A particle of charge q and mass m starts moving from the origin under the action of an electric field E→$$=E0i^$$ and magnetic field B→$$=B0i^$$ with a velocity v→$$=v0j^$$. The speed of the particle will become $$2v0$$ after a time.

Question

A particle of charge q and mass m starts moving from the origin under the action of an electric field  E→$$=E0i^$$ and magnetic field B→$$=B0i^$$ with a velocity v→$$=v0j^$$. The speed of the particle will become $$2v0$$  after a time.

Infinity Learn · Accepted Answer

$$1$$. Magnetic field can’t change speed of charged particle, it will only change the direction of  velocity $$v0j^$$  such that it always lie in $$y-z$$  $$plane.2$$. Since the electric field is in  $$x-direction$$, hence it would not affect the velocity in the $$y-z$$ plane.      After time  t, the electric field will inject some velocityvx in the $$x-$$ direction such that the magnitude of velocity changes to  $$2v0$$.  ∴$$v02+vx2=2v0$$⇒$$vx2=4v02$$−$$v02=3v02$$⇒$$vx=3v03$$. Electric force  $$F=qE04$$. Acceleration  $$=Fm5$$.  The time taken by $$x-component$$ velocity to become from zero to $$3v0$$ due to uniform electric field  $$=vxfinal$$−$$vxinitialacceleration=3v0$$−$$0a$$      here  $$a=E0qmt=3v0E0qm=3v0mE0qTherefore$$, the correct answer is D

A particle of charge q and mass m starts moving from the origin under the action of an electric field E→=E0i^ and magnetic field B→=B0i^ with a velocity v→=v0j^. The speed of the particle will become 2v0 after a time.

Detailed Solution

Get Expert Academic Guidance – Connect with a Counselor Today!