# The balance reading when direction of mag. field changes

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1. Oct 11, 2016

### moenste

1. The problem statement, all variables and given/known data
The diagram shows a rigid conducting wire loop connected to a 6.0 V battery through a 6.0 V, 3.0 W lamp. The circuit is standing on a top-pan balance. A uniform horizontal magnetic field of strength 50 mT acts at right angles to the straight top part of the conducting wire in the direction indicated in the diagram, i. e. into the paper. This magnetic field extends over the shadowed area in the diagram. The balance reads 153.860 g. Calculate (a) the force exerted on the conducting wire by the magnetic field, and (b) the new balance reading if the direction of the magnetic field is reversed.

Answers: (a) 1.3 * 10-3 N, (b) 1.54; 115 g.

2. The attempt at a solution
(a) F = B I L. First we find I = P / V = 3 / 6 = 0.5 A. Plug it into the F formula: F = (50 * 10-3) * 0.5 * (51 / 10 / 100) = 1.28 * 10-3 N. This part should be correct.

(b) This part I have large doubts. First of all, what does "The balance reads 153.860 g." mean? And second of all what does the answer "(b) 1.54; 115 g" mean?

There are two styles of representing numbers: (i) 1,000,000.00 -- one million and (ii) 1 000 000,00 -- also one million. The mass -- what does it mean? .860 is the faction? Or it is 153 860,00? Or 153,86?

I also don't understand the answer. If we need to find the mass, which is 115 g, then what is 1.54?

And finally, how do we find the new balance reading? As I understand, the direction of force was upwards (current is going from +, so current is going from left to right through the wire and field), the field was into the paper, so the force was directed upwards. And now when the field is out of paper, the force is directed downwards. But that's as far as I can go.

Maybe F = m g so m g = B I L? Why do we need the 20 mm though...

2. Oct 11, 2016

### Staff: Mentor

153.860 g are a bit less than 154 gram.
That is a very odd format (scanning+OCR error?). Anyway, you can calculate it!

In which direction does the force point?

Scales measure forces and convert them to mass values using the known value of g. If you know the gram reading of the scale, you can calculate the force. This force will change if you reverse the magnetic polarity - you can calculate the new force and convert back to what the scale will show.

3. Oct 11, 2016

### moenste

That's what is exactly written in the book, perhaps it's a typo.

The force points upwards. We have current from left to right and field directed into the paper. So the force is directed upwards.

F = m g = 0.15386 kg * 10 = 1.54 N the weight force.
FUp = 1.28 * 10-3 N, the force directed upwards.

How can they be equal if they are not?

The new force will be directed downwards. And the weight is directed downwards.

4. Oct 11, 2016

### Staff: Mentor

The magnetic force points upwards, correct.
It is the force on the scale. It is not the weight of anything.
Why do you expect them to be equal?

Did you draw force diagrams?
Correct.

5. Oct 11, 2016

### moenste

We have two forces: FUp = 1.3 * 10-3 N and F? = 1.3 * 10-4 N. What is the relationship between these two forces if they are not F upwards and F downwards?

The new force is directed downwards and will not change and will be equal to 1.3 * 10-3 N. But I'm not sure what to do next.

6. Oct 11, 2016

### Staff: Mentor

Did you draw a force diagram?

There are three forces acting on the circuit. You know two, so you can determine the third.

7. Oct 12, 2016

### moenste

We have the force directed upwards from the field. There is the weight force directed downwards. I don't see any more forces.

8. Oct 12, 2016

### Staff: Mentor

The setup is on top of a scale. There is a force between setup and scale - and you calculated this force already.

9. Oct 12, 2016

### moenste

Like this?

10. Oct 12, 2016

### Staff: Mentor

What does the down-arrow represent?

11. Oct 12, 2016

### moenste

Force on the scale (as you said).

12. Oct 12, 2016

### Staff: Mentor

So the scale is pulling the setup downwards? The force balance for the scale does not matter, this is all about forces on the setup on top of the scale.

And where did gravity go?

13. Oct 12, 2016

### moenste

Gravity you mean fravity force? Maybe it's 1.54 N - 1.3 * 10-3 = 1.5387 N?

14. Oct 12, 2016

### Staff: Mentor

Don't guess, draw your force diagram and calculate it based on that.

You should use more significant digits for the larger force value, otherwise the effect you want to study disappears as rounding error.

15. Oct 13, 2016

### moenste

Situation 1:
(i) Force directed upwards. F = 1.275 * 10-3 N.
(ii) Force on the scale: F = m g = 0.15386 * 10 = 1.5386 N.
(iii) The weight is the force directed downwards. If we have one force directed upwards and one force directed downwards then: 1.5386 + 1.275 * 10-3 = 1.539875 N.

Situation 2:
(i) Force directed downwards. F = 1.275 * 10-3 N.
(ii) Weight directed downwards: F = 1.539875 N.
(iii) Force on the scale is equal to total force, since both forces are directed downwards, therefore we sum the forces up: F = 1.275 * 10-3 + 1.539875 = 1.54115 N.

Now to find the new balance reading: m = F / g = 1.54115 / 10 = 0.154115 kg or 154.115 g. Is this the correct answer?

The book answer is similar, but still these symbols are confusing:
Also, how come we didn't use the 20 mm figure anywhere?

Last edited: Oct 13, 2016
16. Oct 13, 2016

### Staff: Mentor

Correct.

I still think the book had some weird OCR (scan to text conversion) where a . got read as ; and some additional . got inserted due to dust or something like that.

The 20 mm are not necessary to solve the problem. So what?