So, it means that the magnetic force on a current carrying conductor does depend on the shape of the conductor but only on the END POINTS of Conductor.

Corollary: In a current carrying loop, the net magnetic force is zero, this is because there are no end points to the loop. In other words, the starting is itself the ending point (say \(P\)).

So, from above theory, \(F = i \int_{P}^{P} \vec{dx} \times \vec{B} = 0\).

\(\vec{p_m}\) is the same in every coordinate system therefore magnetic moment is also zero under nabla operation! (You can proof this by using the definition of magnetic moment \(\vec{p_m}=\frac{1}{2} \int \vec{r} \times \vec{j} dS\))

Force(here magnetic force) is a conservative quantity. Therefore circulation of force in a closed loop=0.
This is based on vector calculus. The above thing is read out in vector calculus notation as 'del cross F=0'.
Therefore,magnetic force in a closed loop is zero.

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It is clear that force on \(\vec{dx}\) is equal to \(\vec{dF} = i (\vec{dx} \times \vec{B})\).

Integrating it from \(A\) to \(B\),

\(F =i \int_{A}^{B} \vec{dx} \times \vec{B}\).

But since \(\vec{B}\) is constant in both magnitude and direction everywhere. We can write,

\(F =i (\int_{A}^{B} \vec{dx}) \times \bar{B} = \vec{AB} \times \vec{B}\).

So, it means that the magnetic force on a current carrying conductor does depend on the shape of the conductor but only on the END POINTS of Conductor.

Corollary:In a current carrying loop, the net magnetic force is zero, this is because there are no end points to the loop. In other words, the starting is itself the ending point (say \(P\)).So, from above theory, \(F = i \int_{P}^{P} \vec{dx} \times \vec{B} = 0\).

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Same method. Simple and useful. :)

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Thanks.

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Potential energy is:

\(U=- \vec{p_m} \cdot \vec{B}\)

Now we can write the force as:

\(\vec{F}=-\nabla U\)

Which is:

\(\vec{F}=-(\vec{p_m} \times(\nabla \times \vec{B})+\vec{B} \times(\nabla \times \vec{p_m})+(\vec{p_m} \cdot \nabla)\vec{B} +(\vec{B} \cdot \nabla) \vec{p_m})\)

\(\vec{B}\) itself is constant and it is equal to zero under any operation of nabla so we are left with:

\(\vec{F}=-(\vec{B} \times(\nabla \times \vec{p_m}) +(\vec{B} \cdot \nabla) \vec{p_m})\)

\(\vec{p_m}\) is the same in every coordinate system therefore magnetic moment is also zero under nabla operation! (You can proof this by using the definition of magnetic moment \(\vec{p_m}=\frac{1}{2} \int \vec{r} \times \vec{j} dS\))

Then \(\vec{F}=0\).

\((\vec{a}\cdot \nabla)\vec{b}=a_x \frac{\partial b}{\partial x}+a_y \frac{\partial b}{\partial y}+a_z \frac{\partial b}{\partial z}\)

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Great ! and thanks a lot.

Your Physics is good. Can you suggest some books for physics ?

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Thanks!

The most theory i learned from Serbian books but i can recommend you some not so advanced books:

For classical mechanics and intro to special theory of relativity prefer Morin's book.

For EM Irodov Basics laws of electromagnetism And also Morin's book on EM.

These books have a lot of problems so they are good to establishe your knowledge.

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Force(here magnetic force) is a conservative quantity. Therefore circulation of force in a closed loop=0. This is based on vector calculus. The above thing is read out in vector calculus notation as 'del cross F=0'. Therefore,magnetic force in a closed loop is zero.

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