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(15 pts) As shown in the figure, metal bar moves in uniform 2.20 T magnetic field at speed of v 6.00 mls on very low resistance wires_ The arrangement makes loop as...

Question

(15 pts) As shown in the figure, metal bar moves in uniform 2.20 T magnetic field at speed of v 6.00 mls on very low resistance wires_ The arrangement makes loop as shown:4.00(3 pts) This constant magnetic field was created by solenoid with high-current superconducting wires. The solenoid has diameter of 5.00 m with 7200 turns/m Find the current in the solenoid(5 pts) Find the amount and direction of the current induced in the loop(8 pts) Find the magnitude and direction of the force needed t0 k

(15 pts) As shown in the figure, metal bar moves in uniform 2.20 T magnetic field at speed of v 6.00 mls on very low resistance wires_ The arrangement makes loop as shown: 4.00 (3 pts) This constant magnetic field was created by solenoid with high-current superconducting wires. The solenoid has diameter of 5.00 m with 7200 turns/m Find the current in the solenoid (5 pts) Find the amount and direction of the current induced in the loop (8 pts) Find the magnitude and direction of the force needed t0 keep the bar moving at a constant velocity of 6.00 mls



Answers

In Figure $\mathrm{P} 30.23$ , the current in the long, straight wire is $I_{1}=5.00 \mathrm{A}$ and the wire lies in the plane of the rectangular loop, which carries a current $I_{2}=10.0 \mathrm{A}$ . The dimensions in the figure are $c=0.100 \mathrm{m}, a=0.150 \mathrm{m},$ and $\ell=$ $0.450 \mathrm{m} .$ Find the magnitude and direction of the net force exerted on the loop by the magnetic field created by the wire.

Here we assume that the motor of current in the circuit is constant. So if they're currently the constant, then we will get our constant exploration of the bar so fast. We will calculate the exhibition off bar by using the relation. So get off final. It's a bit equal toe so get off initial Issa Pete. But I was doing toe escalation times Distance Goward this use US exhibition Equal toe so good off funnel speed miners so get off initial speed divided by doing do distance covered by the bar is the boss starts its motion from rest so initial speed is equal to zero. So excuse us, it's a nation Equal toe Faneuil Speed of the bar is 20 aunt major parts again minus initial supes zero divided by doing toe distance covered by the body's one meter So disabuse US exhibition that is a magnitude up exploration of the bar Equal toe 300 92 meter parts again secure asked there is no friction so net force on the bar is equal toe the magnetic force acting on the bar due to magnetic field with physical dough I l be scientific IHS Magnetic field is perpendicular. So the angle between guard and amenity feel is 90 diggity this use us bangity force equal toe I l be a sign 90 musical toe one using your 12th love again right Net force on the bar Equal toe mass Times s elation So we can right here I l be equal toe mass times acceleration are we can write This is current through a bar Equal toe mass times extradition divided by Linda Bar Indo said rent off magnetic field now by substituting different well this year Mr Bar is 1.5 gram that is 1.5 times 10 to the power minus three kilogram in tow Exposition of bodies 392 meter par Sickens occur divided by lent our bodies two any two centimeter That is 22 times 10 to the power minus two meter An incident off magnetic field is one point seven isla. So this gives us a motor off current in the bar that it in the circuit equal dough One buyer six I'm here so far The force on the bar Toby in the Dietitian off exhibition Richard is shown in the figure The magnetic field must be Don't so magnetic field must be don't. So we will get a magnet e force in the direction off extradition.

Hi. In the gang problem there is. Are you shaved metallic cream? This is the metallic cream. Well, what a bitch there is on metallic bar which is warming over these rails or to oppose excites off this cream the steed off 1.5 centimeters for second. The worst right here. Like this, This is speed can also begin as 1.5 into 10. Dish part minus two. Niigata, What's second mass off This by has been given us for your 100 g or the 0.5 kg. But we think there is no use off this mass. We'll see it literal. The effective length of this metallic bar within the magnetic field is length. L is well toe 30 centimeter. Or we can say this 0.3 media and the magnetic really be is true. Destler. Yeah, this magnetic release, But lenticular toe the plane off paper and coming out off it like this. This magnetic field is coming out of the Leno paper. Per particularly business dance off the school of the loop has been unis are is equal to one 50. Oh, no. In the first part of the problem, we have to find the value off current induced in the loop between the human by own slow as enough divided by Are there this scheme of absolute is the emotionally mm whose expression is being to be into l divided by Are so putting all these values here, or B, this is two or three. This is 1.5 in tow. Tenders ar minus 2 m per second for lent. This is 0.3 and resistance is 1 50 home. So finally the value of this current comes out to be 6.0 in tow. 10 Daschle par minus five Ampere, This is a magnet. You Doctor Current Indios unit No, we have to find the sense that direction off this current Also we have to determine and use current What margarita in direction and direction of the inducement of the magnetic field, is pointing in tow. In the first case, the draw this metallic roar was moving for which you will use lemmings, right? And the old ask for this planning striking rule the bar the metal bar is moving towards right. So the term off our right and will be kept will be stretched outward and a magnetic release coming out so the forefinger will be stretched out off the plane off paper. In doing this, the middle finger will go down. So this is the direction of currently in use. So overall, if we look into the loo, that direction off the current we become Glock wife. This is the answer for the first part of the problem. No. The second part of the problem, they are saying again, we have to find the direction of the news current. But this time the magnetic brilliance revivals means now the magnetic field is getting into the plane off paper. In doing this? No, the roar is still The metallic part is still moving towards right so the tongue will be kept towards right. But the finger the forefinger will be inward. In doing this, the middle finger we go up. So this time the current is moving upward in the book bar in this metallic bar so that direction off current in whole of the lubie become counter. Look vice. In the third part of the problem, we again we have to find the direction of India's current. If the magnetic field is pointed into the page and the bar moves in War is a magnet for listen, work and the bodies moving in world means the direction of motion off this bottom in it is like this. So in doing so again, now the tongue will be here. The direction of motion of the conductor will be the traction off come the emphasis the forefinger will be in work. In doing this, the middle finger will again go Don't from the direction of current again will become bloke wise. So these are the answers for this getting problem.

Okay, So viciously magnetic field region Andrew and we know LBD which is the lance off a BD 7 45 centimeters which is 0.45 meter when the force the magnetic forces to goto current high intensity left or BD and has been magnetic view. So therefore have we know that the current is given a six point Om Pierre and the land lbd is 0.45 meter and magnetic field with 0.666 Tesla. So therefore, the magnetic force should we go to Ah 6.0 impure times 0.45 meter times 0.666 Tesla which is equal to 1.8 um Newton If you use the right hand rules too applied to the region, you realize that the force should have directions to relapse. So the direction of the force is to the left. And this is the answer for this question

For this problem on the topic of the magnetic field, we're told that a single turn, square loop has an edge length of two centimeters and carries a clockwise current of 0.2 M. P. S. The loop is inside a cell annoyed with the plane of the loop perpendicular to the magnetic field of the solenoid. If the solenoid has 30 tons per centimeter and carries a current of 15 mps clockwise, we want to use this information to find the force on each side of the loop as well as the talk acting on the loop. Now the field produced by the salon, it in its interior is B and B is equal to you. Not times in times I where N. Is the number of transparent at length and I the current. And this is in the minus I had direction. And so this is full pie Times 10 to the -7 because uh meters per M. Pierre into. And which is the number of turns per unit length, which is that he turns Times 10 to the -2 meters times the current I which is 15 And piers. And this is in the -1 had direction. This the this gives the magnetic field to b minus 5.65 Times 10 to the -2 Tesla's. Times I had. Now the force exerted on side A. B. Of the square current loop is given by F. B. At A B. Is equal to the current I times oh, Crosby, which is the current of zero 0.2 mps into the edge land too, Times 10 to the minus two m along unit vector. Jihad crossed with the magnetic field 5.65 times 10 to the -2 Tesla ni minus I had direction. So performing this cross product, we get this magnetic force that is exerted on side A. B. To be too 0.26 Times 10 to the minus for newton's in the K. Hat direction. Now this is 226 micro Newton's. And similarly, each side of the square loop will experience a force lying in the plane of the loop of F is equal to 200 26 micro Newton's away from the center, so that's the force that each side of the loop will experience. Next we want to find the talk. Now from this result, we can see that the net talk exerted on the square. The square loop by the field of the solenoid should be zero. So more formally, the magnetic dipole moment of the square loop is given by mule and mu is equal to I times A. This is equal to zero 0.2 mps times two Times 10 to the minus two m, all squared in the direction of minus I had, which gives the magnetic type of moments to b minus 80 micro mps meter squared in the I direction. Now the talk exerted on the loop, tor is equal to new cross B. And we can see that these two are anti is to a parallel rather minus 80 micro mps meters squared. The I had direction Crossed with -5 point 65 times 10 To the -2 kessler's in the I had direction that since these two are parallel, The talk, which the cross product is equal to zero.


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