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
 A: 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.
A: $$ F = (iL)\times B$$  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 $B$ .
A: 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.)
A: 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 $\int d\vec L = \vec L$
