what is the force per unit length f/l between the two wires? This is a topic that many people are looking for. bluevelvetrestaurant.com is a channel providing useful information about learning, life, digital marketing and online courses …. it will help you have an overview and solid multi-faceted knowledge . Today, bluevelvetrestaurant.com would like to introduce to you Magnetic Force Between Two Parallel Current Carrying Wires, Physics & Electromagnetism. Following along are instructions in the video below:
Current of 30 amps flows north on a five meter long wire and a current current of 50 amps flow south on another wire. Thats five meters long and two wires are three centimeters apart calculate the magnitude and direction of the magnetic force between the two wires. So lets start with a picture now lets go over some basic concepts that you need to know what is the direction of the magnetic force if both currents are flowing parallel to each other if they both go in the same direction.
And whats the direction of the magnetic force if the currents are anti parallel lets say if one current goes north and the other goes south what would you say now what you need to know is if the currents are traveling in the same direction. Then the magnetic force is a force of attraction. Now if theyre traveling in opposite.
Directions. The reverse is true the wires will be repelled by each other so make sure you know that now lets focus on this problem. So.
The first wire has a current of 30 amps and that current is flowing north. Now this wire lets call this current current one this wire will create its own magnetic field. So using the right hand rule.
If you grab a pen and you hold. It with your thumb pointing. North and notice how your fingers curl around that pen.
It should be curl around in this direction. So thats how the magnetic field is moving around this wire. So on the left side.
Its coming out of the page represented by this symbol and on the right side is going into the page so this current will create its own magnetic field b1. Which is mu zero the permeability of free space times i1 divided by 2 pi. R.
Now lets put a second wire right next to it. And we have a current flowing south. And this current is 50 amps and its going in this direction.
So lets call this i. 2 and lets say this wire is separated by a distance of r. So this current will generate a magnetic field.
Which will exert a force on the moving charges on that wire. So lets find out what it is to calculate the magnetic force on a wire with the current flowing through the wire. Its ilb sine theta.
Now because the magnetic field is going into the page in the negative z direction and the current is flowing south in the negative y direction the angle between the z axis and the y axis is 90 so sine. 90 is 1. So we could say that the force is just iob where l.
Is the length of the wire. So now what were gonna do is replace a beam with this term because b is equal to that so this will give us this equation f. Is i this is i two by the way due to this wire.
So its i two times l times b. Which is mu 0. I 1 over 2 pi r so to calculate the magnetic force acting on each wire.
Its gonna be lets start with mu 0. I 1 i 2 times the length of each wire divided by 2 pi r. Where r.
Is the distance between the two wires. Now. Some books may have the same formula.
But using different letters. So make sure you understand what these two values are so in this particular video. Just remember r is the distance between the two wires and capital l.
Is the length of the wire your textbook might use different letters and may have a different meaning for those things so make sure you understand what those variables are earlier. We said that if the two currents and the wires. If theyre moving in opposite directions each wire will fill a force that will basically pull them apart from each other now lets go ahead and confirm that so lets determine the force acting on this wire.
So the first current will generate a magnetic field thats going into the page on the right of the first wire and we have a current flowing in the negative y direction. So lets determine the direction of the magnetic force using the right hand rule. So what you need to do is you need to point your thumb south.
Its like this your thumb represents. The direction of the current and then your four fingers should be going into the page asking me how to draw so this is the direction of i2 and the magnetic field is going into the page and then out of the palm of your hand will be the direction of the force. So.
If you point your four fingers into the page and your thumb going south the magnetic force should be coming out of the palm of your hand. Which is this way and so that shows that the second wire is moving away from the first wire. So anytime.
The currents are anti parallel to each other they will the ys will repel each other theyre going to move apart. Now lets go ahead and finish. The problem museum is 4pi times 10 to the minus.
7. The first current is 30 amps. The second current is 50 amps and the left of each wire.
Is 5 meters. Divided by a 2 pi r. So r.
Is the distance between the two wires. So thats 3 centimeters. Which is point zero 3 meters.
If you divide by 100 so. 4. Pi divided by 2 pi.
Thats going to be 2. So its 2 times. 10.
To minus 7. Times 30 times 15 times. 5.
Divided by point 0 3. And so the magnetic force acting on each wire. Is point zero five newtons and so thats how you can calculate it now lets move on to our second problem.
A current of 50 amps flows east on a stationary wire. So lets draw a picture of that a second wire is one centimeter below. It how much current must be flowing in the second wire.
So that it doesnt fall due to gravity. So lets say this current is called i 1. This is i2 so we need to calculate the magnitude of i2 and also the direction.
Now gravity is going to pull the second wire down in this direction. So we need a magnetic force thats going to lift the second liner. So what direction must occur and be going well keep in mind if the currents are flowing in opposite directions.
We know that these two wires will repel so the forces will be in this direction. And we dont want the magnetic force to bring the wire down. We wanted to bring it up so therefore the currents must be parallel to each other so that the two wires will attract each other so i2 is also flowing east.
So we need to calculate the value of i2 now the first thing we need to do in order to calculate the value of i2 is we need to calculate the force per unit lift. Now the two wires. Theyre separated by a distance of 1 centimeter.
So thats equal to r. Now in order for the wire. Not to fall.
The wire has to be in equilibrium in the y direction. That means the sum of all forces in the y direction must add to zero. So the net force in the y direction is equal to the positive magnetic force minus the downward weight force and if thats equal to zero that we could see that the wave force has to be equal to the magnetic force now we dont have the mass.
But we do have the mass per unit length. So we need to divide both sides by l. So the force per unit.
Length is going to be the mass per unit length times the gravitational acceleration now m over l. We said its five grams per meter. Now we need to convert grams to kilograms one kilogram is equal to a thousand grams.
So five divided by thousand is point two zero zero five. So the mass per unit. Laughs is point zero zero five kilograms per meter and the gravitational.
Acceleration is 98. Meters. Per second.
Squared. So lets multiply those two values point zero zero five times. 98.
Let me. See if i can fit in here. So.
The force per unit. Length is now point zero four nine newtons per meter. So now that we have this value how can we calculate the current flowing in the second wire.
What do we need to do well. We know that the magnetic force between two wires. Its going to be f.
Which equals mu zero. I 1 i 2 times l. Which is the length of the wire.
Divided by 2 pi. R. Where r.
As the distance between the two wires. So lets isolate i 2. In this equation.
So were going to cross multiply. So f. Times.
2 pi r. Is that and that is equal to 1 times mu. 0.
I 1 i 2 times l. Now lets divide both sides by mu 0. I 1 l.
So therefore we could say that i 2 is equal to the force per unit left. So thats going to be f over l. Times.
2. Pi r. And on the bottom.
What we have left over is mu. 0. I 1 so we could use that formula to calculate the current in the second wire.
So the force per unit left. We have that its this value. So.
Thats point 2 0. 4. 9.
Multiplied. By 2 pi. R.
And r. Is 1. Centimeter.
So. Thats if you divided by 100 point. 0 1.
Meters. And mu. 0.
Thats 4 pi times. 10 to the minus 7. And then i.
1. Is this current 50 amps now 2 pi divided by 4 pi thats 1 2. So basically this becomes a 2.
So then i. 2 is simply point zero four nine times point zero 1. And then take that result divided by 2 times 10 to the negative.
7. And then divide that result by 50 and you should get this answer in i. 2.
Is 49 amps and so thats it for this problem. Now you know how to calculate the current in the second wire. .
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