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Question

(D) The force acting on a charged particle moving in both electric and magnetic fields is given by the Lorentz force equation: $F = q(E + v 	imes B)$

Vedclass · Accepted Answer

$v \propto B^0$

Vedclass · Answer

(D) The force acting on a charged particle moving in both electric and magnetic fields is given by the Lorentz force equation: $F = q(E + v 	imes B)$.According to the work-energy theorem,the work done by the net force on the particle is equal to the change in its kinetic energy.The work done by the magnetic force $(F_m = q(v 	imes B))$ is always zero because the magnetic force is always perpendicular to the velocity $(v)$ of the particle.Therefore,only the electric field $(E)$ does work on the charge.The work done by the electric field is $W = F_e \cdot d = (qE) \cdot d$,where $d$ is the displacement.Since the work done is equal to the change in kinetic energy,we have $qEd = \frac{1}{2}mv^2$.From this,$v^2 = \frac{2qEd}{m}$,which implies $v \propto \sqrt{E}$.However,looking at the options provided,the magnetic field $(B)$ does not contribute to the change in speed of the particle.Since the speed $v$ is independent of the magnetic field $B$,we can write $v \propto B^0$ (where $B^0 = 1$).Thus,option $D$ is the correct relationship.

$A$ charge $q$ is released in the presence of an electric field $(E)$ and a magnetic field $(B)$. After some time,its velocity is $v$. Then:

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