Magnetic Fields / Force / Current

[julianwitkowski]

Homework Statement

A magnetic field of 1.4 T [N] is 4.0 m wide. A very long wire crosses the field.

a) If the current in the wire is 2.0 A [W]. find the magnetic force on the wire. Create a diagram.

b) What will happen to the force on the wire if...

i. The wire is rotated slightly in the horizontal plane?
ii. The current in doubled?
iii. the magnetic field is reversed?

Create a diagram...

Homework Equations

F = i B L
F = i B L sinθ
F = sin θ

The Attempt at a Solution

A) If the current in the wire is 2.0 A [W]

F = i B L sin θ
F = 2.0 A ⋅ 1.4 T⋅ 4 = 11.20 N (down)

b) What will happen to the force on the wire if...

i. The wire is rotated slightly in the horizontal plane?

F_net = F ⋅ sin θ = 11.20 sin θ N

ii. The current in doubled?

F_net = 2 ⋅ F = 22.40 N

iii. the magnetic field is reversed?

F = 11.20 N (up)

Should the diagram for question a) be like this?

The a) b) at the bottom isn't relevant.

 

[Simon Bridge]

In (b) you have not been asked to calculate the magnitude of the force. The answers will be short English language statements about what happens to the force. i.e. if you double the current, is the force doubled? halved? Something else? What?

In general, try to answer questions in the terms that they are asked.

You problem statement just sayd "create a diagram"... what of? Is there more to it than that or will any old diagram do?
Certainly you won't be expected to sketch anything so detailed - probably you need to sketch the physical situation described ... the idea is to be able to use the diagram to help you answer the questions. So what would be useful?

 

[julianwitkowski]

Hi Simon, thank you for your reply!

The answers will be short English language statements about what happens to the force.

Ok so more like this?

• The wire is rotated slightly in the horizontal plane?

The force is multiplied by the sine of θ (alignment to the field).​

• The current in doubled?

The force is proportional to the current. When the current in doubled, the force is doubled.​

• The magnetic field is reversed?

When the magnetic field is reversed, the direction of the force will be reversed.
​

Certainly you won't be expected to sketch anything so detailed.

I hope not... And the second question didn't ask for diagrams I added that by accident. But for question a does the diagram above a) in the image look like what it should be.

The wire crossing the field west would be a positive charge if I'm not mistaken?
The force is directed down.
The field is directed North.

So I think this is right, just I'm unsure of myself because this is new...

Thanks again for your help!

 

[Simon Bridge]

Ok so more like this?

The last two are right, the first is not.
"multiplied by the sine of the angle" happens for a big turn as well as a "slight" one ... what is special about the tun being only a slight turn?

Make sure you understand which angle the theta in sin(theta) refers to ... also, does the length of the wire subject to the field change or stay the same, when the wire is rotated horizontally? This is where a good diagram helps more than trying to use equations.

Diagram (a) is correct for a positive charge moving west through a magnetic field pointing north.
A westward conventional current will have a resultant downwards force.

Note: for small values of $\theta$, $\sin\theta \approx \theta$ and $\cos\theta \approx 1$.

 

[julianwitkowski]

Note: for small values of $\theta$, $\sin\theta \approx \theta$ and $\cos\theta \approx 1$.

Given that a slight rotation should be considered a small value for θ, and for small values of θ, sinθ ≈ θ ... Should it be force * θ ?

I threw in some angles for θ and it appears that the force decreases and also reverses until ≈ 12°... After 12° the magnitude decreases but does not reverse.

 

[BvU]

Dear Julian,

As Simon says, you have to take the right $\theta$ . In my (lenient) view, you have done that ("alignment to the field"). Originally perpendicular, so $\theta = {\pi\over 2}$ and the sine is 1. Vary $\theta$, force varies as $\sin\theta$. i.e. its magnitude decreases gently. Your 12 degrees must be a wrong setting on your calculator: the force should go to 0 when $\theta$ goes to 0 and only then change direction.

I also have a different interpretation on what's an acceptable answer (Simon's post #2): If someone asks "what happens if the current is doubled?" then I agree that an answer like "the force is doubled" is probably the expected 'right' answer. "The temperature in the room doesn't change much" is also a correct answer, by the way.

But the answer F_net = 2 F = 22.4 N is at least as good, as the expected 'right' answer, if not better.

This exercise is properly worked out, understood and done with. On to the next ! No reason to get lost in semantics or philosophy. There's lots of interesting physics to explore !

And the diagram is excellent. (if I don't look at the chopped off hand)

 

[NascentOxygen]

When applying the approximation
for small angles, θ ≈[/size] sin θ [/color]
you should be using θ measured in radians. (You'll see what I mean ...)

 

[julianwitkowski]

When applying the approximation
for small angles, θ ≈ sin θ
you should be using θ measured in radians. (You'll see what I mean ...)

That is very good to know... Duhh Moment . Thank you.

 

[Simon Bridge]

Going back t the problem statement:
i. The wire is rotated slightly in the horizontal plane?
... consulting your diagram, does this change the angle to the field? Does it change how much of the wire is inside the field?

And yes always use radians for angles.

 

[julianwitkowski]

Does it change how much of the wire is inside the field?

It doesn't say how much wire is in the field or how they are rotating it.
I assume they mean around the center of the wire's length as if the wire was the diameter of the rotation, as it mentions rotating through the horizontal plane...

So, if that is true, the distance from the wire to the field would change during rotation affecting the force.
And as it rotates back to the origin I guess you would get a force between -11.2 N to +11.2 N depending where it is in the rotation?

Therefore, the force will decrease slightly if it's rotated slightly?

 

[BvU]

Oh, boy, here I go again . My apologies to Simon for sloppily reading his post as well as the problem statement. And to Julian for contradicting Simon who is right.

"Rotated slightly in the horizontal plane means instead of West a bit north of west or a bit south of west (oh how I hate these bearings instead of simple angles like in math class...). And, in order not to confuse anybody, without introducing a new angle, so we look at it at $\theta = {\pi\over 2}$

What Simon patiently points out (kudos, Simon!) is that the 'width' of the area where the field is 1.4 T is limited to 4 m.

The problem statement says that originally the length of the wire that is in the field is equal to this width of 4m. How does that length depend on $\theta$ (exactly ! $\ \$ there is no need for approximations) ?

 

[Simon Bridge]

See how easy it is to be confused... this is why I am being a bit pedantic.

The field points north... it is in the horizontal plane, so rotations in the horizontal are also rotations wrt the field direction.

The field is 4m wide. Taking this to mean that the E-W extent is 4m, the length of wire in the field also changes.

The center of the rotation does not matter for this calculation.

A good diagram to see this would be an overhead view with the field boundaries as dotted lines and the compass rose off to one side.

 

[julianwitkowski]

It sounds like the magnitude would stay the same because the magnetic field would follow the current?

So does the force magnitude stay the same and the force direction change wrt the field?

 

[Simon Bridge]

Can you think of a way to check?

For instance, if you rotated the wire by pi/2 (90deg but we use radians) then the wire would be 100% inside the field (an infinite length!) but the angle to the field would be zero - so is there a force or not? Anyway, pi/2 is not a small rotation - perhaps the force increases before it decreases? Maybe it follows a cosine curve with rotation so, for small rotations the force is approximately unchanged?

You need to know how the length of the wire inside the field changes with small rotations.
I think that's as close to doing the work for you as I can get. If you understand the classwork, then it's actually quite easy but since you don't (not unusual) you will probably have to do some algebra. No pain no gain.

 

[BvU]

And Julian: There's a golden hint in the last line of Simon's post #12 !

 

[julianwitkowski]

I think that's as close to doing the work for you as I can get.

Teach a man to fish :P

I think I'm getting close, I'm drawing some diagrams and reading the chapter again and it's making more sense now...
Thanks for all your help here

 

[BvU]

Hey ho, you're free to go if you want, of course, but you are so close now that (and I am confident I can say "we") we would like to see you wrap this one up !

 

[julianwitkowski]

Hey ho, you're free to go if you want, of course, but you are so close now that (and I am confident I can say "we") we would like to see you wrap this one up !

Ha, we'll see...

This is what I've gathered from my experiments with math... F=qvb sinΘ, where Θ is the angle between the field and the velocity, not the force as I thought before. If the wire is perpendicular to the field, sinΘ=1 and F=qvb... If the angle between the magnetic and the current is 0, sinΘ=0, As the wire rotates slightly it is no longer perpendicular, but the angle > 0.

Thanks guys.

 

[BvU]

I hope I don't surprise you with: your equation F = i B L sinθ is a direct descendant of the Lorentz force we saw in the other thread. We are better off with the descendant (it's in terms of the given current and length).

And yes, $\theta$ is the angle between West and North, and $\sin\theta = 1$ in part a). And it decreases a little in part b). Current i and magnetic field B stay the same.

It's the factor L we have to look at a little better. L is 4 m in part a). Let's rotate the very long wire over 30 degrees towards North, so $\theta$ is now only 60 degrees. What's the length of the part of the wire in the B field ? Simon's drawing is making it all completely clear !

 

[Simon Bridge]

The magnitude of the force is $F=qvB\sin\theta$ ... the actual equation is $\vec F = q\vec v\times\vec B$ so the "sine-theta" comes from the cross product.

In terms of a current $IL = qv$ where L is the length of the wire inside the field. The derivation should be in your notes.
In your case the value of L depends on the angle the current makes to the field.

So you can write, for magnitude: $F=BIL(\theta)\sin\theta$ ... you need to know what $L(\theta)$ is in order to answer the question.

Note: the $\theta$ above is the angle between the current vector and the magnetic field - in this case, that is the same as the angle measured anticlockwise from any N-S axis to the wire. The angle between W and N is pi/2. The only time that is $\theta$ is when the wire lies along the E-W direction.

You should also realize that a small rotation is not a small value for the theta above, but a small change in the value of theta.

Aside:
Basically, this part of the problem is challenging you on fundamental concepts in physics besides the nature of the force on a current-carrying wire. i.e. How you read word problems, how you draw diagrams and use them, the difference between an angle and a rotation ... things that the person setting the problem expects you to already have down as second nature. You should not need to go through all this in so much detail at this stage in your education. What has happened is that you have discovered a problem with your understanding of core physics concepts, but it is good that you found out now though, because you can focus on it and get used to it, now, before it becomes a serious impediment to you. I'm thinking you may need a tutor - someone to sit next to you while you do problems.

 

[julianwitkowski]

Let's rotate the very long wire over 30 degrees towards North, so $\theta$ is now only 60 degrees. What's the length of the part of the wire in the B field ? Simon's drawing is making it all completely clear !

It's longer, when you rotate it, the wire is longer than the field so more wire enters the field...

If I did this right, 4m= L*sin(θ), L = 4m * csc(θ) = 4.6

F = I B L sin Θ = 2.0 A ∙ 1.4 T ∙ 4.6 m sin 60° = 11.5 Newtons (down)

By right hand rule if my fingers are facing the field and thumb facing the charge, my palm is facing the plane so the force is still directed down...

 

[julianwitkowski]

F = I B L sin Θ = 2.0 A ∙ 1.4 T ∙ 4.6 m sin 60° = 11.5 Newtons (down)

I meant 11.15 N (down)...

 

[julianwitkowski]

...someone to sit next to you while you do problems.

I wish but it's not really possible at this moment... Thankfully I'm ahead of where I should be, it's just a correspondence to get into a good school for biology. I had a choice, sounded interesting :p

 

[julianwitkowski]

Thank you so much for your help, you're basically tutors and It's great that you both made me work to get a better understanding, I really appreciate that... Keep it up!

 

[julianwitkowski]

So you can write, for magnitude: $F=BIL(\theta)\sin\theta$ ... you need to know what $L(\theta)$ is in order to answer the question.

In terms of radians, 60° = 1.04 Radians ≈ 1, so I get what you mean about slight rotation is a small change in the value. And as it rotates the wire is adding to the field, while the angle gets smaller, more wire enters the field.

F = i ⋅B⋅L ⋅ (θ) ⋅ sin (θ) = 1.4T ⋅ 2.0A ⋅ (4/sin(1^RADIAN) ⋅ (1) ⋅ sin (1) =11.2 N

The same...

I suppose that L ≠ 4/sin(60°) ... How else would you find L if you don't do it based on Pythagorean ?

I know the question isn't asking to solve this per say, and we already have the right answer, but for my own understanding of the question, I figured the wire could be a hypotenuse in a triangle, and the field length would be the opposite of the triangle? So if you knew the angle of rotation wouldn't that be the length of the wire now in the field?

 

[Simon Bridge]

60deg is written pi/3 and its a big angle.
a change from 59deg would be a small change in that angle although the angle itself is still large.

If the width of the field is D, and theta is the angle to N-S, then L is the hypotenuse of a rt-triangle with D as the opposite side. Use trig. You are correct. Now you know L in terms of theta, you can sub that into the formula.

But what happens when theta is zero?

 

[julianwitkowski]

But what happens when theta is zero?

If θ = 0, should θ = 2π because 0° = 360°?

On the other hand though, the wire is still 90° perpendicular to the field, as it is 0 change, so force would not change and you'de use F=ILB = 11.2 N (down), as F=ILB sin(θ) is used when there is a rotation.

 

[Simon Bridge]

If θ = 0, should θ = 2π because 0° = 360°?

does it matter?
note... theta is angle to the field.
in the initial setup, theta is pi/2
changing theta to 0 or to 2pi requires a rotation of pi/2

Anyway, tour question is answered fir amall angles.

 

[BvU]

It's longer, when you rotate it, the wire is longer than the field so more wire enters the field...

If I did this right, 4m= L*sin(θ), L = 4m * csc(θ) = 4.6 (*)

F = I B L sin Θ = 2.0 A ∙ 1.4 T ∙ 4.6 m sin 60° = 11.5 Newtons (down)

By right hand rule if my fingers are facing the field and thumb facing the charge, my palm is facing the plane so the force is still directed down...

The answer I've been searching for is staring me in the face ! Namely $L = w/ \sin\theta$. (w = 4 m).
That is the theta dependence of L.

So now you can complete the theta dependence of $F$: from the beginning you had

F = i B L sin θ $\quad\Rightarrow\quad$ F = i B (w/sin θ) sin θ = i B w, independent of θ .
Did I now spoil this part of the exercise ?​

(*) I also don't understand where this 4.6 comes from. You mention 60 degrees, but sin(60) = ½√3 ?

 

[julianwitkowski]

Did I now spoil this part of the exercise ?
(*) I also don't understand where this 4.6 comes from. You mention 60 degrees, but sin(60) = ½√3 ?

No heaven forbid you spoil it! ... We're all good, it wasn't asking to place numbers in it thank god, because if it did the person marking it would have known I didn't understand the distinction between radians and regular angles... Hence I got a good grade. This brings me to another point.

In degrees... 4,6 m would be the length of the wire if L = 4/sin(60°) in deg... I don't understand why it is ½√3 in radians, because 60° = π/3... No? This is the edge of my understanding of radians, or lack of it.

If L = 4/sin(½√3) = 5.25 m, F = 1.4 2 (4/sin(½√3) sin(½√3) = 30.82 N (down)

I thought F = i B (w/sin θ) sin θ =1.4⋅2⋅(4/sin(π/3)) ⋅ sin (π/3) = 11.2 N (down)

So I'm confused... want to spoil this?

 

[julianwitkowski]

Oops now I forgot a bracket... F = 1.4 2 (4/sin(½√3) sin(½√3) = 11.2 N (down)

 

[julianwitkowski]

See...

As I understand the distinction between rad and deg is that rad gives you a distance in terms of radians, which gives you an idea of how many times something goes around a circumference...

What else should I know?

 

[BvU]

My bad, sloppy reading (doing other things same time). Of course. And I was the one who threw in this 60 degrees in the first place !

Radians are a natural measure for angles. That way arc length (in a unit circle) = angle. And that way $\displaystyle \lim_{\theta\downarrow 0} \; \sin\theta < \theta <\tan\theta$ etc.

"What else should I know " is a good question. I don't have all the answers...

In degrees... 4,6 m would be the length of the wire if L = 4/sin(60°) in deg... I don't understand why it is ½√3 in radians, because 60° = π/3... No? This is the edge of my understanding of radians, or lack of it.

If L = 4/sin(½√3) = 5.25 m, F = 1.4 2 (4/sin(½√3) sin(½√3) = 30.82 N (down)

I thought F = i B (w/sin θ) sin θ =1.4⋅2⋅(4/sin(π/3)) ⋅ sin (π/3) = 11.2 N (down)

So I'm confused... want to spoil this?

Oh boy, I wish I could turn back time here. ${1\over 2}\sqrt 3$ is the cosine of $\pi\over 3$ (= 60 degrees). (comes from half of an equlateral triangle, sides $1, {1\over 2}, {1\over 2}\sqrt 3$ ).

Think you have it right, have to run.

 

[julianwitkowski]

Think you have it right, have to run.

Well this goes back to what simon said... and we're both technically right. Both π/3 and ½√3 are slight rotations.
We pulled 60° out of nowhere as a way to have an example.

I'm curious as to why a calculator gives the same answer for both expressions when ½√3 = 49.6° and π/3 = 60°...
I guess this goes back to small rotations of θ = 1, as which also gives the same answer.

Is this why my calculator is rounding these things to θ=1?

Maybe I don't understand why θ=1 is used when values of θ are small.
,,,Or why it would be beneficial to use radians for calculations a slight turn, or any turn less than 2π ?

 

[Simon Bridge]

Just some clarifications: $$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$$ ... you can check this on your calculator.
If you make an equilateral trangle where the hypotenuse is 1 unit long and the adjacent side is 1/2 unit long, then the angle between these sides is pi/3 (60deg) and the length of the opposite side is the sine of the angle. In this set up you can find the length of the opposite side without using a calculator. Give it a go.

However, $\frac{\sqrt{3}}{2} = 49.6^\circ$ ... is nonsense.
You need to be more careful when you write down the maths - you've been told about this before (prev in thread) - remember, the people reading this don't know what you mean, they only know what you write. We cannot tell what you've done from what you've said.

If you input "square root of three all divided by two" however your calculator does this, it should output 0.86603 ... maybe more dp than that.
I use gnu octave so I just type in sqrt(3)/2, or what I want to compute, and press enter. I get:

octave:9> sqrt(3)/2
ans =  0.86603
octave:10> pi/3
ans =  1.0472
octave:11> sin(pi/3)
ans =  0.86603

... calculators often need to be placed in a different "mode" to do trig using radians. If you mix them up you can get confused.

Since $\frac{\sqrt{3}}{2} \approx 0.8660$ while $\frac{\pi}{3} \approx 1.0472$ ... this means that 60deg is not a small angle, and it would not normally be considered a small rotation.

Don't forget to always use radians ... when you write $\theta = 1$ you are saying the angle is one radian, which would be about 57.3 degrees. Also not a small angle.

But remember the difference between an angle and a rotation?

Also remember: what counts as small depends on the context.

 

[julianwitkowski]

If you make an equilateral trangle where the hypotenuse is 1 unit long and the adjacent side is 1/2 unit long, then the angle between these sides is pi/3 (60deg) and the length of the opposite side is the sine of the angle. In this set up you can find the length of the opposite side without using a calculator. Give it a go.

That is clever actually, cool :).

However, $\frac{\sqrt{3}}{2} = 49.6^\circ$ ... is nonsense.

What does this mean at wolfram?

Sorry about that I know.. I'm working on it. This is a mysterious adventure for me as I've never done any of this before.
I have a real good streak about learning in trial and error.