Answer:
We know that force on the moving (velocity V) charge q due to magnetic field B given as
[tex]F=q(\vec{V}\times \vec{B})[/tex]
If force act for time t then energy gained by moving charge
[tex]E=t(\vec{F}.\vec{V})[/tex]
[tex]F=q(\vec{V}\times \vec{B})[/tex]
[tex]E=t(\vec{q(\vec{V}\times \vec{B})}.\vec{V})[/tex]
We know that
[tex]For\ vector\ a\ and\ b\\ a.(\vec{a}\times \vec{b})=0[/tex]
So
E=0
Now we can say that total kinetic energy of charge q will become
[tex]K.E.=\dfrac{1}{2}mV^2+E[/tex]
[tex]K.E.=\dfrac{1}{2}mV^2+t(\vec{F}.\vec{V})[/tex]
[tex]K.E.=\dfrac{1}{2}mV^2[/tex]
So
[tex]\dfrac{d(K.E.)}{dt}=0[/tex] (V= constant)
We can say that
K.E.= constant
So the force exerted by a magnetic field does no work on a charged particle.
