Comprehensive Guide to Magnetism: Magnetic Fields, Forces, and Applications

in this video we're going to talk about magnetism perhaps you're familiar with bar

magnets and you know that when you place the North Pole of a bar magnet with another North Pole these two they

repel they're going to push apart however let's say if you were to face

the North Pole of one bar mag with the South Pole of another they won't repel these two will

fill a force of attraction the North Pole is attracted to the South Pole but if you put the

North Pole with another North Pole of another magnet they will repel or if you put the South Pole of one magnet with

the South Pole of another they will also repel every Barb magnet has its own magnetic

field which emanates away from the North Pole and it travels towards the South Pole likewise this one

has a magnetic field em away from the North Pole notice that the magnetic field cancels in the

middle those two they repel each other and in a case of the other example the magnetic field leaves the

north po pole but it enters the South Pole so in the case of Attraction notice

that the magnetic field between the North and the South Pole of the two bar mics notice that

they're in the same direction and so they're additive you have that force of attraction so what causes magnetic

fields magnetic fields are created by moving electric charge and this example can be

Illustrated if you have a wire whenever there's an electric current that flows through a

wire it creates its own magnetic field and the magnetic field created by this uh wire it looks like

this it's circular on the left side the magnetic field is leaving the page so it's represented

by a DOT or a circle and on the right side it enters the page which is symbolized by an X you need to be

familiar with that by the way you can use the right hand rule to figure this out if you take your hand

and if you curl it around a pen with your thumb facing the direction of the current the way your fingers curl around

the pen is the way the magnetic field travels around the wire try that so here's a picture that

describes it as you can see you want the thumb faceing the direction of the current which is upward and as you curl

your hand around the wire notice the way your hand curls around it on the left side

you can see how it's coming out of the page and then on the right side your hand curls into the

page and so the way your hand curls around that wire is the way the magnetic field created by the movement charge

travels around the wire it's out of the page on the left side and it's going into page on the

right side now there's an equation that allows you to calculate the strength of the

magnetic field created by such a wire and here's the equation B is equal to U * I / 2i

R so let's say if you want to calculate the magnetic field some distance point a away from The Wire R is the distance

between the wire and point A B is the strength of the magnetic

field and B is measured in units of Tesla or capital T U or mu0 is equal to 4 Pi * 10-

7 this is known as the permeability of free space and the units are Tesla times meters per

amp now notice that the current and a magnetic field are directly related if you increase the magnitude of

the current the strength of the magnetic field generated by this yre will increase as

well and a reason for that is because the current is on the top of the fraction whenever you increase the

numerator of a fraction the value of the entire fraction will increase now R is on the bottom so that

means that R is inversely related to B if you increase the distance between the wire and a magnetic and a point of

Interest I should say the magnetic field at that point will be

weaker As you move away from The Wire the strength of the magnetic field weakens by the way the number of

magnetic lines that you see in a picture is proportional to the strength of the magnetic

field so for example let's say if the magnetic field and this region looks like this

versus two lines as opposed to three the magnetic field on the left is stronger than the one on the right if

you have more lines that are closer together the strength of the magnetic field is stronger but make sure you know

this relationship so anytime you increase the electric current in a wire the strength of the magnetic field will

increase and as you move away from The Wire the strength of the magnetic field will decrease

let's work on this problem a vertical wire carries a current of 45 amps du sou calculate the magnitude and the

direction of the magnetic field 2 cm to the right of the wire so go ahead and try this problem so let's say this is

the wire and the current is due south so it's going

down and that means that using the right hand rule the magnetic field is going to enter the page on the right side and

it's going to be out of the page I mean it enters the page on the left side but on the right side it comes out of the

page so when it enters a page put a x and when it leaves the page put a circle a closed

circle now our goal is to find the magnetic field two cm to the right of the wire so we already have the

direction of the magnetic field it's out of the page all we need to do now is calculate

the magnitude so we can use this equation B is equal to mu0 * I / 2i

R mu0 is equal to 4 piun * 10 7 and the current is 45 amp R is the distance between the wire

and a point of interest in this case r is 2 cm but we need to convert that met so we got to divide by

100 1 met is equal to 100 cm so that's going to be 02

M so now all you have to do is just uh type this in and you should

get 4.5 * 104 Tesla

so that's going to be the strength of the magnetic field 2 cmers away from The Wire number two a wire carries a current

of 10 amps at what distance from The Wire will a magnetic field of 8 * 10-4 Teslas be

produced so we got to find R in this case we can use the same equation we don't have to worry about

the direction so we don't really need to draw a picture so let's solve for R let's multiply both

sides by R so on the right side it's going to cancel so B * R is equal to mu0 * I / 2

pi now let's multiply both sides of the equation by 1 over B if we do that on the left side B will

cancel so the distance is going to be mu 0 * I / 2 piun * B so it's 4 piun * 10- 7times a current of 10

amps / by 2 pi times the strength of the magnetic field so we can cancel Pi in fact 4 Pi /

2 pi is just 2 so it's going to going to be 2 * 10 - 7 * 10 / 8 * 10 - 4 and so this is equal to

2.5 * 10us 3 and the units is uh meters so if we want to we can convert it to millimeters and to do

that you need to divide actually multiply by th000 there's 1,00 mm per met if you multiply by a00 this will

give you 2.5 mm and so that's the answer that's how

far away from The Wire that you have to be to measure a magnetic field of 8 * 10us

4 by the way if you ever were to place a compass near a wire whenever there's an electric current flowing through that

wire it will cause the compass to deflect you should try it now let's say if we have a current

caran wire what's going to happen if we place this wire inside a magnetic field let's

say the magnetic field is directed East and the current is moving North a magnetic field exerts no force

on a stationary charge however if the electric charge is moving then the magnetic field will exert a force

specifically a magnetic force so whenever you have a wire with an electric current that means you have

moving charges in The Wire the magnetic field will exert a force on the wire you can calculate the strength of

the magnetic force using this equation f is equal to I lb sin th so the strength of the magnetic force

is proportional to the current if you increase the current the magnetic force will

increase the magnetic force is also proportional to the strength of the magnetic field if you increase the

magnetic field the magnetic force will increase as well and it's also proportional to the length of the

wire now depends on the angle as well so here's one example where the current and the magnetic field are

perpendicular and here's another example when the current and magnetic field are at an

angle and here's when that's parallel now Theta is the the angle between the current and the magnetic

field when they are perpendicular sin 90 is equal to 1 and one basically represents

100% so the maximum Force occurs when the current and the magnetic field are perpendicular to each

other now when it's at an angle it's going to be between anywhere from 0 to 100% of its maximum value

so then you could use this equation when they're parallel the angle is equal to 0°

sin0 is equal to Z therefore the magnetic field exerts no magnetic force on a moving charge that

moves parallel or even anti-parallel to magnetic field so for the third example there's

no magnetic force acting on the current they have to be at an angle with respect to each other they can't be

parallel now what about the direction let's go back to our last example so let's say if the current is

due north and magnetic field is directed East in what direction is the force now the force has to be

perpendicular to the current anti magnetic field so if the current is in the north south Direction and if the

magnetic field is in the west east Direction then the force is either into the page or out of the page that is

along the z- axis so how can we figure this out well we have to use the right hand

rule so take your right hand and you want to extend it you

want your thumb to be in a direction of the current and you want your other four

fingers to be in a direction of the magnetic field so this represents

B and this represent the current I so using your right hand look at where your right hand opens towards if you direct

it the way it's presented here it's going to go into the page the force comes out of the palm of your

right hand and so it's going to be directed into the page and that's how you can figure it

out let's try another example so let's say if we have a wire and the current is directed

East and the magnetic field is directed uh into the page and what direction is the force

going to be so if the current is in the east west Direction and well the current is really

west to east and a magnetic field is in the Z direction that is between out of the page and into the page then the

force has to be in a north south Direction these three variables have to be perpendicular to each

other so what you want to do this time you want to direct your four fingers into to the

page and you want your thumb directed East that is in the direction

of the current but you want your four fingers to be into the

page and the force comes out of the palm of your right hand so the force will be directed North if you do that it's kind

of hard to draw the direction of the hand on his video but hopefully you can

visualize it number three a 2.5 M long wire carries a current of 5 amps in the

presence of a magnetic field with a strength of 2 * 10us 3 Teslas calculate the magnitude of the

magnetic force on The Wire using the picture shown below so feel free to try that the the equation that we need is f

is equal to I lb sin thet now let's talk about the angle so the current is directed East and the magnetic field is

directed 30° relative to the horizontal so Theta is always going to be the angle between the magnetic field

and the current so you can also use this angle as well because that angle is between B

and I 180 - 30 is 150 and it turns out that s of 150 and S of 30 they're both equal

to2 so it doesn't matter so whether you choose this angle which is between I and b or if you use this angle the answer

will be the same so just something to know so now let's go ahead and calculate F so it's going to be the current which

is 5times the length of the wire which is uh 2.5 M times the

strength of the magnetic field which is uh 2 * 10us 3 multipli S of 30 so the magnetic force is very small

it's 012 5 Newtons and so that's going to be the

force exerted on this current carrying wire number four a current of 35 amps flows Due West in a wire that

experiences a magnetic force of 75 Newtons per meter what is the strength of the

magnetic field which is directed due s so here's the wire and the current is directed West and the magnetic field is

directed South our goal is to find the strength of the magnetic field we need to solve

for b so let's write the equation f is equal to IB now because the current and the

magnetic field are at right angles to each other because it's 90° s 90 is 1 so we don't need the sign portion of this

equation now we're given the force per meter so that's F / by L if we divide both sides by L we're going to get this

equation F over L is equal to the current multiplied by the magnetic field we have the force per meter that's

75 that value takes care of two of these these variables we have the current which is

35 amps so we got to solve for b so it's just going to be 75 ID

35 and so B is equal to 0214 Tesla so anytime you have the force per

unit left or Newtons per meter make sure you understand that it's F / L the entire thing is f

l so you might see that expression in this chapter in a few problems now what is the

direction of the magnetic force we know the current flows West the magnetic field is South so the force is

either into the page or out of the page so what you want to do is you want to direct your Four Fingers south and

your thumb West and the force should come out of the

page so let's see if I can draw that so you want your thumb facing this way and the four before fingers facing

this way using your right hand if you do that I put inst the page this should be uh

out of the page the force should come out of the palm of your hand so make sure the

current is aligned with your thumb the magnetic field is aligned with your four

fingers and then the force should come out of the Palm beh hand that's out of the

page now let's say if we have have a rectangular metal loop with a current that flows in the

metal Loop clockwise so in this section the current's going up here it's going down

here it's directed towards the right and here it's directed towards the left and only a portion of this Loop is inside a

magnetic field that is only the bottom portion and let's say the magnetic field is

directed into the page so I'm going to put a x everywhere where

will this rectangular Loop move if initially it's at rest will it move towards the right will it begin moving

towards the left up or down what would you say so let's start with this portion of

the wire or of the the metal Loop let's see what the magnetic force on that portion

is directed so what you want to do is you want to place your thumb facing south

you want the four fingers of your hand to be going into the page and the magnetic force should be D Ed E so let's

see if I can draw that so you want your four fingers going into the page you want

your thumb going south so make sure the magnetic field or your forefingers is going into

page and your thumb is in the direction of the current and the force should come out of

your hand and that is out of the palm of your right hand and it should be directed

East if you do it correctly so make sure you try that and make sure you can Master this right hand

rule now for the other side the left side everything is the same except the current because the current is in the

opposite Direction the force has to be in the opposite direction now these two forces are equal

in magnitude and because they're opposite in direction they will cancel out so the loop is not going to move

towards the left or towards the right these forces balance each other out now in the top part of the loop

there's no magnetic field in that region so therefore there's no Force the N force is going going to be

based on this portion of the loop because it's not balanced by this portion of the loop if the entire Loop

was in a magnetic field all the forces will cancel but since it's not this one will create a net

force so now let's draw another picture Point your thumb towards the left and make sure your four fingers are going

into the page if you do it correctly the force should be direct itself so what you want to do is you

want your thumb directed uh West and you want your four fingers going into the

page so if you do it correctly the magnetic force should be coming south out

of the palm of your right hand now let's move on to another topic we talked about how to calculate the

magnetic force on a current carrying wire but what about the magnetic force on a single point

charge because any moving charge will create a magnetic field so let's go ahead and come up with an equation let's

start with this equation f is equal to I lb sin Theta current is defined as the amount of

electric charge that flows per unit time and capital Q is the total amount of charge it can be due to many charge

particles so capital Q is going to be equal to lowercase Q which is the magnitude of each charged particle times

n which is the number of charged particles and that will give you the total charge divide by

T now distance is equal to the speed multipli the time and length can be thought of as

distance they're both measured in meters so we can replace L with VT so let's replace

I with qn/ t and let's replace l with

VT so we can cancel T now if you want to find a magnetic force on a single point charge that

means there's only one charge particle so n is one so when n is one we can get rid of

it so this leaves us with f is equal to bqv sin Theta so that's how you can

calculate the magnetic force on a moving charge particle when it's inside a magnetic

field so let's say if we have a proton and let's say it's moving towards the right and also the magnetic field is

directed towards the right if these two are parallel sin0 is equal to Z so there's

going to be no magnetic force Force they have to be perpendicular now if the proton is

moving at an angle relative to the magnetic field then you can use the equation f is

equal to bqv sin Theta where Theta is the angle between

the magnetic field and the velocity Vector now now let's say if the proton it's moving perpendicular to the

magnetic field that is they're at right angles or 90° relative to each other sin 90 is

one so the magnetic force will have its maximum value at this point and it's equal to Simply

bqv now let's say if we have a proton and it's moving towards the

right and the magnetic force or rather the magnetic field is directed North what is the direction of the magnetic

force it has to be in a z direction if the velocity is in the X Direction and if the magnetic field is in the y

direction the magnetic force have to be in a z direction and you can use the right hand rule

think of velocity as the current so to speak that's where the charged particles are

moving so you can use the the same right hand rule as we've been doing you want your four fingers to be in the direction

of magnetic field and you want uh your thumb to be in a direction of the Velocity so if you Orient your right

hand this way the force should come out of the palm of your right hand and so it should come out of the

page so I'm going to put a circle for that now for a proton the magnetic force is out of the

page but for an electron the magnetic force will be in the opposite direction that is it's going to be going into the

page so for any negatively charged particle simply reverse the direction of the magnetic force if you need to find

it for a positively charged particle it's going to work in the exact same way as the right hand rule directs it number

five a proton moves East with a speed of 4 * 10 6 m/s in a magnetic field of 2 * 10- 4

Teslas directed into the page what is the magnitude of the magnetic force acting on a

proton so if we need to find a magnitude we don't have to worry about the Direction all we need to know is that

the velocity and the magnetic field are perpendicular to each

other the proton is moving East in the X Direction the magnetic field is directed into the page that is in a negative Z

Direction so therefore the magnetic force has to be in the y direction so all we need to do is find

the magnitude so we just got to use the equation f is equal to bqv sin Theta now because the velocity and the

magnetic field are perpendicular that is the Vel velocity is in the X Direction the magnetic field

is in a z Direction the angle has to be 90° and sin 90 is 1 so f is simply equal to

bqv B is the magnetic field in Tesla that's 2 * 10us 4 Teslas Q is the charge of just one

proton the charge of a proton is 1.6 * 109 kums and that's something you just got to

know and the speed of the proton is 4 * 10 6 m/s so we just got to multiply these

three numbers and so you should get 1.2 8 * 10 - 16

Newtons so that's the magnetic force acting on the proton now let's talk about a

proton so let's say if we have a proton and it's moving towards the right and the magnetic

field is directed everywhere into the page so let's say there's an X everywhere what's going to happen where

is the magnetic force so if you direct your thumb towards the right and your four fingers into the page the magnetic

force will be directed North whenever force and velocity are perpendicular to each other the object

will turn so it's going to go this way if force and velocity are in the same

direction the object will speed up if force and velocity are in the opposite direction the object slows down but if

they're perpendicular the object will turn and eventually the particle is going to be moving in a direction of the

force but now that the velocity is directed North the force is going to change if you use the right hand rule

again if you direct your four fingers into the page the velocity North the force will be directed West and so

what's going to happen is this particle the proton is going to move in a circle I haven't drawn a nice Circle

because I ran out of space but you get the picture when it's at the top the force is going to be

directed South as it moves to the left and as the proton moves South the

force will be directed East so notice that for a moving charge particle the magnetic force behaves as a centripetal

force or a censor seeing Force now what if we have an electron if we had an

electron the situation will be opposite as the electron is moving in the same direction

as the proton it's going to f a force in the opposite direction so Theon felt a force

that was directed North the electron will fill a force directed in the opposite direction that is

South so as the proton moves in the counterclockwise Direction the electron will move in a clockwise

Direction so they will move in an opposite opposite direction now how can we calculate the

radius of curvature that a proton or electron might travel in a circle how can we

figure out the radius if you ever get a question like this what you need to do is set the centripetal force equal to

the magnetic force let's call FB the magnetic force FC the cental force the cental force

is provided by the magnetic force in this particular example the centripetal force is equal to mv^2 / R

which is the radius of the circle the magnetic force is equal to bqv now because the magnetic field and

the velocity are perpendicular we don't have to worry about the sign portion of this equation now what we're going to do

is multiply both sides by 1/ V so on the right side V will cancel on the left side because we have a v^2 one of them

will remain so MV / R is equal to BQ so multiplying both sides by R we have uh this

equation so make sure you uh write down this equation MV is equal to bqr because in this format you can solve

for anything so let's say if we want to find the radius of the circle all we need to

do is divide both sides by BQ so the radius of the circle is equal

to the mass of the charged particle times the velocity ID the magnetic field times the

charge number six a proton moves with a speed of 5 * 10 6 m/s in a plane perpendicular to a magnetic field of 2.5

Tesla calculate the radius of its circular path so here's the

proton and if it's moving perpendicular to a magnetic field it's going to move in a circle our goal is to calculate the

radius of that Circle so R is equal to based on the equation that we had

before it's MV / BQ so what is the mass of a proton now the problem doesn't give it to you which

means you can either look it up online or you can look it up in the reference section of your textbook the mass of a

proton is about 1. 673 * 10us 27 kilg the speed which is given that's 5 *

10 6 m/ second the strength of the magnetic field is 2.5

Tesla and the charge of a proton which is the same as that of an electron but the opposite sign it's

1.6 * 109 colums so if we type these numbers into the

calculator we should get 0209

M which is equal to

2.09 CM so that's the radius of the path that the proton is going to travel in and so

that's how you could find it now what about Part B what is the energy of the proton in electron volts how can we find

the answer to that question well first we need to find the energy in jewels

a moving object has kinetic energy any object in motion contains kinetic energy so we got to find the kinetic energy of

the proton which is 12 mv^2 so we know the mass of a

proton it's 1.67 3 * 10 -27 and we also have the speed 5 * 10 6 m/s squared

so this is equal to 2.09 * 1014

jewles now once you have the energy in Jews you can convert it to electron volts electron volts is basically

another unit of energy it's very useful for small particles like protons and electrons one electron volt

is equal to 1.6 * 10 to19 Jews we'll talk about y later but that's what it's equal to and so you just got to convert

it to electron volts and this will give you the answer which is uh

13,000 and 73 electron volts so that's it for this problem so why is it that one electron

volt is equal to 1.6 * 1019 Jew why is that the case electric potential which is

measured in volts is equal to the electric potential energy which is measured in Jews divided

by the charge so the unit volt one volt is equal to one

Jew per 1 colum so therefore an electron has a charge of

1.6 time 10 to19 cols it's negative but let's ignore the negative sign that's a charge of an

electron and an electron that has one Vol or one EV that's one electron volt an electron with one volt will have an

energy of 1.6 * 1019 Jewels volt is basically the ratio

between jewels and clums so if you have a Charged particle that has one Jewel and one clume its

voltage is one volt or the electric potential is one volt voltage is really work per unit charge

electric potential is energy per unit charge now if we have a charge particle that has an energy of 1.6 *

109s and a charge of that many colums then these two will cancel and the electric potential will be 1 V which is

the case of an electron an electron has a charge of 1.6 * 10 to9 colums and if that electron has an

energy of one electron volt its voltage is one volt which means its energy is equal to this number and that's why one

electron volt is 1.6 * 10 to9 J just in case you're wondering now let's say if we have two

wires parallel to each other now let's say that there's a current in wire one and in wire two and these two

currents are in the same direction will these two wires attract each other or will they repel it turns

out that these two wires will attract each other if they have a current in the same

direction now if there's two wires with the opposite current then the situation is

different they will not feel a force of attraction rather they will repel each other so why does that

happen why do we have a force of attraction in the first case but in the second

scenario these two repel what's going on here yre

one creates a magnetic field because it has a moving charge it has a current and that magnetic field exerts a

force on wire two now let's focus on wire one the current is going north if you use the

first right hand rule that we talked about earlier in the video where you curl your hand around the wire and your

thumb is in a direction of the current the magnetic force created by yre one will be out of the page on the

left side but it's going to be into the page on the right

side so y or 2 is on the right side of Y one therefore y or 2 CES a magnetic field that's going into the page now

using the second right- hand rule that we talked about what you want to

do is you want your fingers to point into the page but you want your

thumb pointing North in the direction of the current so here's your thumb it follows

I2 the magnetic field which is B1 that's created by i1 you want that to be in the

page and what's going to happen is the force is going to come out of the palm of your hand and it turns out that force

is directed towards wire one so it's a force of attraction and so anytime you have two

wires with the current going in the same direction it's going to create an attractive Force the two wires will be

tracked to each other if the current is in the opposite direction then the two wires will repel

each other now how can we calculate the magnitude of the force between the two

wires so we said that the first wire wire one creates a magnetic field that causes the second wire to be attracted

to the first if they're moving in the same direction and also the second creates

magnetic field that exerts a force on the first wire causing the first wire to move towards a second if the currents

are in the same direction so let's start with this equation yre one creates a magnetic

field B1 which is U * i1 over 2i R where R is the distance between the two wires so wire one one which produces a

current one will generate a magnetic field that is at yre two and so R is the distance between the two

wires now we need to use the other equation that is the force that acts on a current carrying wire

that's f is equal to IB sin Theta but we're not going to be worried about the angle in this problem so the the force

on wire 2 which we'll call F2 is equal to the current on that wire times the length times the magnetic field created

by y1 so what we're going to do now is replace B1 with mu0 i1 over 2i

R so it's going to be I2 * L time mu0 i1 over 2i R so that's how you can calculate

the force between the two wires F2 and F1 they have the same magnitude number seven what is the

magnitude and direction of the force between two parallel wires that are 30 m long and 2 cm apart each carrying a

current of 50 amps in the same direction well let's draw a picture so we have two wires

and they have the same current so i1 and I2 both equal 50 amps now we have the length of the wire

which is L and the length of both wires is 30

m and the distance between them which is R that's 2 cm which is equivalent to 02

M now for these type of problems there's only two answers for the direction either the force is attractive or they

will repel because the currents are in the same direction we have a force of attraction and so that answers the

direction of the force it's simply attraction now all we need to do is find the magnitude so let's use the formula f

is equal to mu0 * i1 * I2 * L which is the length of the wires

/ 2 pi * R so it's 4 piun * 10- 7 * 50 * 50 which we can write as 50 2 * L which is 30 m / 2 pi time the radius

of 02 m now 4 Pi / 2 pi is going to be 2 so it's 2 * 10 - 7 * 50 2 * 30 /

02 so the force between these two wires is 75 Newtons and a direction is a force of

raction these two forces will be pointed towards each other now the next thing that we need to

talk about is ampers law ampers law describes the relationship

between the current and magnetic field produced by that current perhaps you've seen this

equation the sum of all the magnetic fields that is parallel to any uh segments that the

magnetic field passes through that's going to be equal to mu0 time the current enclosed by the path

that the magnetic field makes and it has to be a closed path a good example is

using a wire so let's say if we have a current that passes through this wire now this current will create a

magnetic field that travels around the wire in a circular path so if we take the magnetic

field and multiply it by the path that is parallel to it let's call it Delta L we can use that to get an equation

that will give us the magnetic field created by this wire so the path that the magnetic field

travels is basically the path of a circle so Delta L is really 2 pi r if we add up

all the small segments so if we add up this segment plus that segment plus that segment we're going to get the

circumference of a circle which is 2i R so it's going to be B * 2i R which equals mu over I I meantimes

I so the magnetic field if you divide both sides by 2 pi r we get this familiar equation new 0 * I over 2i

R now it's important to understand that the current in this equation is the current that is

enclosed by this Loop so that Loop has an area and the current that passes through that area that's this current is

the current enclosed by that uh circular Loop and so that's how you can use amp's law to get the equation of a magnetic

field by a current carrying wire now let's use amp law to come up with an equation for a solonoid a solonoid is

basically a device with many Loops of wire and a reason why it's at advantages to create a solar noid anytime you

create a loop of wire whenever you have a current the magnetic field that is at the center of the wire is very strong

and for every Loop that you add you increase the strength of the magnetic field inside the wire and so solar noids

are very useful for creating powerful magnetic fields outside of the loop the magnetic

field is weak so the magnetic field will travel in a circular

pattern so to calculate the or to derive the formula for a solar noid we need to create a path of the magnetic field so

let's draw a rectangular path so let's say this is a b c and d so what we need to do is add up the

magnetic field that is parallel to each segment so that's going to be the magnetic field times the length of

segment AC plus the magnetic field times the length of segment

CD plus the magnetic field times the length of segment

DB plus the magnetic field times the length of segment

ba now let's focus on segment BD segment BD is outside of the solenoid and the magnetic field is very weak

outside of the solenoid so we can say that the contribution for uh

BL in segment BD is very small so it's negligible now ba and DC they're perpendicular to the magnetic field that

is inside the solenoid that magnetic field is much stronger than than the magnetic field on

the outside so because it's perpendicular to the magnetic field that's inside the

solenoid its contribution is going to be negligable so we can eliminate ba and

DC so therefore all we have is the segment that is parallel to the magnetic field that is inside the solenoid that's

that's AC segment AC is the most important segment because one it's parallel to the

magnetic field that is inside the solar noid and that magnetic field is the strongest one compared to the ones that

are outside of it and so BL in segment AC will have the greatest

contribution towards the sum of all the magnetic fields that is parallel to each segment

so now using ampers law which is basically this equation we can now replace this term with B * L

where L is simply the length of the Sol noord the length of the Sol no being segment AC but we'll just call it

l so BL is equal to Mu 0 * I but this is four if we only have one Loop let's say if you have a wire with a

current of 10 if you have one Loop then the enclosed current is 10 amps but it turns

out that if you add another loop the current enclosed by the magnetic

field it's going to be twice as much even though 10 amps is flowing through the wire the enclosed current is now 20

and if you add another loop the enclosed current is 30 even though 10 amps is still flowing through the same

wi so therefore we need to add n to this equation because the enclosed current increases so the enclosed current is

basically the current that flows in one Loop times the number of Loops so now all we need to do is divide

both sides by L so B is equal to mu0 * n * I / L lowercase n is equal to capital N

L capital N represents the number of terms or Loops L is the length in

meters so lowercase n is the number of Loops or turns per meter so we're going to replace capital

N over L with lowercase n so now we have the equation of a Solar noord B is equal to Mu 0 * n *

I so the magnetic field produced by a solenoid is proportional to the current that passes through it as the current

increases the strength of the magnetic field will increase the second way to increase the

uh magnetic field is to increase the number of turns if you can increase the amount of

turns per unit left or per meter the strength of the magnetic field will greatly

increase so you want to increase the number of turns you want to increase the current but you also want to decrease

the length if you can decrease the length the magnetic fi will increase as well so the magnetic field is directly

related to the current it's related to the number of turns and it's inversely related to

length number eight a solenoid has has a length of 15 cm and a total of 800 turns of wire calculate the strength of the

magnetic field at its Center if the solar Nord carries a current of 5 amps so first let's calculate lowercase

n which is capital N over l so there's 800 turns and the length of the wire is 15

cm but we need to convert that to meters so let's divide by 100 15 ID 100 is 5 or5

m so 800 divided by5 that's equal to

5,333 turns per meter or Loops per meter so now to calculate the strand for

the magnetic field it's mu0 * n * I so it's 4 piun * 10 10 to- 7times the number of turns per meter

which is 5333 times the current which is 5 amps so the strength of the magnetic

field at the center it's about 0335 Tesla and so

that's it for this problem now what's going to happen if we have a current carrying Loop inside a

magnetic field in a magnetic field the loop is going to rotate it's going to produce a

torque so let's say on the left side there's a current and on the right side there a

current so the current travels clockwise in this Loop and it's also a magnetic field that is directed

East now what type of force will we have on the left side of the wire or the metal

Loop so using the right hand rule place your thumb going up and your four fingers in a direction of the

magnetic field and if you do that notice that the palm of your hand opens into the

page and so that's where the force is going to be so the magnetic field will exert a

force on the left side of the loop going into the page therefore on the right side of the loop the current is reversed

so it must be out of the page so therefore this Loop is going to turn it's going to turn this way and then

into the page on the other side now let's talk about how to derive an equation for the torque of this

current carrying Loop f is equal to IB it's I sin Theta but the force is perpendicular both to

the current and magnetic field so sin 90 is one so the force acting on the right

side and on the left side of the loop it's going to be IB there's no force on the top section and the bottom section

of the loop the reason for that is because the current is parallel to the magnetic field and whenever the current

is parallel to the magnetic field there's going to be no magnetic force it has to be perpendicular to it so there's

only a force on the left side and on the right side not on the top or bottom section of the

loop so since we're focused on this side that's where the magnetic force will be exerted on L represents the length of

that section so that's this l b is the magnetic

field the width of the loop let's call it w let's put that

here now we need to calculate the torque this force will create a torque and this force will create another

torque that's additive so the net torque is going to be T1 plus

T2 and just to refresh you on how to calculate torque torque is the product of the force times the L arm so it's f *

R and if the force is at an angle it's going to be f r sin thet so T1 and

T2 it's going to be fub1 * R1 * sin theta plus F2 * R2 sin Theta so what exactly is r in this

example we know that R is the distance between where the force is applied and the axis of rotation that's where the

object moves around so in this particular problem the axis of

rotation is here so which means that R is half of

w so r is W / 2 so now we can replace f with

IB and we can replace r with w/2 and then T2 is going to be the same thing it's the same current times the

same length they both have the same length they're both exposed to the same magnetic field and R is the same R1 and

R2 is the same so we no longer need this picture so what I'm going to do is I'm

going to factor out IB and sin Theta so T is IB sin

Theta W over 2 + W over 2 half plus half is a whole so T is equal to i l w b sin Theta now for the rectangular Loop Loop

that we have which has a length

l and a width W the area of the loop is basically the length times the width so a is L * W so

let's replace LW with a so the torque of a single Loop is the current multiplied by the area times the strength of the

magnetic field Time s of theta now what if we have many Loops if you have two Loops the torque

is going to be twice as strong three Loops three times as strong so we need to add n to this equation for the number

of Loops so it's n i AB sin Theta by the way the quantity niia is known as the magnetic dipole

moment represented by capital M so it's equal to the number of loops time the current time the area that's

the Magnetic diap Moment but this is the equation that we need to calculate the torque it's ni a b

sin Theta now let's say if this is the face of the

loop let's draw the normal line perpendicular to the face of the loop so this is the area of the loop

which we could describe it as a and the red line is the normal line that's perpendicular to

a and sometimes the magnetic field it's not going to be parallel or perpendicular to the

surface it can be an angle the angle Theta is between the magnetic

field and the normal line which is perpendicular to the surface of the coil to make sure you

understand that so Theta is between B and the normal line now instead of drawing a side view

of the current carrying Loop let's analyze it by drawing the top view so

here's this is the top view of it and here's the rest of the loop if you wish to see it this

way and this part is perpendicular to the plane of the loop let's say the magnetic field is at

a angle of 90° when it's at an angle of 90° you're going to get the maximum torque

possible so make sure you understand that but I'm going to rraw like this so let's say B is directed

East we're going to have one force going up and the other Force going down so this part is perpendicular to

the area or to the the surface of the plane of the loop I should say and here's the magnetic field so

it's at an angle of 90° and so you're going to get the greatest torque in a situation

that's when the magnetic field is perpendicular to the normal line which means it's

parallel to the face of the coil now let's dra another picture when it's at an

angle so here is the normal line which is perpendicular to the face of the coil and let's draw the magnetic field

which is directed East so this time the angle Theta is less than 90 here Theta is equal to

90 now fub1 and F2 they're still directed uh North and South but they're much less than the

original values so the torque is less and let's see what happens when the angle is

zero and in this case the normal line is parallel to the magnetic field so therefore the angle between it two is

zero and when this happens the magnetic field is perpendicular to the face of the coil it

passes through the coil and so there's going to be no torque whenever you have this

situation now let's understand why there's no torque if you look at the two forces F1 and

F2 notice that they're parallel to the lever arm anytime you have a force that's parallel to the

lever arm it cannot create a torque the only way a torque can be created if it's perpendicular let's say this is the top

view of a door here's the lever arm R the only way for you to push a door is to apply

Force perpendicular to to the L arm and the door is going to turn if you try to push a door from this

side it's not going to move it won't turn and that's what's Happening Here the axis of rotation is here and the way

the forces are oriented since they're parallel to the Lev arm which is R there's going to be no torque there's no

rotation and so that's why whenever the angle is zero S zero is zero so the torque is going to be zero

so the maximum torque occurs whenever the angle is 90 that is the angle between the magnetic field and the

normal line so whenever the magnetic field is parallel to the face of the coil you're going to have maximum torque

whenever it's parallel to the face of the coil it's perpendicular to the normal line the angle is 90 and whenever

the magnetic field passes through the coil that is when it's parallel to the normal line there's going to be no

torque the angle now notice what happens let's just erase a few

things I think I'll keep that if we look at the first picture on the left F1 will create a torque that

will cause it to rotate in the clockwise Direction and F2 will also create a torque that will

cause CA the system to rotate in the clockwise Direction now once it rotates it's going

to move towards picture two it's going to look like this and as you can see F1 will still create a torque that will

cause the system to rotate clockwise and the same is true for F2 eventually the system is going to

reach this point and F1 and F2 will still be directed North and South which we see that in the third

diagram but once we reach a third diagram where the angle is

zero then the system is at equilibrium the two forces they're opposite and they're equal so they

cancel out there's no net force in fact there's no net force for all pictures in the first one these two forces they

balance each other out however there is a netork for the first and second picture but in the third diagram the net

torque is zero this Force creates no torque and the same is true for the second one so the net torque and the net

force is zero for the third diagram which means it's at equilibrium so basically the loop moves from this

position from an angle of 90 and the magnetic field causes it to rotate to an angle of zero and then it

stops now let's work on some problems number nine a circular coil of wire has a radius of 30 cm and contains 50 Loop

Loops the current is 8 amps and the coil is placed in a magnetic field of five Tesla what is the maximum torque exerted

on a coil by the magnetic field so if we draw a picture we're going to have a a coil of wire that is circular and it

has many Loops 50 loops and so it's going to look

something like that and the radius is 30 cm to calculate the

torque it's equal to n i AB sin Theta now if we wish to find a maximum torque then the angle is going to be 90

and sin 90 is equal to 1 so n is the number of Loops n is 50 the current is 8

amps to find the area it's going to be the area of a circle which is PK R 2 so it's Pi * the radius

squ the radius is 30 cm but we need to convert that to meters 30 cm is3 M so it's Pi

* .3^ s so that's the area the magnetic field is 5

Tesla and sin 90 is 1 so let's start with the area 3^ 2 *

pi which is 9 pi over 100 that's 2827 if we multiply that by 5 * 8 * 50 you should get

5655 Tesla so don't forget to square3 and so that's it for this

problem actually the unit is not Tesla that's for the magnetic field the unit for torque is Newtons time

met so this is it number 10 a rectangular coil contains 200 loops and has a current of 15 amps

what is the magnitude of the magnetic field required to produce a maximum torque of 1200 newtimes

met let's draw a picture and so here's a rectangular Loop and there's 200 Loops but I don't want to

draw it 200 times the area is going to be the width multiplied by the

length and so we have the dimensions so the torque is going to be equal to the number of Loops times the

current times the area times magnetic field and since we're looking for or since we're using the maximum torque the

maximum torque occurs at an angle of 90° so this is going to be sin 90 now in this problem the area of a

rectangular coil is going to be the length times the width and sin 90 is 1 so the maximum

torque is let's put this on the left it's 1200 Newtons time m n is equal to 200

Loops the current is 15 amps and then the area the length times the WID we need to convert 40 cm to met

so that's 4 m time5 m and the magnetic field we're looking

for B so we got to solve for it so let's multiply 200 * 15 * 4 * .5 so that's equal to

600 and so now all we need to do is divide both sides by 600 so these two cancel 1200 over 600 we

can cancel the two zeros so it's 12 ID 6 and that's two so the magnetic field is two

Tesla and so that's the answer so that's it for this video um if you liked this video feel free to subscribe

you can check out my channel for uh more physics videos you can look out for my physics video playlist if you want to

find them all so thanks for watching and uh have a great day