Question on simple static magnetics

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A closed loop with constant current in a uniform magnetic field experiences zero net linear force but can have a non-zero torque due to the distribution of forces around the loop. The confusion arises from the distinction between linear force and angular force, as torque is calculated differently and depends on the position vector. While the total force around the loop cancels out, the forces acting in different directions create a torque that causes rotation. The discussion highlights a concern about the clarity of explanations in textbooks, suggesting that the wording could mislead students. Overall, the physics is understood, but the communication of concepts needs improvement for better comprehension.
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Suppose a closed loop with constant current around the loop. The loop is in a uniform constant B field. The force is

\vec F=I\left (\oint_c d\vec l\right ) \times \vec B.

Being a closed loop \oint_c d\vec l=0. So there is no force acting on the loop.

BUT at the same time, we know there is torque, so there is force! How is that possible?
 
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You calculate the total force, integrated over the ring. This is 0. But it consists of forces in different directions along the ring, which are not 0.

I can give you two apples, and then steal two apples from you. Your total number of apples is the same as before. But certainly your number of apples was not the same all the time.
 
I understand that, I can see the loop is going to turn as one side being push and the other side being pull and the total force cancel out. But isn't it quite dumb for the book to make a statement like this and the very next page start talking about torque?
 
yungman said:
I understand that, I can see the loop is going to turn as one side being push and the other side being pull and the total force cancel out. But isn't it quite dumb for the book to make a statement like this and the very next page start talking about torque?

You're comparing two different integrands though. For torque,

\mathbf{T} = I \oint \mathbf{r} \times \left( d\boldsymbol{\ell} \times \mathbf{B} \right)
 
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By the word "force" we typically mean "linear force", as opposed to an "angular force" (torque). Yes, a single loop of current in a static uniform magnetic field experiences zero net linear force and a non-zero net angular force. That is why little electric motors spin instead of shoot out like a bullet.
 
Born2bwire said:
You're comparing two different integrands though. For torque,

\mathbf{T} = I \oint \mathbf{r} \times \left( d\boldsymbol{\ell} \times \mathbf{B} \right)

I know you use different formulas as torque require an arm to swing. But the circuit loop is the same.
 
chrisbaird said:
By the word "force" we typically mean "linear force", as opposed to an "angular force" (torque). Yes, a single loop of current in a static uniform magnetic field experiences zero net linear force and a non-zero net angular force. That is why little electric motors spin instead of shoot out like a bullet.

I understand this, just the idea is very misleading...at least to me. Angular force starts with linear force, difference is it is on a hinge and it is forced to turn.

Anyway, we all agree on the theory, I just complain about the book. It should say right away about the other possibility. Student read the first time will take it literal.
 
yungman said:
I know you use different formulas as torque require an arm to swing. But the circuit loop is the same.

No, that is the formula for the torque, they are not the same integrals. The infinitesimal force is on a wire in magnetostatics is

dF = I d\boldsymbol{\ell} \times \mathbf{B}

The infinitesimal torque is therefore,

d\mathbf{T} = I \mathbf{r}\times d\boldsymbol{\ell} \times \mathbf{B}

Since the position vector changes depending on where you are on your wire, you cannot take the position vector out of your integral. Hence,

\mathbf{T} = I \oint \mathbf{r}\times d\boldsymbol{\ell} \times \mathbf{B}
 
Born2bwire said:
No, that is the formula for the torque, they are not the same integrals. The infinitesimal force is on a wire in magnetostatics is

dF = I d\boldsymbol{\ell} \times \mathbf{B}

The infinitesimal torque is therefore,

d\mathbf{T} = I \mathbf{r}\times d\boldsymbol{\ell} \times \mathbf{B}

Since the position vector changes depending on where you are on your wire, you cannot take the position vector out of your integral. Hence,

\mathbf{T} = I \oint \mathbf{r}\times d\boldsymbol{\ell} \times \mathbf{B}

I worded it wrong, I meant they both are of different formulas, one for torque and one for linear. There is not disagreement on the physics here, I just complain about the wording of the book. It's all English. If they want to say that, they should clarify more.
 

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