# Homework Help: A rectangular conducting frame in a magnetic field

1. Aug 4, 2015

### kaspis245

1. The problem statement, all variables and given/known data
A frame is made of two firm wires of length $h$ and a rod of length $l$ and mass $m$. This frame can rotate around horizontal axis in uniform magnetic field $B$. In a short amount of time $τ$ a current $I$ passes through the frame. What is the maximum angle that the frame can make with its equilibrium state? Consider that in time $τ$ the frame moved very little.

2. Relevant equations
$F=Bqv$

3. The attempt at a solution
Let's denote the total length of the frame by $L$ so that the length is $L=l+2h$.

The rod is affected by two forces $mg$ and $F=Bqv=Bq\frac{L}{τ}=IBL$.

The rod is moving because it is affected by a force. This force has horizontal and vertical components.

Ox: $ILB+mgsinα$
Oy: $mgcosα-IBLsinα$

So $F=\sqrt{(ILB+mgsinα)^2+(mgcosα-IBLsinα)^2}$.

Now, I don't know if what I wrote above is true. I've been making a lot of sketches and everything seems correct. But if we look into the first drawing, we can easily see that the sum of forces $mg$ and $IBL$ is $F=\sqrt{(ILB)^2+(mg)^2}$. I don't know which expression is correct. Also, I don't know if I'm going the right way. Please help.

2. Aug 4, 2015

### Hesch

I think your calculations are too complicated:

In this short time, τ, only the energy supplied within τ should be regarded ( calculate the angle acceleration of the system ).
The rotational energy ( Ekin ) supplied within the time τ must equal Epot = m*g*h at the maximum angle as for the system.

3. Aug 5, 2015

### kaspis245

So the rotational work done by the rod must be equal to its potential energy.

$W_R=E_p$
$mh^2aα=mgh(1-cosα)$

I don't know how that helps me.

4. Aug 5, 2015

### kaspis245

Here's a solution that I came up with. Is it correct?

5. Aug 5, 2015

### Hesch

Within a short time, τ, ther will be no resulting vertical force, because m*g + Fwire = 0.

The rod will be accelerated horizontally:

Frod = B*I*L.

The vertical acceleration will be: a = Frod / m.

Now calculate the velocity of the rod, v(τ), and its kinetic energy, Ekin. Then: m*g*Δhmax = Ekin.

6. Aug 6, 2015

### kaspis245

I came up with this answer:

$a=\frac{BIL}{m}$
$v=\frac{BILτ}{m}$
$Δh=h(1-cosθ)$

$mgΔh=\frac{mv^2}{2}$
$θ=cos^{-1}(1-\frac{(BILτ)^2}{2gh})=cos^{-1}(1-\frac{(BIτ(l+2h))^2}{2gh})$

7. Aug 6, 2015

### rude man

1. what is force on the bar? Note that you're told the angle will be small. Note also there is the mag. force and gravity. Resolve the force components in the θ direction.
2. write diff. eq. relating torque to rotational inertia and angular acceleration.
Solve this eq. for the angle the bar makes with the vertical as a function of time (linear, const.-coeff. 2nd order ODE).
My answer looks like θ(τ) = ai(1 - cos bτ), a and b constants. i is the current, in lower case to distinguish from I, rotational inertia.

PS Lots of things I don't like about your approach. For starters, there is no B force on the wires, and no g force either since they are assumed massless.

Last edited: Aug 6, 2015
8. Aug 6, 2015

### kaspis245

1. I've tried to find this force in my first post, but I don't know which variant is the correct one.
2. Say the rod is affected by the force $F$, then
$τ=Iα$
$Fh=mh^2α$
$F=mhα$
$\frac{d^2θ}{dτ^2}=\frac{F}{mh}$

Of course the wires are not affected by any force. Well, they are affected by the magnetic field when the angle $θ$ appears, but since the current in each wire is opposite the forces cancel out. The reason why the length of the wires appear in $F=BIL$ is because the rod is affected by the force $F=Bvq=B\frac{Length of the frame}{τ}q=B⋅Length of the frame⋅\frac{q}{τ}=BLI$.

9. Aug 6, 2015

### Hesch

Correction:
The horizontal acceleration will be: a = Frod / mrod.

10. Aug 6, 2015

### rude man

That is correct. But you're on the wrong track getting F.
F = BiL is incorrect. See below why you should be using l (lower-case L) instead of L. And then you are neglecting gravity.
Why is F associated with the length of the frame? The small force on the wires at a finite angle is in the wrong direction to give torque about the horizontal axis. You can't just add force scalars when the corresponding vectors are orthogonal to each other!

Why introduce v at all? And where do you get v = L/τ? The units aren't even right. The left side units are LT-1 whereas the right side is L/ML2T-2 = F-1. One of your main goals in physics should be to check units; it's a very powerful checking tool.

Bottom line: find the correct expression for F comprising both the mag. and gravity forces, and solve your (correct) torque equation.

P.S. You're mixing up I = rotational inertia with I = current. Use i for current.
EDIT: you're also mixing up τ = time with τ = torque, so scratch what I said about v above. Again it's length of the bar, not length of the frame.

Last edited: Aug 6, 2015
11. Aug 7, 2015

### kaspis245

Ok, let's say that the force caused by the magnetic field on the rod is $F=Bil$ (even though I don't understand what's wrong with the units). Then:

$F_{net}=\sqrt{(mg)^2+(Bil)^2}$
$\frac{d^2θ}{dτ^2}=\frac{\sqrt{(mg)^2+(Bil)^2}}{mh}$

Is it ok?

12. Aug 7, 2015

### rude man

There is nothing wrong with your units. F = Bil is correct. I did edit my last post about that. You were using tau (τ) to mean both time and torque, which threw me.

But - your second formula is wrong still. The grav. force mg is down, not along θ. What is the grav. component along θ?

So OK, let's get total F along θ right. There is a mag. component and a grav component. Actually, both are functions of θ, but the mag. component is simplified to Bil. You should understand why. (Hint: you were told θ will stay "small" until time = τ). The grav. component will also be simplified, but must remain a function of θ. Draw force diagrams for the mag. and grav. components on the rod along the θ direction, with a small but finite angle θ to get and understand the force components acting along θ.

Hint #2: for small θ,
cosθ ~ 1 and sinθ ~ θ.

13. Aug 8, 2015

### kaspis245

Again, the question how to express the net force acting on the rod confuses me the most in this problem. As I've highlighted in my first post, I have two variants of how to find the net force. From your last answer I assume that $F=\sqrt{(mg)^2+(Bil)^2}$ is incorrect, so the right answer according to the diagram below must be $F=\sqrt{(ILBcosθ+mgsinθ)^2+(mgcosθ−IBLsinθ)^2}$ .

14. Aug 8, 2015

### kaspis245

Please look into my problem statement. The problem's diagram shows that $g$ is directed upwards. And I believe that you've drawn $Bil$ incorrectly, it should always remain horizontal.

15. Aug 8, 2015

### rude man

I had to delete my last post since I misdefined the angle theta. Here again:
OK, my turn for a diagram. Yours is making me seasick! See attachment below. The blue circle is the rod, looked at sideways. The long thin line are the two wires, looked at sideways. Try to determine the components of the two forces mg and Bil on the rod along θ using this diagram.
You keep using the Pythagorean theorem which is not relevant here. You need to find two forces acting on the rod in the same direction so they will just add, not root-sum-square.
g is directed downwards. ( I don't know, is it upwards in the southern hemisphere? ) And Bil is always orthogonal to B and to l.
EDIT: Bil direction shown wrong. What's shown is the component of Bil along θ.

#### Attached Files:

• ###### Presentation1.pdf
File size:
30.9 KB
Views:
43
Last edited: Aug 9, 2015
16. Aug 8, 2015

### rude man

Yeah, g is shown upwards in the diagram you were given. I believe that to be a mistake.

17. Aug 8, 2015

### kaspis245

Even if it is a mistake, I think I should try to solve this problem with the information that I was given and who knows, maybe it will be correct.

18. Aug 8, 2015

### rude man

Boy, do I need to eat crow! You are right, the Bil force is always horizontal, being orthogonal to B and l! However, fortunately it changes nothing. The component of Bil along θ is still Bil since cosθ ~ 1 here.

If you make g point up your bar would take off in the up direction like a balloon, wouldn't it?
Anyway, sorry about my misdirection of Bil and good luck with the problem.
r m

19. Aug 8, 2015

### rude man

Updated pic in case you change your mind ...

File size:
30.9 KB
Views:
47