Why in $F = iLB$, $L$ is a vector but $i$ is not?

I learned $F = iLB$ recently. However, I don't understand why $L$ is marked as a vector but $i$ is not.
For a normal rod, how should I define the direction of length vector $L$? And if I reverse the current in it, the force exerted on it by the magnetic field would reverse direction, correct?
So I think in this formula, $i$ should be the vector but not $L$. Am I right?

I'm using the Physics II by Halliday Resnick and Krane

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I believe that in that text, $i$ refers to the magnitude of the current (a scalar), which is assumed to be in the same direction as the length vector $\vec{L}$ (a vector).

There's no need for both $i$ and $\vec{L}$ to be vectors. Think of current flowing through a wire—if $i$ were a vector ($\vec{i}$), then the direction of $\vec{i}$ would always be the same as the direction of the wire, because current always flows along a wire. The direction of the wire is already captured by $\vec{L}$, so it's not necessary to make $i$ a vector quantity also.

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This seems very much reasonable to me ;-) –  Waffle's Crazy Peanut Mar 21 '13 at 4:21

$$F = (iL)\times B$$ Here $B$ is a vector and $(iL)$ is also a vector. Direction of $(iL)$ is that of flowing current along the lengh $L$. $F$ is cross product of $(iL)$ and $B$ .

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And this also solves the doubt that current is vector or scalar –  ABC Mar 21 '13 at 4:02
It's the other way around, though, $(iL)\times B$. –  David Z Mar 21 '13 at 4:16
@DavidZaslavsky Thanks , done! –  ABC Mar 21 '13 at 4:18

Well, in theory - We've taken the element of length $l$ which carries current $I$. Hence, the vector belongs to the whole product, which is named as the current element $\vec{Il}$. Strictly speaking, current $I$ is a vector quantity. It's not like voltage or energy. It has a direction, which we say - "It's flowing from here to here".

(Just like every theory, where we consider a small element of length or area or volume so that we can work our calculations in it.)

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Simply put, current doesn't add like a vector. If I have a star-junction:

with currents $i_1$ and $i_2$ entering from the bottom and $i_3$ leaving the top, $-_3=i_1+i_2$, which is scalar addition. If we try to add the corresponding vectors, we get $\vec i_1+\vec i_2=\sqrt3(|\vec i_1|+|\vec i_2|)\hat i_3 \neq \vec i_3$.

On the other hand, $d\vec l$ is a vector. So, force on a small element of a wire = $id\vec l\times\vec B$. For a rod in a uniform magnetic field, we can integrate to get $\vec F=i\vec L\times\vec B$ since the other terms are independent of the position on the wire, and $\int d\vec L = \vec L$

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