5

A uniform, $458-mathrm{g}$ metal bar $75.0$ Figure P7.47 $mathrm{cm}$ long carries a current $I$ in a uniform, horizontal $1.25-mathrm{T}$ magnetic field as shown i...

Question

A uniform, $458-mathrm{g}$ metal bar $75.0$ Figure P7.47 $mathrm{cm}$ long carries a current $I$ in a uniform, horizontal $1.25-mathrm{T}$ magnetic field as shown in Fig. P7.47. The directions of $I$ and $overrightarrow{oldsymbol{B}}$ are shown in the figure. The bar is free to rotate about a frictionless hinge at point $b .$ The other end of the bar rests on a conducting support at point $a$ but is not attached there. The bar rests at an angle of $60.0^{circ}$ above the horizontal. What is the

A uniform, $458-mathrm{g}$ metal bar $75.0$ Figure P7.47 $mathrm{cm}$ long carries a current $I$ in a uniform, horizontal $1.25-mathrm{T}$ magnetic field as shown in Fig. P7.47. The directions of $I$ and $overrightarrow{oldsymbol{B}}$ are shown in the figure. The bar is free to rotate about a frictionless hinge at point $b .$ The other end of the bar rests on a conducting support at point $a$ but is not attached there. The bar rests at an angle of $60.0^{circ}$ above the horizontal. What is the largest value the current $I$ can have without breaking the electrical contact at $a$ ? (See Problem 7.77.)



Answers

A thin, 50.0-cm-long metal bar with mass 750 g rests on, but is not attached to, two metallic supports in a uniform 0.450-T magnetic field, as shown in $\textbf{Fig. E27.37}$. A battery and a 25.0-$\Omega$ resistor in series are connected to the supports. (a) What is the highest voltage the battery can have without breaking the circuit at the supports? (b) The battery voltage has the maximum value calculated in part (a). If the resistor suddenly gets partially short-circuited, decreasing its resistance to 2.0 $\Omega$, find the initial acceleration of the bar.

Okay, Gordon. Chapter 29 from 29. So, in this problem, uh, bar here on this is all in the field that is pointing into the pidge earned the calls for seeds for the and along this there is a bar and that is moving along at a velocity V in this direction. Okay, You are also tells us that links here is 50 centimeters. Can you do magnitude? B is zero point a Tesla. Okay, So says are a find the magnitude of the meth induced in the rod when it is moving towards the right with the speed of 7.5 meters per second to the eagle, 7.5 years. Second. So we want to use the equation here we've been using at the university, Miss and we know the field and tr 90 degrees apart. So since all the equation needed on, he's signed hard, so we know that this is going to give a 7.5 years. The second terms you're appointed, Tess are times Europe for five years. This is 3.0 votes room. So now, for part B says in what direction is a curve blowing rock. So we know that B is into page, and as it moves to the right, A gets bigger right? The area because this is our loo. Great years. Our group, the is getting bigger than be is constant, so that means our hooks B is into the page and it's increasing. So using Lindsay's law, the induced the must be out of the cage. Okay, so now, using the right hand rule does a counterclockwise or clockwise pointed out of the page. Do that and you'll find it's counterclockwise. So that means in the Rod. If it's a moving counterclockwise, it's moving from the there. Cool. So part see asks if the resistance of the circuit is 1.5 homes and it's constantly on the force required to keep the rod moving to life. The constants 73735 years per second. Okay, so we can find the current in this circuit now by taking the he England and fined by the resistance. They're not kooks have to do two points there amps. Okay, so we also know that to keep this maintain this constant speed force external must be equal in opposite the magnetic force that's exerted on the run Okay, so it's kind of a lot words. So base you say the external force equal and opposite this other force on Iran were called. Okay, So force on how we can calculate because, you know, this is just a magnetic force on a rod. So this is the current times the length of the Rod times making field stream time's sign of the angle between the current in the field, which in this case, is won. So this is two hams times 0.5 meters times your appointment a Tesla's. Then this comes out being zero point a Newton's. Okay, so that means the external force, the magnitude of the external force required to keep this moving is exactly very 0.8 Mittens could We wanted to be at a constant speed, so they went to zero. That force. Awesome. Then this must be in the direction of the right because l is in the direction of the left. The current is pushing, trying to make this smaller, but we're trying to keep it awesome. Gets enough apart, D says, compare the rate at which mechanical work is done by the force. So for mechanical work, this is F times the velocity. This is the rate of mechanical work and that's one of these. You go from eight noons tons 7.5 meters per second and it's come down to exactly 6.0. What's if you want to compare this to the rate at which thermal inversions developed in the circuit and that is given by ours, are I square are so that is given by two squared turns 1.5 Guess what? This is also 600 watts. So they're both the same, which is exactly what they need to be is required because of conservation of energy. So this whole problem is working out great.

This problem. We have a conducting rod sighting on rails in the uniform magnetic field. Argo is gonna be to find various things about this setup and on this depiction, we have our magnetic field that going into the page, it's values 0.8 Tesla. The route is moving to the right at 7.5 meters per second. The height of this from CD is 50 centimeters and I'm denoting and the length from D to be as l. It's the first first question is defined the induced e m f. So we know that the induced the IMF is gonna be given by our change in flux over a change in time. We can also write our flex as B times a where you don't have to deal with any co sign because the perpendicular of this square region of area and the magnetic field er in the same direction So that coastline is one R b feel this constant so we can pull that out, be changed area over changing time. Our area we see that that is our age Times or L so we can put in e engine edge l And here are height is not changing its only the length from D to be that's changing. So this is B age changing my over changing time. I'm not changing length over changing time is just our velocity that were given. So we have all these numbers and we can see what this value is. Darby is zero point a Tesla. Her height is 50 centimetres, but that in the meters are changing. My cover changing time is 7.5 meters per second. But if you plug all of this in, we get three votes. Now that we have our induced him. If the question is what is the direction of the induced current here, we're gonna need lenses loss. The lenses law tells us that the induced magnetic field establishes to mitigate the change in net magnetic flux. So we see here, since this bar is moving over to the right, our magnetic flux from this external field into the page is growing In order, Thio mitigate that change. To reduce that change, the induced Byfield from this little square region must be pointing out of the page. So I'm gonna draw that in red Red will be our abused field. The induced and using our right hand roll here re point our thumb in the direction of the induced the field in the direction that our fingers curl around that vector gives us the direction of the current. So we see that our current was going counterclockwise, counterclockwise induced. Okay, The last part is asking us to find the direction and magnitude of the force required to maintain this uniform motion. So we know that in the setting of a magnetic field, a rod feels a force given by I l B. And one thing I forgot to mention for this part is that we consider a resister in this circuit so resistance of destroying era to that consider a resistance now in the circuit of the 1.5 pumps. So our force on a rod is given by I l B. And we know from homes law that the current is given by our voltage over our resistance. Here are voltages are induced enough. We are in a over What is this thing's times l r. And here are l. The length of the Rod were denoting by age. So let me just correct that really quick. I h b Well, there, there. And now I can see what magnitude this gives us. We have all of these pieces. So we have three over 1.5 are high, is 50 and our magnetic field is point. And this gives us 0.8 news. Now, to think about the direction of this force, let me blow up our bar here. So we know that the current is going up. We know the magnetic field is going into the page, so if we consider a actually make this a little bit bigger. So if we consider a proton in here, this proton, if we use our right hand drawer for the magnetic force, we point our index finger along our velocity. And I'm just gonna you know, that right there would point our index finger along our velocity, the point, our middle finger into the direction of the magnetic field. And our thumb is pointing to the left. So the protons are feeling a force to the left and weaken. Do the same argument for the electrons. That will be the same because the velocity is reversed. But also the charges reversed. So this is telling us the broad is feeling a force from this magnetic field to the left, which means that we must supply force the right and it must be you're a 0.8 noons. If we have this force, it's 0.8 names. The Ron feels no net force, so it can continue at a constant velocity of 7.5 meters per second.

For the given problem, we were asked to find Ah, the terminal voltage across the battery here we had also for a DUI which is connected to the resistor. And this is connected me to hear, uh, with the road here sported on a wire and we have ah, man tick field that goes into the cage. So be it on the current in the role feels the lifting force when the current moves produce it with experience lifting force. And we can write that this will be f b physical to current times length times be and which is equal to the weight off the road so mg then from here we can write a current So current will be mg divided by lb subsidiary invaluable 0.750 is diplomacy The kg storms 9.8 meters per second squared Is there a 0.500 times 0.45 And this gives us a current all for 30 to 17 and peers from here. Then we can use the arms law to complete the terminal Walters across Petrie, which is musical toe I r eyes a 32.7 and Pierre times are 25 homes. This will give us a 17 potential difference of worlds

So in this problem, we are given a diagram with this contraption. We have a magnetic field that is going into the page. As noted by these Blue X is the magnetic field has a magnitude of 0.8 Tesla's, and we also have this red bar that is able to slide back and forth along this track of wire along the track of wire. We have a resistor on the left hand side with a resistance of 1.5 homes, and we assume that this bar is going to be moving to the right at a velocity of 7.5 meters per second. And we also assume that the bar is gonna be half a meter long. And so this first part of the problem here we are asked what the electro motive force is that is being induced by this moving rod. And so to calculate the MF, we're going, Teoh put Faraday's law to use. So Faraday's law states that we have an electro motive force that is induced. It is going to be induced by a change in magnetic flux. And so I will denote the changing magnetic flux as this Delta Phi and It's a changing magnetic flux over time. And so Faraday's law, we have negative end, and it's simply the number of loops. In this case, we have a singular loop, so end is going to be one and can be ignored. And then we have a derivative. We have the change in magnetic flux over the change in time. So let's simplify this. We have negative changing magnetic flux. Well, what is the change of magnetic flux? First of all, the magnetic flux is going to be the amount of magnetic field that is flowing through this area. So the magnetic field is going to stay the same. That will be be. And what is area or area is simply with times hyped. We know the height is l so we can write that in. And what is the width? Well, the width is going to be a function off velocity. And so if you multiply velocity by time, we will get a whip. So we carry the denominator over and you will see that the Times cancel out. And so now we have the e. M F. That is induced is gonna be equal to the magnetic field times the height of the bar times the velocity off the bar moving that will be equal to We know that B is play Tesla's. We know that l is going to be 0.5 meters. It's a really bad five there, and we know he is gonna be 7.5 meters per second. We multiply this out, we get the e. M. F is equal to three bolts. So I will write that here three bolts. And it is also worth noting that there is a native sign here. Ah, part to the problem will make use of the state of sign. Voltage really can't be negative. But this negative sign will help in part two. And so in part to the questions asking Which way is the current flowing around this loop? Well, the negative sign that is in Faraday's law is showing that a current will produce a magnetic field of its own that will counter act this current change in magnetic field magnetic flux. Excuse me. And so, as our bar is moving to the right, are areas increasing, which means the magnetic flux is increasing, so we need a magnetic field that counter acts this increase. And so we need a magnetic field that goes out to the page to counter act the amount of magnetic field that is going into the page. Well, how do we think of which way the magnetic field is going to flow? We put to use our right hand rule, and so we want the magnetic field to flow out of the page towards us. And so, if you were to put your thumb pointing upward such that it lines with the red bar and then curl your fingers towards you, you will see that the magnetic field will point out the page towards us if the current flows counterclockwise is the that negative and the problem is telling us that the current will be flowing counterclockwise in this scenario. So I will draw a black arrow showing up but currents flowing counterclockwise. Part three of the problem asks us how much force is needed to continually move this rod at a constant 7.5 meters per second. And so to do this, we will use what is known as work. So work is simply force times distance and in part three were wanting to solve for a force. And so I can rearrange this. Such that forces on the left hand side of the equation. And we have force equals work divided by distance. Well, what is work? Work is simply a measurement of energy. It's units are jewels. And in this problem, we know the energy cannot be created or destroyed. And so we think to ourselves. Well, how is energy entering this diagram? Well, it is entering when we put a force on the rod moving it to the right. Well, where is energy leaving the diagram? Energy is leaving the diagram through this resistor, which is dissipating energy in the form of heat. And so this work right here can either be the amount of energy we put into the rod or the amount of energy dissipated. I'm going to calculate the amount of energy dissipated in this case. And so simply amount of energy dissipated. Would be the power times time. I guess I should be adult a t change in time. Over distance. Well, what is Power? Power is simply voltage times current and carry over the delta T We already solved for distance. We know the distance is the velocity multiplied by the change in time and a look at this. We have our changes in time that cancel out and we're after the voltage times the current divided by the velocity. Well, we don't know what the current I is yet, so let's solve for that. We know that voltage equals current times resistance solving for current. We rearrange and we see that it's going to be voltage divided by resistance. And so we know horrible to just three volts divided by resistance, 1.5 homes and we get our current is gonna be two amps using this top equation here. If we were toe plug those numbers an unsolved We have three bolts multiplied by the current, which is to camps divided by the velocity, which is 7.5 meters per second. We get a force f of 0.8 Newtons, so I will write that in here. For part three, the answer will be 30.8 Newtons of force to move the bar at a constant velocity to the right. Now in part four of the problem. It is asking which rate of which which ranges greater, and I sort of alluded to this in part three. Already we were talking about energy conservation where energy cannot be created or destroyed. And so here we're talking about rates of energy entering and leaving the diagram. So by putting to use energy conservation, we can kind of assume that the rates will be the same. But to make this a bit more intuitive, let's calculate these rates to figure out if they are the same. So the rate of work on the rod is going to be work divided by the changing time that right there is a rate and that will be equal. Teoh. Really no work is force times distance over time and we knew really solved for distance, which is going to be velocity times that change in time, divided by the change in time. So those cancel out we're left with force times velocity, which is simply 0.8 new tens multiplied by the velocity, which we know is 7.5 meters per second and we have a rate of work of six jewels per second. Now let's go clear. The rate of energy dissipated. Uh, this will simply be power. That's that's units of jewels per second. So what is power power is voltage times current. I should have left current up here, but we already complete that. It is two amps. Voltage is gonna be three bolts, so two times three is simply going to be six. So we have six Jules per second of power and you see that these numbers are the same, meaning that we have six joules per second entering the diagram, six joules per second of energy entering. And we have six joules per second, leaving through the resistor in the form of heat. And so this will do it for the four parts in this.


Similar Solved Questions

5 answers
21-25 Use the form of the definition of the integral given in Theorem 4 t0 evaluate the integral_ 24. 6 (x r')dx 35-40 Evaluate the integral by interpreting it in terms Of areas_40.S1 2x Idx
21-25 Use the form of the definition of the integral given in Theorem 4 t0 evaluate the integral_ 24. 6 (x r')dx 35-40 Evaluate the integral by interpreting it in terms Of areas_ 40. S1 2x Idx...
5 answers
Find y' for y = (cotx)*
Find y' for y = (cotx)*...
5 answers
Point) Fcr ? = f(8) = gi2(8) - 1: (A) Find the Erea contained witbin {(8).(B) Finc tbe slope of #ne targent Iine # f(8) st & = 0,%5,
point) Fcr ? = f(8) = gi2(8) - 1: (A) Find the Erea contained witbin {(8). (B) Finc tbe slope of #ne targent Iine # f(8) st & = 0,%5,...
5 answers
Tot 2-pentanone Please show (structure the question pentanone: mechanism 9) that undergoes results McLafferty the fragments rearrangement miz-58 due (MR) to MR in 2
Tot 2-pentanone Please show (structure the question pentanone: mechanism 9) that undergoes results McLafferty the fragments rearrangement miz-58 due (MR) to MR in 2...
5 answers
Dal 88.2gDal2 88.2gMass empty flask Mass of flask and condensed vapors after cooling88.96g89.0g0.76g0.80gMass of condensed vapor98%98%Boiling water_temperature371.15371.15Boiling water_temperatureatmatmBarometric Pressure (atm)150 mL150 mLVolume of Flask (mL)0.1500.150Volume of Flask Molar mass of unknown mRT MMIsopropyl alcoholIsopropyl alcoholIdentity of Unknown% ErrorShow calculation of Molar Mass_ (Molecular Formula: C3H8O)Show calculation of percent error:
Dal 88.2g Dal2 88.2g Mass empty flask Mass of flask and condensed vapors after cooling 88.96g 89.0g 0.76g 0.80g Mass of condensed vapor 98% 98% Boiling water_temperature 371.15 371.15 Boiling water_temperature atm atm Barometric Pressure (atm) 150 mL 150 mL Volume of Flask (mL) 0.150 0.150 Volume of...
5 answers
Show the Root Test and the Ratio Test cannot be used to determine the convergence of (In()) 7 = where € is a constant
Show the Root Test and the Ratio Test cannot be used to determine the convergence of (In()) 7 = where € is a constant...
5 answers
Question 15 (1 point) At what point(s) (#1-4) in the figure is the partial pressure of oxygen the lowest?Question 16 (2 points) The blood circulation of crustaceans is(faster/slower) than that of fish primarilybecause of lower(resistance/pressure)
Question 15 (1 point) At what point(s) (#1-4) in the figure is the partial pressure of oxygen the lowest? Question 16 (2 points) The blood circulation of crustaceans is (faster/slower) than that of fish primarily because of lower (resistance/pressure)...
5 answers
Which one of the following has Lewis formula most similar to that of NO- ?02 NO NO"
Which one of the following has Lewis formula most similar to that of NO- ? 02 NO NO"...
5 answers
2 0 Prove that * +2y = nSa a oy3.13. 2=x2ind 4 u for the functions:9.44.13.cus3 Task 6. Investigate the following functions for an extremum:9.6.13_ 2 = e21+3y (8x2 6xy + 3y2);
2 0 Prove that * +2y = nSa a oy 3.13. 2=x 2 ind 4 u for the functions: 9.44.13. cus 3 Task 6. Investigate the following functions for an extremum: 9.6.13_ 2 = e21+3y (8x2 6xy + 3y2);...
5 answers
Choose the best answer to each of the following. Explain your reasoning with one or more complete sentences.The charge of an antiproton is (a) positive. (b) negative. (c) neutral
Choose the best answer to each of the following. Explain your reasoning with one or more complete sentences. The charge of an antiproton is (a) positive. (b) negative. (c) neutral...
5 answers
2428u[4BM 8118014 %p 184} JABY pagexapjje 2q SUOIIap 34? Plnoqs ajuajagip [Bquanod J8qM qanoIL "unltz 0 > &q Ppnoqs SUo1njaja 341 jo 4#Sua[ABM 3[8018 %p a1} a1OjJj4L (toje 018u1S B JO 3Z1S 31} Sau? O01z nnoqe) urltz 0 anoqB IJIOUep JO S2p-41ed Apnzs 01 YSLM DM adoDSOJI UOIp? ue 4 [squjod &]
2428u[4BM 8118014 %p 184} JABY pagexapjje 2q SUOIIap 34? Plnoqs ajuajagip [Bquanod J8qM qanoIL "unltz 0 > &q Ppnoqs SUo1njaja 341 jo 4#Sua[ABM 3[8018 %p a1} a1OjJj4L (toje 018u1S B JO 3Z1S 31} Sau? O01z nnoqe) urltz 0 anoqB IJIOUep JO S2p-41ed Apnzs 01 YSLM DM adoDSOJI UOIp? ue 4 [squjod...
5 answers
Fill in the missing reactants {rLagentsRejiction _Roric tlonRocbon=ReictlonWenchotRepctionRa,ic(ion
Fill in the missing reactants {rLagents Rejiction _ Roric tlon Rocbon= Reictlon Wenchot Repction Ra,ic(ion...
5 answers
Question 241ptsThe solution set of 2x + > <3 includesthe area above the line_ including that linethe area above the line_ excluding that linethe area below the line; including that linethe area below the line excluding that lineQuestion 251ptsThe solution set ofx +1 includesthe area above the line; including that linethe area above the line. excluding that linethe area below the line_ including that linethe area below the line; excluding that line
Question 24 1pts The solution set of 2x + > <3 includes the area above the line_ including that line the area above the line_ excluding that line the area below the line; including that line the area below the line excluding that line Question 25 1pts The solution set of x +1 includes the area...
5 answers
Find the missing number in the following frequency table where the sample size is 20.15Canttbe determined with given information
Find the missing number in the following frequency table where the sample size is 20. 15 Canttbe determined with given information...
5 answers
Prove the identity cos(Θ)/ 1-tan Θ≡ cos Θ+sin Θ
prove the identity cos(Θ)/ 1-tan Θ≡ cos Θ+sin Θ...
5 answers
12. (6') Find parametric equations for the line tangent to the curve of intersection of the surfaces atthe given point: Surfaces: xe+ Zy 2z = 4 Y = Point: (1, 1,Yz)
12. (6') Find parametric equations for the line tangent to the curve of intersection of the surfaces atthe given point: Surfaces: xe+ Zy 2z = 4 Y = Point: (1, 1,Yz)...
5 answers
Find the general solution t0 the differential equation;~4xy.
Find the general solution t0 the differential equation; ~4xy....

-- 0.020854--