected through a 25.0-$\Omega$ resistor ($\textbf{Fig. E29.30}$), so the apparatus makes a complete circuit. Ignore the resistance of the bar and the rails. (a) Calculate the magnitude of the emf induced in the circuit. (b) Find the direction of the current induced in the circuit by using (i) the magnetic force on the charges in the moving bar; (ii) Faraday's law; (iii) Lenz's law. (c) Calculate the current through the resistor.

## a. $2.438 \mathrm{V}$, $2.4 \mathrm{V}$b. (i) $\overrightarrow{\boldsymbol{F}}=q \overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}}$(ii) $\varepsilon<0$(iii) counterclockwise sensec. $0.09752 \mathrm{A}$, $98 \mathrm{mA}$

#### Topics

Electromagnetic Induction

Inductance

### Discussion

You must be signed in to discuss.
##### Andy C.

University of Michigan - Ann Arbor

##### Farnaz M.

Simon Fraser University

##### Aspen F.

University of Sheffield

Lectures

Join Bootcamp

### Video Transcript

for the first item off. This question we use for it is law which states that induces electro motive Force is given by minus that derivative off the flux of magnetic field with respect to time. The area that we have to use them for these law is this area is the area off the circuit, and that area is changing because this part of the circuit is moving to the right. Then we can calculate the flux as follows. The flux off magnetic field is given by the integral off D A, which is an area element times the magnetic field. Be now notice the following the orientation off that surface A is given by your vector that is normal to the surface. A factor that is normal to the surface is a factor that is pointing either our ports or in words. We can choose the orientation of that factor. It is conventional to choose, eat as pointing outwards. So the vector e should be something like this. Then I'm going with this convention. The reform. You can see that A and B are untie parallel so we can write it as minus the integral off D A times be where now we are working with magnitudes The magnitude of the magnetic field is constant So we have miners b times the integral off the A which is just the area. So the flux is minus B times a The area off that rectangle is given by the product off its size. So the flux is my news. Be times l which is these side times is more l which is disorder site Now we can take the derivative respect Time to get the following Mine has been times l They are constant times the l divided by the teeth The LTTE is the reach off change off l which is the same as the velocity off this bar search results in minus b times l a times we there afford induce it e n f is minus minus beat LV which results in B times L A times. We therefore plugging the numbers we get 0.75 times 0.65 times five, which results in an e m F off 2.4375 which can be rounded to 2.4 votes. And this is the answer to the first item before proceeding. Let me clear my board. Okay. The second item has three super items in all of them. We have to do to remind what is the direction of the current, but using different methods, The first item would have to use the magnetic force that is acting on the current. So you let us suppose that there is a charge seating right here? Discharge is a charge Q. Then the magnetic force acting on these charge is given by cube times V cross beat. We can see that in the situation V is pointing to the right while B is pointing inside the board like this. Then using the right head who we conclude that the magnetic force we will be pushing the charge upwards. So the magnetic force pushes the charge in this direction, which means that they induce a current is a counterclockwise current like this. Now, in the second soup item, we have to use foreign Desslok Well, you noticed that the flux is negative and it's negative and decreasing the reform. We have the derivative off the flux off magnetic field beings. Moledet Zero before to compensate for that we need an e M F that is bigger than zero. It means that, according to the orientation that we had chosen that the electric current should run counter clockwise. So this is how you use for a these lot to solve these questions. Now for the last item, we have to use lenses Low Liz Low is very similar to far these law, but instead of treating the E. M F, it treats the magnetic field directly. So you you have to think about the same things. Something has to happen in order for the flex. Two becomes motor. Two things can happen. They are quite equivalent. From a general point of view, these two things are there. Should there must be a positive IMF more. On the other hand, some magnetic field should be induced in the opposite direction off the doctor on magnetic field in order to lower the floods off magnetic field lines. In order for this to happen, these inducing magnetic few to shoot points to an opposite direction with respect to the background magnetic field. So we have a background magnetic field pointing in wards, the reform, these reducing magnetic future point outwards and the magnetic field pointing our parts is induce it by a current that goes in this direction in that circuit so that outwards pointing magnetic field is endings that by a counterclockwise current, and this is how you can use lenses low to justify the answer for the second item. Finally, we have to calculate the current from the resistor. For that we can use orders. Law homes Low tells us that the voltage difference is equals to the resistance off that resistor times the current. So the current is the voltage divided by the resistance on these results. In chu 0.4375 divided by 25 noticed that I'm using the voters. We wrote approximations to get a more accurate results and then the results in a current that is zero 0.975 and this could be rounded to 0.0 98 which is 98. Milli amperes uses the answer to the last item

Brazilian Center for Research in Physics

#### Topics

Electromagnetic Induction

Inductance

##### Andy C.

University of Michigan - Ann Arbor

LB
##### Farnaz M.

Simon Fraser University

##### Aspen F.

University of Sheffield

Lectures

Join Bootcamp