The balance reading when direction of mag. field changes

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Homework Help Overview

The problem involves a conducting wire loop connected to a battery and a lamp, placed on a balance, within a uniform magnetic field. Participants are tasked with calculating the force exerted on the wire by the magnetic field and determining the new balance reading if the magnetic field direction is reversed.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of current and magnetic force, questioning the meaning of the balance reading and the format of the answers provided. There is uncertainty about the interpretation of the mass reading and how it relates to the forces acting on the setup.

Discussion Status

Participants are actively exploring the relationships between the forces involved, including the magnetic force and weight force. Some guidance has been offered regarding the calculation of forces and the importance of drawing force diagrams, but no consensus has been reached on the interpretation of the balance reading or the final calculations.

Contextual Notes

There are concerns about the clarity of the problem statement, particularly regarding the format of the balance reading and the relevance of certain measurements, such as the 20 mm figure. Participants are also navigating the implications of reversing the magnetic field direction on the forces acting on the setup.

moenste
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Homework Statement


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.

2223ebe7c663.jpg


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...
 
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153.860 g are a bit less than 154 gram.
moenste said:
(b) 1.54; 115 g.
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.
 
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mfb said:
That is a very odd format (scanning+OCR error?).
That's what is exactly written in the book, perhaps it's a typo.

mfb said:
In which direction does the force point?
The force points upwards. We have current from left to right and field directed into the paper. So the force is directed upwards.

mfb said:
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.
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.
 
The magnetic force points upwards, correct.
moenste said:
F = m g = 0.15386 kg * 10 = 1.54 N the weight force.
It is the force on the scale. It is not the weight of anything.
moenste said:
How can they be equal if they are not?
Why do you expect them to be equal?

Did you draw force diagrams?
moenste said:
The new force will be directed downwards. And the weight is directed downwards.
Correct.
 
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mfb said:
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.
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.
 
moenste said:
What is the relationship between these two forces if they are not F upwards and F downwards?
Did you draw a force diagram?

There are three forces acting on the circuit. You know two, so you can determine the third.
 
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mfb said:
Did you draw a force diagram?

There are three forces acting on the circuit. You know two, so you can determine the third.
We have the force directed upwards from the field. There is the weight force directed downwards. I don't see any more forces.
 
The setup is on top of a scale. There is a force between setup and scale - and you calculated this force already.
 
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mfb said:
The setup is on top of a scale. There is a force between setup and scale - and you calculated this force already.
2e82a5b3ab81.jpg


Like this?
 
  • #10
What does the down-arrow represent?
 
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  • #11
mfb said:
What does the down-arrow represent?
Force on the scale (as you said).
 
  • #12
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?
 
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  • #13
mfb said:
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?
Gravity you mean fravity force? Maybe it's 1.54 N - 1.3 * 10-3 = 1.5387 N?
 
  • #14
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.
 
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  • #15
mfb said:
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.
c2b1c29522a8.jpg


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:
moenste said:
(b) 1.54; 115 g.

Also, how come we didn't use the 20 mm figure anywhere?
 
Last edited:
  • #16
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?
 
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