# Bar movement, find angle to keep the bar stationary

1. Oct 10, 2016

### moenste

1. The problem statement, all variables and given/known data
The diagram represents a cylindrical aluminium bar AB resting on two horizontal aluminium rails which can be connected to a battery to drive a current through AB. A magnetic field, of flux density 0.10 T, acts perpendicularly to the paper and into it. In which direction will AB move if the current flows from A to B?

Calculate the angle to the horizontal to which the rails must be tilted to keep AB stationary if its mass is 5.0 g, the current in it is 4.0 A and the direction of the field remains unchanged.

(Acceleration of free fall, g = 10 m s-2.)

2. The attempt at a solution
"In which direction will AB move?"
Is this question asking about the direction of the force? If yes, what direction of field should be taken? On the left side the field is directed towards the paper (since current goes from A to B) and on the right side the current goes out of the paper.

I tried to calculate the angle but I didn't get the correct answer.

Φ = AB cos θ, where AB is the length of the cylinder.
Φ = B A, where Φ is magnetic flux, B is the magnetic field and A is the area of a circle.
B = m g / I L = [(5 / 1000) * 10] / [4 * (5 / 100)] = 0.25 T, we found the magnetic field first.
A = π r2, this is the formula for the area of a circle, we still lack the radius.
B = μ0 I / 2 π r → r = μ0 I / 2 π B = (4 π * 10-7 * 4) / 2 π * 0.25 = 3.2 * 10-6 m, we found the radius from the magnetic field formula.
A = π * (3.2 * 10-6)2 = 3.2 * 10-11 m2, we found the area of a circle.
Φ = 0.25 * 3.2 * 10-11 = 8.042 * 10-12 Wb, we found the magnetic flux.
cos θ = 8.042 * 10-12 / 0.05 → θ = 89.9 °, we found the angle.

What's missing?

2. Oct 10, 2016

### Staff: Mentor

It's the external field that you need to consider (which is given as into the paper), not the field created by the current-carrying wire.

3. Oct 11, 2016

### moenste

What does it mean to consider the external field? Could you please elaborate more on that... By looking for "AB movement direction" we are looking for force? F = B I L?

4. Oct 11, 2016

### Staff: Mentor

You are told that a uniform field acts into the page. That's the field that produces the force on the bar.

Right.

When you put the track at an angle, you'll want to make yourself a free body diagram of the bar.

5. Oct 11, 2016

### moenste

But how do we find the force direction? We have a bar with current pointed downwards. If we grab the bar the field will enter into paper at the left side and go out of paper at the right side.

F = B I L sin θ
m g = B I L sin θ
(5 / 1000) * 10 = 0.1 * 4 * (5 / 100) sin θ
sin θ = 0.05 / 0.02
θ = math error.

6. Oct 11, 2016

### Staff: Mentor

You are still focused on the field created by the current. But we don't care about that field!

In your diagram, the field of interest is uniform and points into the page. That's the 0.10 T field given in the problem statement. (It's not the field due to the current in the wire.)

To find the direction of the magnetic force on that current you must use a right hand rule. $\vec{F} = \vec{I}L \times \vec{B}$

7. Oct 11, 2016

### Staff: Mentor

8. Oct 11, 2016

### moenste

But how do you know it acts into the page?
Maybe it's my English, but I understand that the field acts into the paper and out (?) of paper. I'm not sure about "perpendicularly to the paper" phrasing.

If we assume that the field acts into the paper, then the force is going to the right (towards the circuit). The first finger on the right hand is poiting towards B (current), middle finger is pointing downwards (field) into the paper and the thumb is pointing to the right.

F = 0.1 * 4 * 0.05 = 0.02 N.

9. Oct 11, 2016

### Staff: Mentor

Because it tells you!

Perpendicular to the paper just means that its either going straight in or straight out of the paper. (Not at some other angle.) And we are told that it's going into the page.

Right!

10. Oct 11, 2016

### moenste

Ha, so: perperndicularly to the paper and into it = into the paper, perperndicularly to the paper and out of it = out of it?

OK, so we have:
It will move to the right, or to the circuit.

We found F = 0.02 N, which is the force to the right. Now we need the force downwards: F = m g = 5 / 1000 * 10 = 0.05 N.

tan α = 0.02 / 0.05 → α = 21.8 °. This is the angle of the triangle of the two forces (weight force and the movement to the right force). And so we need to tilt the bar to the left on 21.8 ° to stop the bar from moving to the right.

I think this should be correct.

11. Oct 11, 2016

### Staff: Mentor

Right.

Right.

Right.

I would have done it by treating it just like a box on an incline. There is a component of weight acting down the incline (mg sinα) and a component of the magnetic force acting up the incline (F cosα, where F is the magnetic force already calculated). Set them equal and solve for α.

But your method is fine, as long as you're clear on it.