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Straight horizontal wire with length-to-mass ratio of 17 mlkg is carrying a current of 3.0 A to the west: The wire is placed in a uniform magnetic field B. With the...

Question

Straight horizontal wire with length-to-mass ratio of 17 mlkg is carrying a current of 3.0 A to the west: The wire is placed in a uniform magnetic field B. With the weight of the wire pointing down to the south; what minimum field B (in T) is required for the wire to achieve static equilibrium? (gravitational acceleration = 9.8 m/s2)Select one: 55.53 north0.19 into the page 0.19 out of the page 55.53 into the page 55.53 out of the page

straight horizontal wire with length-to-mass ratio of 17 mlkg is carrying a current of 3.0 A to the west: The wire is placed in a uniform magnetic field B. With the weight of the wire pointing down to the south; what minimum field B (in T) is required for the wire to achieve static equilibrium? (gravitational acceleration = 9.8 m/s2) Select one: 55.53 north 0.19 into the page 0.19 out of the page 55.53 into the page 55.53 out of the page



Answers

In Fig. $30-10$, the magnetic field is up out of the page and $B=0.80$ T. The wire shown carries a current of 30 A. Find the magnitude and direction of the force on a $5.0 \mathrm{~cm}$ length of the wire. Fig. $30-10$ We know that $\Delta F_{M}=I(\Delta L) B \sin \theta=(30 \mathrm{~A})(0.050 \mathrm{~m})(0.80 \mathrm{~T})(1)=1.2 \mathrm{~N}$ By the right-hand rule, the force is perpendicular to both the wire and the field and is directed toward the bottom of the page.

For the first part of this question. We're trying to figure out what the minimum current is needed flowing through the rod so that the bar can be in the position as shown in the picture. So it's supported against gravity. So we know that the gravitational force you he called you G and we know the worst would cringe in a magnetic field is equal to i o press be. And in this example since the l and B are both perpendicular to each other, this cross product goes away and this becomes I'll be So what we want to do is we want to set the gravitational force equal to the four from the current. So we have. MGI is equal to i o b and we're searching for the current. So we just sold for I by dividing both sides by elope. So I is equal to G over. Oh, and then we're giving all these values so massive. The rod is 18 grams and remember, to put this into kilograms, gravity will use 9.81 meters per second squared Well, it's 20 centimeters and the magnetic field Wait 15 Tesla. So if you multiply that out. We get a minimum current value of 5.9 amps and then for part B were asked which direction the current needs to be flowing. So if we look back at part A, we have that cross product, eh? Physical io across B and for it to equal, I'll be for that too. True, it means that all of the opponents have to be perpendicular. So if we use our right hand, have we know the direction of the forests on the bar? I want the force to be going up, so it will be opposing gravity. So if we kind of use the right hand rule in reverse a little bit, we know the direction of the Byfield. We know the direction of the force and then we can find that direction of the current will be going this way from A to B.

Well. Consider straight wire with link. Hell is equal to do. 0.5 meter. Which gate is a current off? I is equal to 1.5 ampere in the location where the Earth's magnetic field is 0.55 cows. So arts magnetic field B is equal to 0.5 five goes and its direction is from south south to north. We need to find the magnitude and direction off the force acting on the wire due to this field. Let's do party first. If the current in the wire is running from west to east, the angle between the magnetic field and the current is 90 degrees. So five is equal to mine two degrees so the magnitude of the force is given by we know the F equals hi L B time sign If I Well, in this case, five is equal to 90 degrees before if becomes I l b no. We insert values. Then we have F is equal to 1.5 m Pierre into 2.5 meter into magnetic fuel efficiency 2.55 times 10 to the dollar, minus four Destler minus four Gasela. One girls equals 10 to the power minus for Destler. So force is equal to 2.1 times 10 to the power minus four Newton. Right? Well, the magnetic field points to the north and the current forms to the east. Right? So that's CNBC's. These are the directions, right? So this is north. This is south. Uh, it's east, and this is the best. So here, the magnetic fields points points to the north. Right? So BDs pointing in this direction be and the current points to the east, you know, electrical and eyes pointing in the east direction right according to the right and grew the direction off the forces out of the beach. Right. So eighties out of the beach. So for out off the beach. Re troll. Undaunted here, right? No, let's do part b. Well, if the current in the wire is threatening vertically upward bangle between the magnetic field and the current is equal to 90 degrees bicycles to 90 degrees. So the magnitude of the foresees f is equal to I times at all times. Be right. No, we're plugging numbers. Then we have force equals 2.1 times 10 to the power dentally power minus four. Gutin Well, here the magnetic field points to north and the current points out of the beach. So you're magnetic field points, uh, to north, Mrs B, an electrical and eyes pointing out off the beach, so eyes pointing out of the bait. It's coming out of the beard according to the right and with the direction of the forces from east to west. So this is the direction, or four staff from east to best. Now let's talk about part C. Well, if the current in the wire is running from north to south in this case, the angle between the magnetic field and the current ease 180 degrees. So in this case, angle is 180 degrees. So the force zero, because sign off 1 80 degrees, is equal prosciutto. So force is equal to Zito

It's probably we're gonna talk about the magnetic force on a wire, So suppose that we have a wire that has a current hi, carries a current I and has a length l Yeah, and also assume that we have a magnetic field. Well, let's say that the magnetic field this pointing in this direction here. So the magnetic force, the total magnetic force f that's exerted on the wire is equal to be cross I Time's up. Yeah. Where the cross here is the vector product. Okay, so we have to suppose in our problem that we have wire that has a link of 2 m and a mass off 150 Gramps on. And this we have that at this point, the magnetic field of the earth. That is it. Which is 0.55 goals. That is 0.55 times 10 to the minus fourth Tabla. This is the magnetic field of the earth, and it is horizontal. So I'm going to assume that this magnetic field is pointing from west to east like this. Okay, this is magnetic field. Mhm. And we have a wire. I'm gonna assume that the wire is also on the horizontal but perpendicular to the magnetic magnetic field. So it is entering the ah, this queen and we want the force on the wire to be enough to counterbalance the gravitational force so that the wire stays put stays arrest. So basically, we want the, um, magnetic force to be equal to M g. Diplomatic magnetic forces be times deep current I times l since the magnetic field in the current i r. Political and are going to find what is the current necessary. So the current I is equal to, um mg divided by B l. Yeah. Now M is 150 g so 1.15 kg times gravitational constant off 9.8 m per second squared divided by B. That's 5.5 times 10 to the minus five Tesler times the length l of 2 m. So I is equal to one point 34 times 10 to the fourth compares. Okay, this is the electric current necessary, and exercise asks us if, um, it seems likely that the why're will be able to support this current and the answer is no eso. It's unlikely because this is a very large current. Many question be We have to, uh, say where this current should be pointing to. Just remember, we need the force caused by the magnetic field to be pointing outwards so that it counterbalances the, uh, the wait that the earth exerts over. Ah, the wire. Yeah. So for that should be true. Yeah, I That is the vector current must be pointing towards that is that it must be, uh, entering the screen as shown here in the figure, so that when we make the right hand rule and make the cross sector the cross product, I'm sorry between be and I we obtained that the f points upwards. So we want I, um to be pointing towards the screen, at least in this, um, set up that we created here. That Yeah. And this concludes our exercise. Mhm.

Mhm. This problem covers the concept of the magnetic force on a current carrying buyer by external magnetic field. And to solve the problem, we are going to use the situation. And first of all, we will find the direction of the magnetic force on the fire. So from right and room a free point our fingers along the direction of the current and the palm along the direction of the magnetic field. Then we can find out the direction of the magnetic force on the wire. The direction of the magnetic force will be along the direction that is along the left. Okay now calculate the magnitude of the force so F equals the current is eight mp. Uh The current is eight MPO. And the length of the wires 0.5 m. The magnetic field is for Tesla. And to sign off the angle between the Via and the magnetic field is 90°. So the magnitude of the forces. 16 Noten advocacy. The force is 16 northern to the left.


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