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44 Scholarships Urv44 University of Houst _Ben EgglestonA 22-cm length of wire is held along an east-west direction and moved horizontally to the north with a speed...

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44 Scholarships Urv44 University of Houst _Ben EgglestonA 22-cm length of wire is held along an east-west direction and moved horizontally to the north with a speed of 3.8 m/s in a region where the magnetic} field of the earthis 66 micro T directed 19* below the horizontal Whatis the magnitude ofthe potential difference between the ends ofthe wire? Write your answer in micro-volts:117.96ptsQuestion 6bar of negligible resistarce slides along conducting arrangement shown shown F 4ohm In the rails

44 Scholarships Urv 44 University of Houst _ Ben Eggleston A 22-cm length of wire is held along an east-west direction and moved horizontally to the north with a speed of 3.8 m/s in a region where the magnetic} field of the earthis 66 micro T directed 19* below the horizontal Whatis the magnitude ofthe potential difference between the ends ofthe wire? Write your answer in micro-volts: 117.96 pts Question 6 bar of negligible resistarce slides along conducting arrangement shown shown F 4ohm In the rails connected as horizontal parallel friction-lessconduotimg mentioned in the figure for the this value Ignore the ohm the plane of the paper: resistor (use field is perpendicular - uniform 1.1-T magnetic in the resistor at the resistance} generated thermal energy being IfL 41 Cm, at what rate equal 8 m/s? instant the speed = of the bar B45PM Wnoad DeEC 200 342 S Aneat Delksdr



Answers

Figure $32-35 a$ shows the current $i$ that is produced in a wire
of resistivity $1.62 \times 10^{-8} \Omega \cdot \mathrm{m} .$ The
magnitude of the current versus
time $t$ is shown in Fig. $32-35 b$ . The magnitude of the current versus
time $t$ is shown in Fig. $32-35 b$ . The
vertical axis scale is set by $i_{s}=10.0$ A, and the horizontal axis scale is
set by $t_{s}=50.0 \mathrm{ms}$ . Point $P$ is at
radial distance 9.00 $\mathrm{mm}$ from the
wire's center. Determine the magnitude of the magnetic field $\vec{B}_{i}$ at point $P$ due to the actual current $i$
in the wire at $(a) t=20 \mathrm{ms},(\mathrm{b}) t=$
$40 \mathrm{ms},$ and $(\mathrm{c}) t=60 \mathrm{ms},$ Next, assume that the electric field driving
the current is confined to the wire.
Then determine the magnitude of
the magnetic field $\vec{B}_{\text {id}}$ at point $P$ due to the displacement current $i_{d}$
in the wire at $(\mathrm{d}) t=20 \mathrm{ms},(\mathrm{e}) t=$
$40 \mathrm{ms},$ and $(\mathrm{f}) t=60 \mathrm{ms}$ , At point
$P$ at $t=20 \mathrm{s},$ what is the direction
(into or out of the page) of (g) $\vec{B}_{i}$
and $(\mathrm{h}) \vec{B}_{i d} ?$

In this problem on the topic of magnetism, we are shown in the figure the current that is produced by a wire of relativity, 1.6, 2 times 10 to the -8 or meters. The manager of the current versus time is shown in the next figure. The vertical axis in that figure is set by I. S. Which is 10 mps. And the horror horizontal axis scale is set by tears, which is 50 milliseconds. The point P. Is that a radio distance nine millimeters from the center of the wire? And we want to find the magnitude of the magnetic field B. I. At point P. Due to the actual current i in the wire firstly at 20 then 40 and then 60 milliseconds. Next, we'll assume that the electric field driving the current is confined to the wire. And we want to find the magnitude of the magnetic field. The idea point P, which is due to the displacement current in the wire again at 2040 and 60 milliseconds. And then we want to find a 20 seconds the direction of B. I. And B. I. D. Now from the figure we can see that I. Is equal 24 mps. When the time T. Is 20 milliseconds. And so from here, the magnetic field due to the actual current B. I. Is you're not I over two pi R. And using the information given this is zero point zero 89 millie Tesla's. Yeah. Next. If we are given. Yeah. The time of 40 milliseconds. The current is eight piers. And so using the equation above, we get the magnetic field B. I. To be 0.18 molly Tesla cars. Yeah. And then for part C. We have at 60 milliseconds. The current I. is equal to 10 and piers since the current is 10 mps. When T is greater than 50 milliseconds, which means that the displacement current or rather the magnetic field B. I. Is equal to 0.22 molly Tesla's now for part D. The displacement current, in terms of the time drought of the electric field, I. D. Is equal to absolutely not A. D. E. D. T. Yeah. Yeah. And the electric field E. Can be written as the resistive. Itty role times I over A. And this is in terms of the real current. And so therefore the displacement current I. D. Can be written as absolutely not times row time's D. I. D. T. Now for a time of less than 50 milliseconds we have from the graph that D. I. D. T. Is equal to 200 mps per second. And so therefore the magnetic field due to the displacement current B. I. D. Is equal to mu note i. d. over two pi r. And this is 6.4 Times 10 to the -22 kessler's for part E. As we've done in B, we find the I. D. Is you note I'D Over two Pi R. Which is 6.4 Times 10 to the -22. Okay, Tesla's for part F. Yeah. Yeah. We have the I. D. T. Is equal to zero. And so we can see that by the reasoning we've done before the magnetic field, B. I. D. Is equal to zero. Okay, next. For part G. We can use the right hand rule to find the direction of B. I. And so at T. is equal to send 20 seconds be I. Is out of the page And at T. is equal to 20 seconds. The magnetic field due to the displacement current B. I. D. Is also out of the page.

So here we know that four part A. We want to find the total resistance and the total length of the closed loop formed by the two Rady I. Plus the art would be two times the radius. Plus our times fada are times data would be equal to the length of the ark and so this would be equal to our times two plus data. And so the are of course, would be the resistance. And I thought it would be the angle of the ark and so we can find the total resistance by this would be equal to the resisted Vitti a row times the total length divided by a the cross sectional area in this case. So this would be there is a festivity times the radius multiplied by two plus Fada. So we're simply plugging in l. A and then divided by a And at this point, we can solve. So this would be 1.7 times 10 to be negative eighth hums meters and then multiplied by 0.24 meters plus rather to rather times two times data to plus data. This would be divided by the area of 1.20 times, 10 to the negative sixth meters squared and so the resistance is going to be equal. The total resistance would be equal to 3.4 times 10 to the negative. Third multiplied by two plus Fada Um And then the units here would, of course, be bums. So this would be our answer for the total resistance in terms of the angle of the ark. For part B, we know the area of the loop to be 1/2 our square times the angle, so the magnetic flux through the loop would be equal to B times A or simply 1/2 be r squared Fada on. And then we can solve so 1/2 times 0.5 point 15 zero Tesla's times fada times point 240 meters quantity squared and we find that the magnetic flux is going to be equal to 4.32 times 10 to the negative third fada. And then the units here would of course, be Webber's. So this would be the magnetic flux in terms of Fada or, in this case, the angle four part See, we know that the induced E M. F. Is going to be equal to the negative change in the magnetic flux. With respect the time, this is simply according to Faraday's law, with an end equaling one. So this would be equal to the negative derivative with respect the time of 1/2 b r squared data, this is gonna be equal to one half b r squared times the change in the angle with respect to time. Ah, according to angular, uh, Kenna Matics, we can we know that this is simply going to be equal to negative 1/2 times B r squared and then defaced the change in angular displacement with respect the time here, this is simply a kn angular velocity. So we can say lower case, Omega lower case Omega, of course, denoting the angular velocity. And so we can say that this is going to through arms law, we can find the current. This would be the induced E m f uh, divided by our and this would be equal to be r squared Omega, divided by two are and so I would be equal to two times 3.4 times 10 to the negative. Third multiplied by two plus data because we already had found the resistance. Um, in part A. So we're simply just plugging in. And then it would be the magnitude of the magnetic field times the radius squared times omega, the angular velocity. This would be equal to again two times. 3.4 times, 10 to the negative, Third times two plus data. Ah, we know that here it'll be on the for the numerator. It'll be the magnitude of the magnetic field times r squared. And then per the definition of the angular velocity, this would be the numerator would be multiplied by substituting in for the angular velocity Alfa. The angular acceleration multiplied by time, knowing that, of course, the angular velocity is equal to the angular acceleration multiplied by time. And so essentially, we can then find a derivative of the current with respect to time. And this would be equal to, uh, b R squared Alfa multiplied by four minus alfa t squared. And then this would be divided by 3.4 times 10 to the negative. Third times four plus Alfa T squared, quantity squared. Uh, we know that here the induced current is at a maximum when four minus Alfa t squared equals zero. So we know that I Max, occurs that, uh, four minus alpha t squared equals zero. Or we can say t is gonna be equal to the square root of four. Divided by Alfa The angular acceleration. At this instant, we can say that data would be equal to 1/2 times the angular acceleration times the time squared. This is simply angular, Kenna Matics. This would be equal to one over two times. Alfa T squared would simply be equal to four. Divided by Alfa Alphas. Cancel out and we have that data is equaling 2.0 radiance. So there's going to be a maximum current at data equaling 2.0 radiance. This is our answer for a part. See, now, finally, for part D we can say that when the current is Max, I'm Max at, uh, angular, velocity equaling Alfa T equaling outfit times the square root of four divided by Alfa. Or we can simply say, equaling the square root of four Alfa. Thus I max calculating for the maximum current we have B R squared Alfa, divided by two are aah! This would be B R squared multiplied by the square root of four Alfa, simply substituting in for the angular velocity divided by two times the resistance. Ah, we can then substitute, So this would be equal to B R. Squared times for Alfa, divided by two times 3.4 times 10 to the negative, third times two plus data. And then we can solve by plugging in Fada and the angular acceleration and essentially Oliver Constance. So the maximum current occurs. Rather, the value for the maximum current will be 0.150 Tesla's multiplied by 0.24 meters Quantity squared, um, multiplied by the square root of four times 12 radiance per second squared Aah! This would be divided by two times, 3.4 times 10 to the negative. Third multiplied by two plus 2.0, and we find that the maximum current is gonna be equal to 2.20 And pierce, This would be our final answer for Part D. That is the end of the solution. Thank you. For once

Hi. In a given problem, there are two parallel metallic rails which are connected at they're right and and there is a metallic roar which is allowed to move over these two rails. Words left with velocity V. In this set up there is a uniform magnetic field which is directed out of the plane of paper, but particularly like this, this is the land off. A gap between the whales will serve as the length of this metallic roar and that is given us then Centimeter, which may also be written a 0.10 m. The velocity with which the metallic rod is moving towards left is given as 5.0 m or second. And magnetic field is having a magnitude 1.2. Yes, lock in the first part of the problem. We have to find the EMF induced across the ends of this metallic roar and that will be emotional mm induced growth. This metallic Lord be given by the expression be into V. Into L. So plugging in all the non values here for magnetic well this is 1.2 Tesla or velocity of the Lord. This is 5.0 meter per second and four length. This is 0.10 m. So finally the CMF induced emotionally induced across the ends of the road From south to be 0.6 fold, which is the answer for the first part of this problem. Now in the second part of the problem, Resistance of the circuit is given a 0.40 home. So using arms law, current passing through this metallic rod will be given by the IMF in dues divided by resistance means this is 0.60 old Divided by 0. 0 mm. Finally this current passing through The Metallic Rod comes out to be 1.5 NPR. This is one of the answer For the 2nd part and is the magnetic flux is outward and that will be increasing when this role will be moving towards. Left flux will be increasing as the area is increasing. So we can say as the magnetic flux which is responsible for the current in dues, this is outward increasing. So as her lenses law induced magnetic plugs could be opposing in nature and opposing in nature. So it should be into the plane of paper against the outward magnetic locks and we know magnetic field enters into the south pole only to hear the direction of current could be black, white direction of quarantine. Dues should eat block twice. It is a clockwise current whose magnitude is given as 1.5 ampere which combine lee becomes the answer for the second part of this problem. No. In the 3rd part of the problem we have to find rate of thermal energy being added to the road means the power, electric power consumed in the road that will be given us B. H. That would be. And having an expression as per usual slow fitting. This is given as I square into our oh for this P. H. For I. This is 1.5 mm peer to the whole square into our Which is given as zero point for all. And finally this rate off Thermal energy being added to the road will be will come out to the 0.9. What which is answer for the third part of this problem Then in the 4th part of the problem course required to keep this role moving will be equal to magnetic force experienced by the road. Yeah, course. Why he is roll moving Will be actually magnet force experienced by eight which is having an expression be into I. Into L. So, plugging in unknown values. Again here, 1.2 Tesla. For the magnetic field or current induced. This is 1.5 ampere. And for the length of the road this is 0.10 meter. So finally this force here comes out to be 0.18 newton. And that should be two words that should be acting words right, sorry towards left. It should be. This is the answer for the 4th part of this problem. Finally, in the last part of the problem, rate of work done by external agent on the road means power instant in his power and doomed by external agents to move the road. That will be given as technical power E. M. Which is equal to the product of the force applied on it with the instantaneous velocity this P. M. There will be given by 0.8. Newton multiplied by velocity five me too per second. So interestingly here, this power In June by the external agent also comes out with 0.9 what? Hence we can say power consumed power converted rate of conversion of energy as a thermal energy and rate of doing work by the external agenda, who are same as PM. Musical to PH And that is equal to 0.94. This is our conclusion. Thank you.

For this problem. On the topic of induction, we have shown a wire that has been bent into a circular arc of radius 24cm. The wire is centered at Oh and a straight wire O. P. Can be rotated about oh and make sliding sliding contact with the R. P. We then have another straight wire O. Que that completes the conducting loop. The cross sectional area of the three wires is common to be 1.2 sq mm And they all have a resistive itty of 1.7 times 10 to the -8 or m. The entire apparatus slides in a uniform magnetic field of magnitude 0.15 Tesla is that is directed out of the figure. Now why? Opie starts from rest at an angle detail is equal to zero and has constant angular acceleration of 12 radiance per square second. We want to find the loops resistance, the magnetic flux through the loop and the angle theater for which the induced current is a maximum. And lastly we want to find that maximum. Now the total length of the closed loop that is formed by the two radio plus the ark is L. And our is equal to to our plus our peter Which is our into two plus data. Where are here is the radius. The total resistance is capital are and this is the resistive. Itty wrote times the length L over the area. A And so this is rho times are into two plus theta. All over A. And so we can find this resistance in terms of detail, we get this resistance to be 1.7 Times 10 to the minus eight um meters times zero point 24 m into two plus theta divided by The area 1.2 Times 10 to the -6 square meters. And so this gives us the resistance in terms of theater to be 3.4 Times 10 to the -3 into two plus theater Holmes. For part B we have the area of the loop A. To be half R squared peter. The steam in eric flux to the loop. five B is equal to the field strength b times the area a. So we can write this as a half B. R squared peter, this is a half times 0.15 Tesla's times zero point 24 meters squared times theater. And so this gives us steaming like flux through the loop to be fool 0.32 Times 10 to the -3 data. Weber's okay. Yeah, for part C. The induced MF epsilon is minus the time rate of change of flux defy B D. T, which is minus the by D. T. Of uh huh, B R squared data which is minus a half B R squared the data DT since B is constant in time. And so this is minus a half B r squared omega omega is depleted T. And so the current i is by bombs low absalon over our which is B R squared omega over two times the resistance. And so this is B R squared mega divided by to into 3.4 Times 10 to the -3 into two plus theater. And so this becomes B R squared alpha times time T Over two into 3. four times 10 To the -3 into two plus alpha T squared over to Now this is as the magnitude of the induced current. No, we've let omega here be alfa T and theater a half alpha T squared for constant angular acceleration ALPHA. If we differentiate I with respect to T, we get the I D. T. To be B R squared alpha into for minus alpha, T squared over 3.4 Times 10 to the -3 into four plus alpha t squared whole squid. Now in the induced current is at a maximum when four minus alpha, T squared is equal to zero or the time T is equal to the square root or for over alpha. And so at this instant the angle theater is a half alpha T squared, which is a half alpha times four over alpha. And so theater is too radiance and for party when the current is at a maximum we have omega equal to alpha T. is equal to alpha into the square root of four over alpha, which we can right as the square root of four alpha. And therefore the maximum current I max is able to be r squared omega over to our which is B r squared times the square root of for alpha Over two times the resistance, which we can write as B r squared times the square root of four alpha, Divided by two into 3.4 Times 10 to the -3 times two plus theta. And so putting in our values. This is 0.15 Tesla as times 0.24 m squared times the square root of four times 12 radiance per square second, all divided by to into 3.4 times 10 to the minus three Into two plus theater, which is to radiance. And so calculating, we get the maximum current To be 2.2 and piers.


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