# Bar movement, find angle to keep the bar stationary

• moenste
In summary: It's a bit of a weird way to phrase it, but it's correct.In summary, the diagram shows a cylindrical aluminium bar AB resting on two horizontal aluminium rails with a uniform magnetic field of 0.10 T acting perpendicularly into the paper. The question asks which direction AB will move if a current flows from A to B. To keep AB stationary, the rails must be tilted at an angle of 21.8 °, found by calculating the magnetic flux, magnetic field, and area of the circle. The force acting on AB due to the external magnetic field is found by using the right hand rule, with the force being directed towards the right (towards the circuit).
moenste

## Homework Statement

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?

moenste said:
"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.
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.

moenste
Doc Al said:
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.
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?

moenste said:
What does it mean to consider the external field? Could you please elaborate more on that...
You are told that a uniform field acts into the page. That's the field that produces the force on the bar.

moenste said:
By looking for "AB movement direction" we are looking for force? F = B I L?
Right.

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

moenste
Doc Al said:
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.
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.

moenste said:
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.
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}##

Doc Al said:
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.)
But how do you know it acts into the page?
moenste said:
A magnetic field, of flux density 0.10 T, acts perpendicularly to the paper and into it.
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.

Doc Al said:
To find the direction of the magnetic force on that current you must use a right hand rule.
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.

moenste said:
But how do you know it acts into the page?
Because it tells you!

moenste said:
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.
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.

moenste said:
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.
Right!

moenste
Doc Al said:
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.
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:
moenste said:
In which direction will AB move if the current flows from A to B?
It will move to the right, or to the circuit.

moenste said:
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.
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.

moenste said:
Ha, so: perperndicularly to the paper and into it = into the paper, perperndicularly to the paper and out of it = out of it?
Right.

moenste said:
It will move to the right, or to the circuit.
Right.

moenste said:
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.
Right.

moenste said:
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 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.

moenste

## 1. What is the purpose of finding the angle to keep the bar stationary?

The purpose of finding the angle to keep the bar stationary is to ensure that the bar remains in a fixed position and does not rotate or move when subjected to external forces.

## 2. How is the angle to keep the bar stationary calculated?

The angle to keep the bar stationary is calculated using trigonometric principles, specifically the sine, cosine, and tangent functions. The angle can be determined by analyzing the forces acting on the bar and using the appropriate equations.

## 3. What factors can affect the stability of the bar?

There are several factors that can affect the stability of the bar, including the weight and distribution of the load on the bar, the materials used to construct the bar, and the external forces acting on the bar.

## 4. How can the angle to keep the bar stationary be adjusted?

The angle to keep the bar stationary can be adjusted by changing the distribution of the load on the bar, altering the materials used to construct the bar, or by adding additional supports or braces to the bar.

## 5. What are some real-world applications of finding the angle to keep the bar stationary?

Finding the angle to keep the bar stationary is important in a variety of fields, including engineering, architecture, and construction. It is used to design and construct stable structures such as bridges, buildings, and cranes.

• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
14
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
190
• Introductory Physics Homework Help
Replies
7
Views
909
• Introductory Physics Homework Help
Replies
8
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
282
• Introductory Physics Homework Help
Replies
12
Views
261
• Introductory Physics Homework Help
Replies
26
Views
2K
• Introductory Physics Homework Help
Replies
11
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
422