2.0 The Magnetic Force on a Moving Charge

  Magnetics

In order to define the Magnetic field $\vec B$ we deduce an expression for the force on a moving charge in a magnetic field.

Consider a positive charge $q$ moving in a uniform magnetic field $\vec B$ with a velocity $\vec V$ let the $θ$ be the angle between $\vec V$ and $\vec B$

i. The magnetic of force $\vec F_m$ experienced by the moving charge is directly proportional to the magnitude of the charge.$$i.e.,\quad {\vec F_m} \propto \;q$$

ii. The magnetic of force $\vec F_m$ is directly proportional to the component of velocity acting perpendicular to the direction of magnetic field$$i.e.,\quad {\vec F_m} = v\sin \theta $$iii. The magnetic of force $\vec F_m$ is directly proportional to the magnitude of the magnetic field applied.$$i.e.,\quad {\vec F_m} = B$$Combining the above factors we get,$${\vec F_m} \propto \;q\;B\;v\sin \theta $$ or$${{\vec F}_m} = kq\;B\;v\sin \theta $$Where k is constant of proportionally it’s value is found to be $k = 1$

$\therefore\ \ $ the magnitude of $$\vec F_m=q\ v\ b\ \sin \theta\ \ \ ....(1)$$

The magnetic force $\vec F_m$ on a charge $q$ moving with the velocity $\vec v$ in a magnetic field $\vec B$ is given by, $$\vec F_m=q(\vec v \times \vec B)\ \ \ ....(2)$$Special cases :-

i. If $\theta =0^\circ$ or $180^\circ,$ then $\sin \theta =0$

From equation $(1)$ $$\vec F_m=q\ v\ b(0)=0$$It means a charged particles moving along or opposite to the direction of magnetic field, does not experience any force.

ii. If $v=0,$ then $$\vec F_m=q(0)b\ \sin \theta=0$$Its means if a charged particles is at rest in a magnetic field, it experience no force

iii. If $\theta =90^\circ$ then $\sin \theta =1$$$\therefore\ \ \vec F_m=q\ v\ b(1)=q\ v\ b\ \text{(maximum)}$$

iv. Unit of $\vec B.$ The unit of $\vec B$ can be obtained from $\frac{F}{qv}$

$S.I$ unit $B$ is tesla $(T)$ or $\frac{{{\text{weber}}}}{{{{\left( {{\text{meter}}} \right)}^{\text{2}}}}}$

i.e., $\left( {\frac{{wb}}{{{m^2}}}} \right)$ or $NsC^{-1}m^{-1}$$$\therefore\ \ 1T=10^4$$

v. The magnetic force is always perpendicular to the direction of motion of charged particle, and therefore magnetic force does no work on it. Hence the kinetic energy of moving charged particle in magnetic field remains constant.

vi. In using the formula $\vec F_m=q\ v\ b\ \sin \theta,$ put the value of the charge $q$ without sign.

vii. In using the formula $\vec F_m=q(\vec v\times \vec B),$ put the value of the charge with the proper sign.