14.3.1 Derivation of F=Bqv

Why does a magnetic field exert a magnetic force on a current-carrying conductor? A current is nothing more than a flow of electric charges. Could it be that the magnetic field is actually exerting magnetic forces on the moving charges? If that’s so, we should be able to derive the formula for the magnetic force acting on a moving charge, starting from the \displaystyle {{F}_{b}}=BIL formula!

Let’s consider a section of a copper wire of length L. Let the total amount of mobile charges in this section be Q.

If v is the drift velocity of the electrons, then the time taken for Q to completely pass out of this section would be \displaystyle L\div v. So

\displaystyle \begin{aligned}{{F}_{b}}&=BIL\\&=B\frac{Q}{t}L\\&=B\frac{Q}{{L\div v}}L\\&=BQv\end{aligned}

BQv is actually the summation of the magnetic forces acting on each of the individual mobile charge making up Q. So the magnetic force acting on a single charged particle carrying charge q is given by the formula

\displaystyle {{F}_{b}}=Bqv

This is of course for the case when B and v are perpendicular. If they are not, we simply replace B with \displaystyle \displaystyle {{B}_{\bot }}, the component of B perpendicular to v.

\displaystyle \displaystyle {{F}_{b}}={{B}_{\bot }}qv

As for the direction, we can use the Fleming’s Left Hand Rule for \displaystyle {{F}_{b}}=Bqv similar to how we use it for \displaystyle {{F}_{b}}=BIL. The thumb and index fingers still point in the direction of Fb and B. The middle finger points in the direction of conventional current produced by the movement of the charge. If the moving charge is positive, point your middle finger in the direction of v. If the moving charge is negative, point your middle finger in the direction opposite to v.

Concept Test



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