Yeah I used to _hate_ it when the prof would write "Left To Students" on the board for us to figure out the nearly 8000 steps he skipped. For this problem, even the textbooks would skip over dozens of math steps. To be honest, it took me almost an hour to figure out this problem without any shortcuts, and another 6 hours to plan, record, and edit the video. I'm glad I did though 'cuz it seems to help lots of people. Best of luck to you!
+Sandor Fogassy Thank you for your comment! I truly appreciate it! I'm testing a new approach to recording these videos, and I'll be sure to use more color-coding when substituting in the new videos.
A little fast but easy to listen to and entertaining enough to keep you interested, with a few views over and some good notes this should definitely help me :D
I just wanted to point out that this part: (1/a) * [ sin(theta1) + sin(theta2) ] is also a useful integral in computing the y-component of the Electric field at the same point.
I am missing the why in this process. Why do you go through each step like this? If I were to see a problem like this, how would I know to proceed in this way?
You're most welcome! This is a tricky problem, with lots of ways to accidentally make the math harder depending on how you set it up. Good luck with your course!
If you follow the right hand rule, the direction of B is k at one point on the wire's plane. To get the direction at every point, shouldn't you define and angular unit vector instead of k?
Great question! This comes up a lot. The direction of the magnetic field comes from ds X r hat by definition. In this setup, ds is always along the x axis (so i hat) and r hat is always in the xy plane so r hat only has i hat and j hat components. Taking ds X r hat then gives (i hat) X (i hat + j hat) which works out to k hat at every point along the wire. Thinking about it in terms of everyday objects, if you place a straight wire on a horizontal page, with current going to the right, then the current (ds) is in the plane of the page, and the r hat vector is also always on the plane of the page. Unless the wire is infinitely long, r hat will never be parallel to the wire, and then the only direction perpendicular to both ds and r hat is perpendicular to the page - no matter what point you're considering on the wire. Since B is perpendicular to both ds and r hat, B must be in either the +k hat direction or the - k hat direction -- and that depends on whether you choose a point above (+y) or below (-y) the wire (assuming the wire is along the x axis). I hope that helps! Scott
Redmond Physics Tutoring Still, if you choose cylindric coordinates you would get no field in both R and Z axis and the field would be in the angular hat vector.
Guido Diforti I believe you've got it. It's been many years since I worked with cylindrical coordinates, but yes that works if you put z hat along the wire in the direction of current, r hat towards the point of interest, perpendicular to z hat, and then the angular hat vector perpendicular to both z hat and r hat, you would get the direction of the magnetic field. You're right - it would always give you the correct direction of B for any orientation of the wire.
Ampere's law only works for steady currents (where no charge accumulates anywhere in space). In a finite wire, charges accumulates at its ends, making Ampere's Law unsuitable.
On all others videos that solve this problem, the solution multiplier is always the sum of the sines of the two angles. Here, you answer contains [sin(theata_1) + sin(theata_2)], but your first angle is (- theata_1). So, according to other answers, your solution should have [sin(-theata_1) + (sin(theata_2]. Note the "-" sign. I'm no a bit confused...Thanks much
Good questions! 1. ds or dx is arbitrary - it doesn't really matter. They just specify a tiny length along an axis. 2. I played around with different approaches and this was the simplest derivation I could find at the time. If you have time, I'd encourage you to try different derivations so you can get a sense of the tradeoffs. Good luck!
I'm not 100% sure I understand what you're asking... When I set up the angles for this solution, I tried quite a few different approaches before choosing what I presented here. IMHO, there's a bit of an art to setting up these problems, and the best way to understand it is to take the time to try other approaches yourself and see what you prefer. Good luck!
For this one I used Gimp (free) for the drawing and Camtasia for screen recording and video editing. Later I switched from Gimp to Ink2Go which is simpler but easier to use for switching pen colours.
This was one of my earliest videos, and I used GIMP for the visuals with a bunch of layers for what look like slides. I used Camtasia Studio for screen recording and video editing, and I used Audacity to record and edit the audio. Later I replaced GIMP with Ink2Go because I found it easier to use for screencasting, especially for rapidly changing the pen colour.
This was an informative video which I will study more closely. However, what I really want to know is the Force between two finite current carrying wires. I may be able to struggle through and figure it out but do you have an example of how to solve that problem? I have found at least one solution online that I do not think is correct.
Everything in this video appears correct up to the final evaluation at the end. Just a minor correction needs to be made to end result for the integral's result. The evaluation of the integral results in the difference of the sin(theta) terms rather than the sum. This example is also resolved in "Introduction to Electrodynamics" (4th edition--David J. Griffiths) on page 225.
+Maverick Acoustics Solutions You're right, and that _is_ actually considered in this video but I skipped that step. It should be "sin(theta2) - sin(-theta1)" inside the square brackets, but since sin(-x) = -sin(x) that simplifies to "sin(theta2) + sin(theta1)". I try hard not to skip steps, but every now and then something like this sneaks by. When I was putting this solution together, I found it was a bit of an art to get a simple integral. Another way to consider this is to look at the magnetic field contributions from each segment of wire - the segment from the y-axis to the left end and another segment from the y-axis to the right end. Each of these 2 segments would contribute to a magnetic field coming out of the page, so I expect that these two portions _should_ add up. I don't think any correction is necessary, but please let me know if you disagree. Thanks for your comment! Scott
Nope - the dot product uses cosine of the angle between the two vectors; the cross product uses the sine of the angle. Here it is on Wikipedia: en.wikipedia.org/wiki/Cross_product#Definition
It's a unit vector and has been defined with a magnitude of 1 for convenience. This way we can multiply it to specify a direction without affecting anything else in the problem. For more info, I'd suggest starting with something like this: physics.bu.edu/~redner/211-sp06/class03/comp_vectors.html There's a wikipedia page on unit vectors, but it goes into way more detail than we need here. Thanks for the question, and good luck!!
At time-stamp 3:23, when you introduced the 2nd theta, shouldn't these parameters actually be distinct. In other words--would it be correct, instead, to use theta1 AND theta2 at this introduction point. The reason I ask is that these two angles don't appear to necessarily be equal here (likewise for the "r'"s). I want to under the Biot-Savart law fully, so please forgive the picking.
Hmm... The point I was trying to make is that the substitution of sin(gamma) = cos(theta) is valid for the entire length of the wire, which lets me continue the derivation as a single expression. The two angles are _not_ necessarily equal (and neither are the "r"s); in fact, theta ranges from 0 to its maximum in each direction. At 3:23, the green thetas and rs merely correspond to some arbitrary point on the wire. Using theta1 and theta2 _would_ be correct. It would split the derivation into two pieces, from each end of the wire to the point on the wire that's closest to P, but there's nothing wrong with that! In fact, it might lead to a cleaner derivation. Nice suggestion!
I don't see any reason why you can't use sin(gamma) for the entire range. I don't see why you'd need to split it up. Perhaps the trig substitutions wouldn't be as nice?
+Arabinda Biswas I believe that the short answer is to use -theta for the angle of the end closest to P. This is what you'd get if you split it into 2 parts: first, extend the wire so it stops at the closest point to P and integrate from the far end. This wire is longer than the actual wire, and the resulting magnetic field at P will be stronger than the actual magnetic field. In the second step, find the magnetic field from the extra section you added in part 1 --- and then _subtract_ that magnetic field from what you found in part 1. This is a neat question - thank you!! Maybe I'll create a video of it to show more clearly how/why it works...
+Arabinda Biswas Thanks again for the question! I used it as the basis for my latest video, which explains it better than I can in words: ua-cam.com/video/oCfM-a1aE24/v-deo.html
Arabinda Biswas Cool! I'd love to visit India some time, maybe when my kids are a bit older... In the meantime, if you have any questions or topics for a future video, just let me know and I'll work it into my schedule.
You can write it as Sin(theta2) - Sin(theta1), if you just make sure to explain that when substituting theta1 it should be a negative value. After all, we're measuring the angle relative to "Down" in this case. Or, better yet, to be consistent with your presentation, You could write: [sin(theta2) - sin(-theta1)] It's nice how sin (-x) = -sin(x)
Arkadaş devamlı Tetha’ya takmış Durmuş, bir adam akıllı bir şey öğrenelim diye geldik. You just keep traveling around the Tetha, we came here to learn something not your f...ing tetha equations!
Dear o dear, you only calculated the length of a piece of wire! You must have buried your physical intuition some years ago, right? Using this stupid unit-vector in the B-S Law in this case is just a foolish thing to do.
Haha, you're not alone in commenting on this. I'd suggest using a slower playback speed in your UA-cam settings and/or pausing the video more often. Best of luck to you!!
@@RedmondPhysicsTutoringVideo thanks sir but it was just a suggestion because all you have to know is that you are the best in making things clear and easy ..but note that the majority of your channel followers are the beginners in their Faculties or whatever .. so you have to slow down so that we can able to install the logic of the particular problem you solving as many as we can
In depth descriptions, exceptional pacing, considerate formatting, with a touch of humor; brilliant work!
Bobbac Kashani Thanks for the feedback!
and at last brilliant comments like this.
I love that you didn't leave any of the math concepts out. it truly help me follow along better.
Yeah I used to _hate_ it when the prof would write "Left To Students" on the board for us to figure out the nearly 8000 steps he skipped. For this problem, even the textbooks would skip over dozens of math steps. To be honest, it took me almost an hour to figure out this problem without any shortcuts, and another 6 hours to plan, record, and edit the video. I'm glad I did though 'cuz it seems to help lots of people. Best of luck to you!
Thank you very much for explaining how this equation is derived from the Biot-Savart law.
My god, it took me 4 hours trying to solve one exercise almost exactly like this and I still couldn't. Finally I went here and did it in 7 min.
Thank you for sharing your insight into the Biot-Savart law with us :-)
You also have a very nice profile picture. It clearly shows your interest in the field of mathematics, and how it relates to the universe.
Best video on this. Good job switching colors when substituting.
+Sandor Fogassy
Thank you for your comment! I truly appreciate it! I'm testing a new approach to recording these videos, and I'll be sure to use more color-coding when substituting in the new videos.
A little fast but easy to listen to and entertaining enough to keep you interested, with a few views over and some good notes this should definitely help me :D
+CruznKnFuzn
Awesome! Thanks for your comment, and good luck with your course!
Yo this guy makes me want to become a master in Physics
I just wanted to point out that this part: (1/a) * [ sin(theta1) + sin(theta2) ] is also a useful integral in computing the y-component of the Electric field at the same point.
I am missing the why in this process. Why do you go through each step like this? If I were to see a problem like this, how would I know to proceed in this way?
Yo thank you so much, this is incredibly helpful
why I can't use ampere's law in that problem?
I could finally understand it, thank you
You're most welcome! This is a tricky problem, with lots of ways to accidentally make the math harder depending on how you set it up. Good luck with your course!
Really love and appreciate the beating the dead horse with the angles. Angles confuse everyone and no one ever knows how to ask
You're awesome and you've saved me.
If you follow the right hand rule, the direction of B is k at one point on the wire's plane. To get the direction at every point, shouldn't you define and angular unit vector instead of k?
Great question! This comes up a lot. The direction of the magnetic field comes from ds X r hat by definition. In this setup, ds is always along the x axis (so i hat) and r hat is always in the xy plane so r hat only has i hat and j hat components. Taking ds X r hat then gives (i hat) X (i hat + j hat) which works out to k hat at every point along the wire.
Thinking about it in terms of everyday objects, if you place a straight wire on a horizontal page, with current going to the right, then the current (ds) is in the plane of the page, and the r hat vector is also always on the plane of the page. Unless the wire is infinitely long, r hat will never be parallel to the wire, and then the only direction perpendicular to both ds and r hat is perpendicular to the page - no matter what point you're considering on the wire. Since B is perpendicular to both ds and r hat, B must be in either the +k hat direction or the - k hat direction -- and that depends on whether you choose a point above (+y) or below (-y) the wire (assuming the wire is along the x axis).
I hope that helps! Scott
Redmond Physics Tutoring It did. Thanks a lot!
Redmond Physics Tutoring Still, if you choose cylindric coordinates you would get no field in both R and Z axis and the field would be in the angular hat vector.
Guido Diforti I believe you've got it. It's been many years since I worked with cylindrical coordinates, but yes that works if you put z hat along the wire in the direction of current, r hat towards the point of interest, perpendicular to z hat, and then the angular hat vector perpendicular to both z hat and r hat, you would get the direction of the magnetic field. You're right - it would always give you the correct direction of B for any orientation of the wire.
Thank you very much, you were too clear.
+Ahmed Raafat It's my pleasure - I'm glad you liked it!
so if the field is in theata^ direction and its value its constant amoung a circle, why cant we use ampers law???
Ampere's law only works for steady currents (where no charge accumulates anywhere in space). In a finite wire, charges accumulates at its ends, making Ampere's Law unsuitable.
On all others videos that solve this problem, the solution multiplier is always the sum of the sines of the two angles. Here, you answer contains [sin(theata_1) + sin(theata_2)], but your first angle is (- theata_1). So, according to other answers, your solution should have [sin(-theata_1) + (sin(theata_2]. Note the "-" sign.
I'm no a bit confused...Thanks much
fyi biot savart law will make it absolute so that it remain
positive
I think this is the hardest formula to master in physics but it looks fairly easy once you have mastered how it works.
Why is it ds instead of dx and why do you convert everything to angles instead of just converting cos theta?
Good questions!
1. ds or dx is arbitrary - it doesn't really matter. They just specify a tiny length along an axis.
2. I played around with different approaches and this was the simplest derivation I could find at the time. If you have time, I'd encourage you to try different derivations so you can get a sense of the tradeoffs.
Good luck!
@@ScottMRedmond Yours is the most common, but after I wrote that, I tried converting the sine to a ratio of distances and it worked that way too.
Great video!
Thank you so much.
+Tal Girhish
It's my pleasure. Thanks for the feedback!
Keep up the good work!
can u give a link when n how the angles becomes positive n negative.
I'm not 100% sure I understand what you're asking... When I set up the angles for this solution, I tried quite a few different approaches before choosing what I presented here. IMHO, there's a bit of an art to setting up these problems, and the best way to understand it is to take the time to try other approaches yourself and see what you prefer. Good luck!
Which software was used by you in making this video ..?
For this one I used Gimp (free) for the drawing and Camtasia for screen recording and video editing. Later I switched from Gimp to Ink2Go which is simpler but easier to use for switching pen colours.
@@RedmondPhysicsTutoringVideo
Thankyou for replying ...and yess you did explained the concept well
what application did you use for doing that? thank you
This was one of my earliest videos, and I used GIMP for the visuals with a bunch of layers for what look like slides. I used Camtasia Studio for screen recording and video editing, and I used Audacity to record and edit the audio. Later I replaced GIMP with Ink2Go because I found it easier to use for screencasting, especially for rapidly changing the pen colour.
This was an informative video which I will study more closely. However, what I really want to know is the Force between two finite current carrying wires. I may be able to struggle through and figure it out but do you have an example of how to solve that problem? I have found at least one solution online that I do not think is correct.
Unfortunately I don't, and I'm no longer working in this area... Good luck!!
Everything in this video appears correct up to the final evaluation at the end. Just a minor correction needs to be made to end result for the integral's result. The evaluation of the integral results in the difference of the sin(theta) terms rather than the sum. This example is also resolved in "Introduction to Electrodynamics" (4th edition--David J. Griffiths) on page 225.
+Maverick Acoustics Solutions You're right, and that _is_ actually considered in this video but I skipped that step. It should be "sin(theta2) - sin(-theta1)" inside the square brackets, but since sin(-x) = -sin(x) that simplifies to "sin(theta2) + sin(theta1)". I try hard not to skip steps, but every now and then something like this sneaks by.
When I was putting this solution together, I found it was a bit of an art to get a simple integral. Another way to consider this is to look at the magnetic field contributions from each segment of wire - the segment from the y-axis to the left end and another segment from the y-axis to the right end. Each of these 2 segments would contribute to a magnetic field coming out of the page, so I expect that these two portions _should_ add up.
I don't think any correction is necessary, but please let me know if you disagree.
Thanks for your comment!
Scott
Isn't the dot product the product of the magnitude of the vectors and the sin of the angle between them? How are you using this as the cross product?
See 1:15
Nope - the dot product uses cosine of the angle between the two vectors; the cross product uses the sine of the angle.
Here it is on Wikipedia: en.wikipedia.org/wiki/Cross_product#Definition
It's a common mix-up though. Much better to sort it out here so you can get full marks on your exam!! :-)
Can I ask a question? Why is the value of |ř| 1?
It's a unit vector and has been defined with a magnitude of 1 for convenience. This way we can multiply it to specify a direction without affecting anything else in the problem.
For more info, I'd suggest starting with something like this:
physics.bu.edu/~redner/211-sp06/class03/comp_vectors.html
There's a wikipedia page on unit vectors, but it goes into way more detail than we need here.
Thanks for the question, and good luck!!
@@RedmondPhysicsTutoringVideo thank you so much for your answer.
thank you. it really helped me. :) easy explanation.
You belong to which country
At time-stamp 3:23, when you introduced the 2nd theta, shouldn't these parameters actually be distinct. In other words--would it be correct, instead, to use theta1 AND theta2 at this introduction point. The reason I ask is that these two angles don't appear to necessarily be equal here (likewise for the "r'"s). I want to under the Biot-Savart law fully, so please forgive the picking.
Hmm... The point I was trying to make is that the substitution of sin(gamma) = cos(theta) is valid for the entire length of the wire, which lets me continue the derivation as a single expression. The two angles are _not_ necessarily equal (and neither are the "r"s); in fact, theta ranges from 0 to its maximum in each direction. At 3:23, the green thetas and rs merely correspond to some arbitrary point on the wire.
Using theta1 and theta2 _would_ be correct. It would split the derivation into two pieces, from each end of the wire to the point on the wire that's closest to P, but there's nothing wrong with that! In fact, it might lead to a cleaner derivation. Nice suggestion!
I don't see any reason why you can't use sin(gamma) for the entire range. I don't see why you'd need to split it up. Perhaps the trig substitutions wouldn't be as nice?
Thank you sir!!
sir if point under consideration sayp is away frm the wire what will be theta
+Arabinda Biswas
I believe that the short answer is to use -theta for the angle of the end closest to P.
This is what you'd get if you split it into 2 parts: first, extend the wire so it stops at the closest point to P and integrate from the far end. This wire is longer than the actual wire, and the resulting magnetic field at P will be stronger than the actual magnetic field. In the second step, find the magnetic field from the extra section you added in part 1 --- and then _subtract_ that magnetic field from what you found in part 1.
This is a neat question - thank you!! Maybe I'll create a video of it to show more clearly how/why it works...
+Arabinda Biswas
Thanks again for the question! I used it as the basis for my latest video, which explains it better than I can in words:
ua-cam.com/video/oCfM-a1aE24/v-deo.html
This is so good
Thanks so much :)
what a good video!!
+Paolo Andreozzi Thanks for the feedback!!
Great scott
Thanks a lot
Thanks
why is sin(gamma)=sin(pi-gamma)
finally no more bullshit equations
really good!
Take your thumb up good man.
Thank you Sir.....this really helped!
very useful || viewers read this with example 5.5 of introduction to electrodynamics , third edition by david j griffiths
Thank you very much :)
sir thanks u live in america sir?
+Arabinda Biswas
You're welcome! I'm actually based in Montreal, just a couple of hours north of the USA.
thanks sir i live inagartala stae tripura india u are wl come to come at my house
Arabinda Biswas
Cool! I'd love to visit India some time, maybe when my kids are a bit older... In the meantime, if you have any questions or topics for a future video, just let me know and I'll work it into my schedule.
why is sin(y) = sin (y - pi)
Check out mathcentral.uregina.ca/QQ/database/QQ.09.10/h/janet3.html
Thank u!
thanks!
thank you so muchhhhhh 😭❤
Glad I could help! Good luck!!
Thank you 🙃😊😀
amazing
Amazing vid, but at the end it should be sin theta 1 minus sin theta 2 not plus.
Edit: NVM, I just realized you called theta 1 negative theta 1.
Yeah, that's something that I don't love about this derivation. But the other approaches I saw for this problem are more complex in other ways...
You can write it as Sin(theta2) - Sin(theta1), if you just make sure to explain that when substituting theta1 it should be a negative value. After all, we're measuring the angle relative to "Down" in this case.
Or, better yet, to be consistent with your presentation, You could write: [sin(theta2) - sin(-theta1)] It's nice how sin (-x) = -sin(x)
how is sin(gamma)=sin(pie-gamma)
try it out with real numbers. let's say gamma = 30 deg. sin(30) = 1/2. now, evaluate pi - gamma: 180 deg - 30 deg = 150 deg. so, sin(150)=1/2. HTH!
thanks
Thank you :)
Arkadaş devamlı Tetha’ya takmış Durmuş, bir adam akıllı bir şey öğrenelim diye geldik. You just keep traveling around the Tetha, we came here to learn something not your f...ing tetha equations!
Pie IS yummy
Definitely! :-) Thanks for watching!
Koi hindi wala hain kya
Dear o dear, you only calculated the length of a piece of wire! You must have buried your physical intuition some years ago, right? Using this stupid unit-vector in the B-S Law in this case is just a foolish thing to do.
what the fuck am i looking at
Very non intuitive, not gonna lie.
you are very fast sir, this can cause the lose of your channel followers...please take it it slow
Haha, you're not alone in commenting on this. I'd suggest using a slower playback speed in your UA-cam settings and/or pausing the video more often. Best of luck to you!!
@@RedmondPhysicsTutoringVideo thanks sir but it was just a suggestion because all you have to know is that you are the best in making things clear and easy ..but note that the majority of your channel followers are the beginners in their Faculties or whatever .. so you have to slow down so that we can able to install the logic of the particular problem you solving as many as we can
Thanks
Thank u!
+Mayara Cordeiro França you're welcome! I'm glad that I could help!!
Thanks