Derive the expression for the magnetic force acting on a moving charge $q$ with velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$.

(D) Consider a small segment of a conducting wire of length $dl$ and cross-sectional area $A$,containing $n$ charge carriers per unit volume,each with charge $q$ moving with drift velocity $\vec{v}$.
The current $I$ in the wire is given by $I = nAq v$.
The force $d\vec{F}$ on a current element $I d\vec{l}$ in a magnetic field $\vec{B}$ is $d\vec{F} = I d\vec{l} \times \vec{B}$.
Substituting $I = nAqv$,we get $d\vec{F} = (nAqv) d\vec{l} \times \vec{B}$.
Since the velocity $\vec{v}$ is in the direction of $d\vec{l}$,we can write $v d\vec{l} = (dl) \vec{v}$.
Thus,$d\vec{F} = nA(dl) q (\vec{v} \times \vec{B})$.
The total number of charge carriers in the volume element $dV = A dl$ is $N = n A dl$.
The force on a single charge $q$ is $\vec{F}_m = \frac{d\vec{F}}{N} = \frac{nA dl q (\vec{v} \times \vec{B})}{nA dl}$.
Therefore,the magnetic force on a moving charge is $\vec{F}_m = q(\vec{v} \times \vec{B})$.