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Voltage power dissipated Pz? 11) (42 points) What the magnetic field at the dot?strength and direction12) (12 points) An antiproton (same properties prolon except t...

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Voltage power dissipated Pz? 11) (42 points) What the magnetic field at the dot?strength and direction12) (12 points) An antiproton (same properties prolon except that 4 = -)is moving thc combined clectric and magnetic fields shown Whai afe the magnitude and direction of the antiproton' acceleralion al this instani? ma 1.673 10" kg;13) (15 points) _ metal rod forced nore with constant vclocity along paralle] metal rails connected Wth strip of metal = one ende magnetic ficld of magnitud

voltage power dissipated Pz? 11) (42 points) What the magnetic field at the dot? strength and direction 12) (12 points) An antiproton (same properties prolon except that 4 = -)is moving thc combined clectric and magnetic fields shown Whai afe the magnitude and direction of the antiproton' acceleralion al this instani? ma 1.673 10" kg; 13) (15 points) _ metal rod forced nore with constant vclocity along paralle] metal rails connected Wth strip of metal = one ende magnetic ficld of magnitude 0350 - points ofthe page: If the rails scparated by 25.0 cm; the specd ofthe rod is 0.550 mls and the loop his resislance of [8.0 0, whal - magnitude and direction of the current in the rod? 14) (15 points) In this problem you will use Ampere' Law find expression for the magnetic field inside toroid: Conccptually, describe the direction of the magnetic field Writc down Amper' law When selecting an Amperian loop, what relationship between the magnetic field and the loop are you looking for? Select an Ampcrian Loop: Solve for the magnetic field 15) (10 points) Two loops of wire are moving in the vicinity of a very long straight wire currying steady current shown Find the direction of the induced current in each loop points) Maxwell'$ cquations arc the four fundamcntal equations clectromagnetism and were listed 'grcatest cquations cler' in the October 200 ! issue of Physics World_ Although Maxwell s equations arctcnlre simplc, they daringly reorganizc our perccption of naturc, unifying clectricity endmagnclism and linking Ecometzy, topole? physics. They Arc csscntial- understanding thc suounding #ok Andr tirt field cquations; they nOt only showed scientists NeW Wa} of approaching Physis but also took thcm on thc first = Jcp - InwaTds unification of tbe fundamental fores of nature Write down the equation that has do with the value of the magnctic flux through cloxd surface:



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The Meissner effect. Compare this problem with Problem 65 in Chapter $26,$ on the force attracting a perfect dielectric into a strong electric field. A fundamental property of a Type I superconducting material is perfect diamagnetism, or demonstration of the Meissner effect, illustrated in Figure $30.35,$ and described as follows. The superconducting material has $\mathbf{B}=0$ everywhere inside it. If a sample of the material is placed into an externally produced magnetic field, or if it is cooled to become superconducting while it in a magnetic field, electric currents appear on the surface of the sample. The currents have precisely the strength and orientation required to make the total magnetic field zero throughout the interior of the sample. The following problem will help you to understand the magnetic force that can then act on the superconducting sample.
A vertical solenoid with a length of 120 $\mathrm{cm}$ and a diameter of 2.50 $\mathrm{cm}$ consists of 1400 turns of copper wire carrying a counterclockwise current of $2.00 \mathrm{A},$ as in Figure $\mathrm{P} 32.79 \mathrm{a}$ . (a) Find the magnetic field in the vacuum inside the solenoid. (b) Find the energy density of the magnetic field, and note that the units $J / \mathrm{m}^{3}$ of energy density are the same as the units $\mathrm{N} / \mathrm{m}^{2}$ of pressure. (c) Now a superconducting bar 2.20 $\mathrm{cm}$ in diameter is inserted partway into the solenoid. Its upper end is far outside the solenoid, where the magnetic field is negligible. The lower end of
the bar is deep inside the solenoid. Identify the direction required for the current on the curved surface of the bar, so that the total magnetic field is zero within the bar. The field created by the supercurrents is sketched in Figure P32.79b, and the total field is sketched in Figure P32.79c. (d) The field of the solenoid exerts a force on the current in the superconductor. Identify the direction of the force on the bar. (e) Calculate the magnitude of the force by multiplying the energy density of the solenoid field times the area of the bottom end of the superconducting bar.

For this problem. On the topic of inductions were given an introduction into the mess no effect. And we are told to consider a vertical solenoid that has a length of 120 cm and the diameter of 2.5 cm. It consists of copper wire with 1400 turns carrying a current in the counterclockwise direction of two amperes. As shown in the figure, we want to solve various, solve the various quantities of the solenoid. Firstly, we want to find the magnetic field in the vacuum inside the solenoid. Now we know the magnetic field B is equal to new note. Times in times I divided by L. And this is the magnetic constant unit, which is four pi times 10 to the minus seven Tesla meter. Her ampere multiplied by the number of turns In the wire, which is 1400 times the current of two and piers divided by the length 1.2 meters. So this gives us the magnetic field to be too 0.93 times 10 to the minus three kessler's. And the right hand rule gives us the direction which is upward. So that's the magnetic field inside the solenoid. For part B of the problem, we want to find the energy density of the magnetic field. Now the energy density is given by you and we know you is the magnetic field strength squared, which is B squared over two times more. Not so now that we have the magnetic field, we can calculate this energy density. This is too 0.93 times 10 to the -3. Tesla has calculated above squared, divided by two times The magnetic constant four pi times 10 to the -7 Tesla meter. For NPR which gives us the energy density to be three 0.42 jules a cubic meter Which is equivalent to three 0.42 paschal's in part C. Of this problem, we're told that a superconducting bar that has an amateur of 2.2 centimeters is inserted part way into the cell, annoyed it's up end is far outside the solenoid where the magnetic field is small and the lower end is deep inside the solenoid. We want to find the direction required for the current on the curve surface of the bar so that the total magnetic field is zero within the bar. Now the field is sketched in the figure. Now to produce a downward magnetic field, the surface of the superconductor must carry a clockwise current. And so by the right hand rule we have a clockwise current giving us a downward magnetic field. For party, we're told that the field of the solenoid exerts a force on the current in the superconductor. And we want to find the direction of the force on the bar. Now the vertical component of the field of the solenoid exerts an inward force on the superconductor. The total horizontal forces zero over the top end of the solenoid. Its field diverges and has readily outward horizontal component. This component exerts upward force on the clockwise superconductor current and the total force and the core is therefore upward. And it can be thought of as a force of repulsion between the solenoid with it's not end pointing up and the core with its north end pointing down. And lastly for part of the problem, we want to calculate the magnitude of this force by multiplying the indeed density of the solenoid by the area of the bottom end of the bar. So this force is equal to. We can think of the indie density as a pressure and we can multiply that by an area. So this is three point for to paschal's, like we calculated above times the cross sectional area of the bar, which is pi R. Let's pie into 1.1 Times 10 to the minus two m squared. This gives us the force exerted on the bar to be one 0.3 Times 10 to the -3 newton's.

Okay, certain Chapter 27 Problem 47 says the long problem here. And it says a whole probe used to measure magnetic field strength consists of a rectangular slab material with free electron density and within thickness, tea and carrying current I along its lots of variables here. And it says it's immersed in a magnetic fueled be that is perpendicular to the rectory rectangular face of the whole probe and so that the whole E N f. You know each year is produced across is with you. The probe's meg meant magnetic sensitive sensitivity is defined by K H given by e m f of the whole probe over I be This indicates the magnitude of hall the whole IMF achieves were given applied magnetic field and current. So a slab with a large cage is a good candidate, for they hope Rube. In part, A says show that K H equals one over e and T. Thus, a Goldhawk has small values for both and okay, so if we look back at Equation 27 14 an equation 25 13 weaken right current in terms of drift velocity from this one. So that means our current here we can write in terms of velocity as e in times the distance it goes across, which would be tee times d times d d for the drift velocity. Now we can also write from 27 14 that the death of the whole probe is given by the drift velocity times B times that d which is the width of this law. Okay, so let's plug both of these equations and to this equation for the sensitivity and see that K H becomes the D e d over e and T B B B B lots of things VD, VD is all appear and it's all down here. This is just one over eat and deep. Is that what we want to show? Semporna? Okay, so what is part B? Ask heart Be now asks as possible candidates for the material used in a hall probe. Consider a R. I. A typical matter medal just in a approximately one times 10 to the 29 per cubic meter and another type of material a doped semiconductor in three times 10 of the 22 per square meter. So it says that for the first case, we can get the sickness t me of 0.15 millimeter. How thin should a metal slab be to yield a K H value equal to that of a semiconductor slab? Compare this metal slab thickness with the 0.3 Ganymede size of a typical metal. Adam. Okay, so we want to set the magnetic sense inherited use for each of these cases come equal to each other. Right? So we do K h for the metal equaling K H for the silicone. Then we get an expression here of one of he in metal. T meddle equals one over E in silicone pieces. Of course, the Z's cancel out and we can solve now for the thickness of the metal and that's given by N S Over and and t us. So let's plug these values in. They give us these ends. So we're three years of the 22 and one each of the 29 and the thickness we use 0.15 millimeters. So if we plug all that in, we get a thickness of the metal slab to get the same sensitivity for our whole probe must be five times 10 to the negative 11 meters which is less than 16 to tie its eyes. The typical metal. Adam. So this obviously doesn't make sense to use metal for a whole probe, as we need it to be very high. Sensitivities. That's why we'll use Don't silicone instead. Okay, so let's move on the part. See here, heart. See? Now asks for the typical semiconductor slab described in part B. What is the expected value for when I equals 100? No, IMS be a 0.1. Test us. Okay. So weakness. Rearrange our equation for sensitivity. And see that the Yuma of the whole probe is given by this sensitivity times I times be okay so we can just plug an equation for sensitivity. I mean, I be over e t. So her e h now is ah 100 No. AM's times zero point one. Tesla. Just realize that should be about male. Right? Um, this is divided by e negative. 19 cools are in which we used three times 10 of the 22 inverse meters cute and our thickness, which was 0.15 millimeters. Awesome. So double truck. We're all in s i u nits everywhere, and we can then get our answer out. And it is roughly 14 levels. Oh,

Scag Matic diagram for the masses, Spectrograph having charge Q equals to 1.602 Moncler by 10. To the power minus 19 column of gold and molybdenum. Okay. And electrical E. Of magnitude 1.789 10 to the powerful world per meter is pointing in inside the velocities, pointing towards the top of the face. And magnetic field of the magnet or B is equal to one point triple zero Tesla. And pointing out of the face as further diagram. So for the part of the question we have to draw the electric force and magnetic force vector acting on the iron inside the velocity selector. So this is our iron which is moving in this direction with the speed we note as for the question and the magnetic field is pointing in the direction that is outward direction. And electrical vector is pointing in the port direction that is towards the top of the pace. So this is the electric field direction. Okay, so as the charge is positive. Okay, so electric force vector will be in this direction. Okay, Since the this is velocity vector, so the magnetic force vector will come out in the opposing direction. So this will be the F. B. Vector. Or we can determine this direction of magnetic force from the right and root and taking by charge is positive. Okay, So this will be the direction for the electric force and magnetic force. Okay, now moving to the party of the question in which we have to calculate the velocity we not of the iron that make it through velocity selector. So from the theory we know that we equals two E by B. So equals two 1.789 Multiplied by 10 to the power four world per meter. And magnetic field is equal to one point triple zero Tesla. So from here, Stevie comes out to be 1.789 But year by 10 to the power four m per second. Okay, so this is the required the speed of the iron to make it velocity selector. So now moving to the party in which we have to calculate, we have to write the equation for the radius of the semicircular path. So radius R. Is given by M. We not divide by Q. B. Two. Okay, so magnetic field will be here used as B. Two. Okay because in the magnetic field B. Two the semi circular parties taking place. Okay. And charge will be taken as model S. So this will be the expression for the party. Now moving to the next part D in which we have the gold irons represented by the dark grey spears exit the particle at a distance. D. Two equals to 40.0 centimeter and molybdenum particles are exiting at a distance. Divan equals to 19.81 centimeter. Okay and mass of the gold M gold M G equals to 3.271 But clearly it into the power managed 25 kg. So we have to determine the mass of the molybdenum. I am. Okay so we can see that from the formula of the radius. The radius is given by r equals two Mv by QB. So from here we can see that radius is directly proportional to the mass. So we can write that radius or diameter. This is also same. So this is directly proportional to the diameter. Okay so diameter of gold iron is taken as to so we can take this mass of the gold as M two and mass of the molybdenum as M one. So there is no confusion in the data so we can write that D too bad even D to buy Divan, this will be equals to M to buy em one. So from here we get em one equals two embodies mass of the malevolent of iron. So it is equals two driven by D two, multiplied by empty. Okay, so substituting values, so Divan is equals to 19.81 centimetre divided by D two. It is 40.0 centimeter Medical herb I. M two, which is this value of 3.271 particular by 10 to the power managed 25 kg. So from here after solving mass and one mass of molybdenum iron comes out to be 1.6, 1.6 to 0, multiplied by 10 to the power minus 25 kg. Okay, so this is the answer for the part of the of the problem. Okay, this is the mass of the molybdenum iron. Okay.

In this problem. We're thinking off a fish. Bye. We obstructed tow a cylinder. Now we know that the cylinder has length ofthe 20 centimeters, which is they don't want to meters has And in India Krasic malaria off one centimeter square and has a voltage difference of three world. Now we want to find out. We also know that it is basically made off. See order on the the festivity ofthe sea water. Is there a 0.13 o meter for party? If you want to find out the current passing through deficient for that we need to go. I know they're resistant first so that we can use home slaw that we buy Our is our existence, as we know, is ruined by simple attacked on role is 0.13 Lent Is there 12 meters and area is one centimeter squared. Remember that a centimeter is 100 off major that is once individuals to four minus two meters on the whole square which is zero point 0 to 6. Divided by 10 4 minus four. It is 260. Oh, now that we have this current is simply what is by assistance. That is three World by 2 60 which is 11 point 5,000,000 ampere. Now question bees is interesting. We have another cylinder off half the land parle into it. Andi also has half the festivity now we want on. This replaces this half off the original cylinder, so we have the original cylinder for half the land on the new cylinder for the rest of them. So now let's say the resistance off this whole cylinder was R, which is to 60 home now. The resistance off this half surrender will obviously be our guy, too, because resistance is proportional relent and half the length would help have to resistance. Now, as you can see, these two cylinders are identical. They have the same length and area and only the rhesus. Studious changing. He resisted the off. This one is half. There isn't Studi off this cylinder, and their assistants off the cylinder will Also we have the resistance off this cylinder, giving us the resistance off the cylinder. Toby are, too. I do, which is already full, and this system is equal in tow. A toast to our sisters in Siri's with resistance is our by two on our rightful hands. The total assistance. Now we comes our way to place our way for because they're in serious which is three are for which is three by four times to 60 which is 195. Hence, In this new guest, the current will be wi I I be by our side which is three world by 195 which is 15 point for me. Happier be Lord Resistance increases the gallant and the change in current is one fifteen 15 point 4,000,000 amperes minus 11 point 5,000,000 years which is so you find 9,000,000 years. The current increases by three point 9,000,000 years.


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