Write the expression for the magnetic force `vecF` acting on a charged particle q moving with `vec upsilon` in the presence of the magnetic field `vec B` in a vector form. Show that no work is done and no change in the magnitude of the velocity of the particle is produced by this force. Hence define the unit of magnetic field.

Explanation

Work done by a magnetic force on a charge particle. The magnetic force ` vecF=q(vecupsilonxx vec B)` always acts perpendicular to the velocity `vecupsilon` or the direction of motion of charge q. Therefore,
`vecF.vecupsilon=q(vec upsilonxx vecB).vec upsilon=0`
According to Newton's second law,
` vecF=mveca=m(dvecupsilon)/(dt)`
` therefore m(dvecupsilon)/(dt).vecupsilon=0`
or ` (m)/(2) [(dvecupsilon)/(dt).vecupsilon+vecupsilon.(dvecupsilon)/(dt)]=0`
or ` (m)/(2)(d)/(dt) (vecupsilonxx vecupsilon)=0`
or `(d)/(dt) ((1)/(2)mv^2)=0`
or `(dK)/(dt)=0`
or` K ="constant"`
Thus, a magnetic force does not change the kinetic energy of the charged particle. ` F=q upsilonB sin theta`
From above relation if,
`q=1C,upsilon1ms^-1, 0=90^@,F=1N`
then, `B=1`Tesla
i.e., magnetic field at a place will be 1T if a charged of 1C moveing in a perpendicular magnetic field with unit velocity experience a force of 1N.