What's the difference between $f= i (l × B)$ and $f= (i∫dl )× B$?

Question

for any current carrying conductor of any arbitrary shape in uniform magnetic field experiences a force given by F = i (l × B) where l is length between end and start points of the conductor and in f= (i∫dl )× B, ∫dl is vector sum of all lengths elements from initial to final end

Answer 1

$\int{\bf dl=l}$ for $\bf l$ and $\bf dl$ being vectors, so they are the same.

Answer 2

Your wording is confusing; the word "length" along a curve is usually used to refer to a scalar quantity, but ${\bf l}$ must be a vector if it goes into a cross product.

I assume you mean an arbitrary 1D conductor, in order for the notion of "current" to be unambiguous. If you also mean that ${\bf l}$ is the displacement vector that connects the ends of the conductor, then you can pull the constant ${\bf B}$ field out of the integral to get $\int d{\bf l} = {\bf l}$, so the two expressions are equivalent. But this only works if the ${\bf B}$ field is spatially uniform along the conductor.

Answer 3

In a uniform field B can be taken out of the integral. If you have a straight current carrying conductor, the first equation will hold. For a ring for instance, you will apply the second equation, you will find that the forces on the ring will try to orient the ring so that its axis points along the direction of B.

Answer 4

The correct formula is $$\vec{F}=I\int \left(d\vec{l}\times\vec{B}\right).$$ In this expression, $\vec{F}$ is the total force on a length of current-carrying wire, and the line integration runs over this length. The magnetic field $\vec{B}(\vec{r})$ remains inside the integral, because the field is generally a function of the position along the wire.

If the magnetic field is actually constant over the physical length of the wire, then it cam be pulled out of the integral, yielding something like the expressions in the question. However, the notation in the question suggests some confusion about the vector nature of the integral. The product in the force law is a vectorial cross product, and even if $\vec{B}$ is constant, so we can factor it out and have $$\vec{F}=I\left(\int d\vec{l}\right)\times\vec{B},$$ the quantity $\left(\int d\vec{l}\right)$ is still a vector. It is not the length of the wire, but rather the vector displacement between the two ends of the wire. Only for a straight wire is the magnitude of the vector equal to the arc length.