At first I didn't realize you were deriving the formula and thought "this looks so complicated!" but then you got to the simple final formula. Thank you for explaining the derivation so well! My professor isn't requiring my class to know it, but this was super helpful and interesting anyway! I love when a derivation is explained so clearly with such detail!
You are welcome! That is correct. The key to learning is to UNDERSTAND the equations, not just memorizing them. (Although it is also good if you can memorize them).
dear sir when calculating for torque shouldn't dt be Rsin(theta)? if we take the angle between vector r and vector dfx a pi then pi is pi=360-theta thus sin(pi)= sin(theta) making the magnitude of torque with sin(theta). what am i missing?
for whom didn't understand this video, you should watch this video again. Because the angle and the point of view of the shape probably misguiding you. Drawing is not that proper so you should try to imagine the location of the loop with respect to the magnetic field.
Dear sir 4:03 you explained that dTorque = dFx × R cos(θ). I still don't understand with this because something that i know is Torque is the cross product between force and distance. But in your explanation is using cos(θ) instead of sin(θ). What am i missing?
hmm i think that when you perform the derivative operation to move from "d" torque to simply torque this will be resolved. He defined the length of the X component as R cos(θ) and if you look to the top right corner of the board he has the definition of torque shown with the correct formulation. So i think dont worry that the derivation has cos(θ) for several steps before coming back to the familiar form
Essentially you have to take each tiny piece and calculate the vertical distance (as you indicate) and the horizontal distance and then add them all up. The vertical distance (when adding up all the pieces) will be equal to the diameter of the circle.
@@MichelvanBiezen now I understand you I confused first because you didn't take two forces But finally I realized that you multiple dt by 2 So I wanna say thank very much ❤🌸😅
the lorentz force is formally for magnetism: q(dL x B), and i was wondering, because in this problem, you set the current loop vertical, so each of the dL's has a different angle with relation to B; at the top of the ring, it's 90 degrees, and at the sides of the rings, it's exactly the tilt of the ring.
hello Professor, can you give a quick explanation how do you decide to put df dT? Few videos before I saw you write dw=Fdtheta. Why do not put dF as well?
Hello sir Sir if we have to find the torque experienced by a current carrying semicircular ring in a uniform magnetic field about one of its EDGE(or free end corner) , how do we find it.sir please help.
Hi Michel, I have a question about the previous video and this video. When you did the force in the semi-circle wire carrying current you said that only the y component of F will be effective because the x component will cancel out but in this video, we are just saying that the x component of F is effective. Could you please explain more on that? I don't understand the difference. Thank you for your amazing videos
It turns out that the torque depends on the loop's area and not its shape. If the number of turns are the same, the torque will be the same if the area of the circular and square loops are the same.
First sir with due respect...I don't think that there will be any torque acting on the loop because the direction of magnetic moment and magnetic field is parallel so according to the formula τ=M×B ...it comes out to be zero ...
You cannot look at the equation, without drawing a diagram of what is happening and then referencing the angle on the diagram. Take a careful look at the angles that are referenced and you will see why the equations are written as they are.
The distance is wrong, the perpendicular distance is Rsintheta, the force is acting horizontally. The distance should be perpendicular to the force not parallel or in the same direction. The distance is horizontal, the force is horizontal too, so I am sure that this is not true as a man who got A in statics
It depends on where the axis of rotation is. In this case the axis is vertical and goes through the middle of the loop. It doesn't matter what the shape of the loop is, only the strength of the magnetic field, the cross sectional area of the loop, and the relative direction of the magnetic field and the normal to the loop area, determine the magnitude of the torque.
The only wrong step was to suppose that the total torque is 2 times the torque in the right. It isn't because if you look at the torque generated in the left is opposite to the torque generated in the right
In this orientation, the torque is indeed zero as the force is directed in the plane of the loop. But that was not the purpose of this video. It is to show that the maximum torque (independent of the orientation) is IBA (and independent on the shape of the loop).
+leen dagher Look in the playlist: PHYSICS 43 MAGNETIC FIELDS AND MAGNETIC FORCES All physics videos are placed in playlists that have been numerically ordered per topic as text books would cover them.
At first I didn't realize you were deriving the formula and thought "this looks so complicated!" but then you got to the simple final formula. Thank you for explaining the derivation so well! My professor isn't requiring my class to know it, but this was super helpful and interesting anyway! I love when a derivation is explained so clearly with such detail!
It's amazing how a lot of these complex problems simplify into simple equations.
Best video on UA-cam
Thank you. Glad you found it.
Thanks for showing us how to derive the equation. It is the difference between memorization and learning.
You are welcome! That is correct. The key to learning is to UNDERSTAND the equations, not just memorizing them. (Although it is also good if you can memorize them).
Sir van biezen , you video is splendid. It clear my concept
dear sir
when calculating for torque
shouldn't dt be Rsin(theta)?
if we take the angle between vector r and vector dfx a pi then pi is
pi=360-theta
thus sin(pi)= sin(theta) making the magnitude of torque with sin(theta). what am i missing?
look at other comments below. You will find the answer.
Hey Hey Hey, wait a minute. Should we use the Left Hand Rule here and not the Right Hand Rule? I mean this is an example of the motor effect no?
The standard assumption (by convention) is that the current is positive (positive charges moving).
for whom didn't understand this video, you should watch this video again. Because the angle and the point of view of the shape probably misguiding you. Drawing is not that proper so you should try to imagine the location of the loop with respect to the magnetic field.
Dear sir
4:03 you explained that dTorque = dFx × R cos(θ). I still don't understand with this because something that i know is Torque is the cross product between force and distance. But in your explanation is using cos(θ) instead of sin(θ).
What am i missing?
hmm i think that when you perform the derivative operation to move from "d" torque to simply torque this will be resolved. He defined the length of the X component as R cos(θ) and if you look to the top right corner of the board he has the definition of torque shown with the correct formulation. So i think dont worry that the derivation has cos(θ) for several steps before coming back to the familiar form
It depends on the reference angle. See the torque videos in the mechanics section
hi mr Michel van Biezen
i have a question
why the Vertical distance Is not equal 2 R cos (theta)
Essentially you have to take each tiny piece and calculate the vertical distance (as you indicate) and the horizontal distance and then add them all up. The vertical distance (when adding up all the pieces) will be equal to the diameter of the circle.
@@MichelvanBiezen but the diameter of the circle = 2R
@@MichelvanBiezen ohh u took a one force i thougt that u took another focre with the first so i sad the Vertical distance is 2R cose(theta)
thank you
@@MichelvanBiezen now I understand you
I confused first because you didn't take two forces
But finally I realized that you multiple dt by 2
So I wanna say thank very much ❤🌸😅
what happened to the cross product of dL x B in the dF_x equation? the angle between the dL vector and B shouldn't be 90 degrees at all times.
Sorry, but I am not sure what you are asking. What part of the process shown are you not sure about?
the lorentz force is formally for magnetism: q(dL x B), and i was wondering, because in this problem, you set the current loop vertical, so each of the dL's has a different angle with relation to B; at the top of the ring, it's 90 degrees, and at the sides of the rings, it's exactly the tilt of the ring.
@@NotLegato I am also thinking about that. Torque on a circular loop seeming extremely complex.
I think then that will be
dFx=IdLBsin(theta)cos(theta)
In this scenario there is no torque because the magnetic momentum of the circular current loop is already parallel to the magnetic field B
sir have u posted the lecture on the tension produced in a circular loop carrying current. in a uniform magnetic field .if yes plz tell the link
Not tension, but torque yes. See this playlist: PHYSICS 43 MAGNETIC FIELDS AND MAGNETIC FORCES
hello Professor, can you give a quick explanation how do you decide to put df dT? Few videos before I saw you write dw=Fdtheta. Why do not put dF as well?
Neat and clean
Thank you 😊
If the question asked for a magnetic field created by a current Do we just solve the equation in the video for B?
Yes, if the geometry is the same as your problem. (Otherwise look at the other videos for examples.)
Hello sir
Sir if we have to find the torque experienced by a current carrying semicircular ring in a uniform magnetic field about one of its EDGE(or free end corner) , how do we find it.sir please help.
Can we use torque=magnetic moment×B?
You would need to integrate. Take a small length (dl) and find its torque (dT) and then you integrate around the half loop.
Thanks a lot sir.
Hi Michel,
I have a question about the previous video and this video. When you did the force in the semi-circle wire carrying current you said that only the y component of F will be effective because the x component will cancel out but in this video, we are just saying that the x component of F is effective. Could you please explain more on that? I don't understand the difference.
Thank you for your amazing videos
Watch video # 8 in this playlist: PHYSICS 0.5 STANDARD UNITS IN PHYSICS
If the square loop is changed into a circular shaped loop with same number of tums, will there be a change in torque? Will it increase or decrease?
It turns out that the torque depends on the loop's area and not its shape. If the number of turns are the same, the torque will be the same if the area of the circular and square loops are the same.
First sir with due respect...I don't think that there will be any torque acting on the loop because the direction of magnetic moment and magnetic field is parallel so according to the formula τ=M×B ...it comes out to be zero ...
The torque is not zero. The direction of the current and the direction of the matnetic field are perpendicular to one another.
why d(perpendicular) is not R Sin(theta), Sir? thank you
d(perpendicular) is adjacent to the angle theta. cos(theta) = adjacent side / hypothenuse
Hi. Thank you for videos. Before 2 video (13 of 26 ) you said dF = I.dl.B.sin(theta) but now it turns dF = I.dl.B.cos(theta). Why ?
You cannot look at the equation, without drawing a diagram of what is happening and then referencing the angle on the diagram. Take a careful look at the angles that are referenced and you will see why the equations are written as they are.
Michel van Biezen Thanks. I will look at more carefully
A big thanks
why is the perpendicular distance for the torque Rcostheta? shouldnt it be Rsintheta?
It depends on the location of the angle.
Michel van Biezen but isnt Rsintheta along the line of dFx? Shouldn’t it be perpendicular to dFx?
The distance is wrong, the perpendicular distance is Rsintheta, the force is acting horizontally. The distance should be perpendicular to the force not parallel or in the same direction. The distance is horizontal, the force is horizontal too, so I am sure that this is not true as a man who got A in statics
The distance is correct. Look carefully at which angle we are using to find the perpendicular distance.
@@MichelvanBiezen the force is pulling in the x axis, the distance should be vertical? Otherwise this a sphere that is 3 dimensions
i thought that net torque should be zero since all the force vectors cancel out.. this is very strange and complex and not intuive for me. :(
It depends on where the axis of rotation is. In this case the axis is vertical and goes through the middle of the loop. It doesn't matter what the shape of the loop is, only the strength of the magnetic field, the cross sectional area of the loop, and the relative direction of the magnetic field and the normal to the loop area, determine the magnitude of the torque.
@@MichelvanBiezen I got it. Thank you so much for your speedy reply. Best regards!
Isn't d perpendicular wrong. D perpendicular is parallel to Fx.
d perpendicular represents the perpendicular distance from the axis of rotation.
The only wrong step was to suppose that the total torque is 2 times the torque in the right. It isn't because if you look at the torque generated in the left is opposite to the torque generated in the right
The net torque is zero . I t contradicts your video 11 of 26 (The first image). I am totally sure .
In this orientation, the torque is indeed zero as the force is directed in the plane of the loop. But that was not the purpose of this video. It is to show that the maximum torque (independent of the orientation) is IBA (and independent on the shape of the loop).
@@MichelvanBiezen Oh . Thanks for your answer !!! , now its all clear . I see all your videos btw .
greetings from Peru
Hello,
where can i find the rest of the videos up to 26?
+leen dagher
Look in the playlist:
PHYSICS 43 MAGNETIC FIELDS AND MAGNETIC FORCES
All physics videos are placed in playlists that have been numerically ordered per topic as text books would cover them.