Questions

When a proton is released from rest in a room, it starts with an initial acceleration ${a}_{0}$towards west. When it is projected towards north with a speed ${\upsilon }_{0}$it moves with an initial acceleration $3{a}_{0}$ toward west. The electric and magnetic fields in the room are

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a
ma0eeast,3ma0eυ0up
b
ma0eeast,3ma0eυ0down
c
ma0ewest,2ma0eυ0up
d
ma0ewest,2ma0eυ0down

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detailed solution

Correct option is D

when proton is stationary force is only due to electric field  i.e, E⇀ =ma0e along westforce due to magnetic field must be   2ma0along west  its velocity is along north (j) force is along  west (-i)F→=q(V ⇀ xB⇀)    (-i=j x-k)  hence magnetic field is  vertically downwards direction and having a magnitude 2ma0e

A charged particle of mass m, carrying a charge Q is projected from the origin with a velocity $\stackrel{\to }{\mathrm{V}}={\mathrm{V}}_{\mathrm{o}}\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}\right)$. A uniform magnetic field $\stackrel{\to }{\mathrm{B}}=\mathrm{B}\stackrel{^}{\mathrm{i}}$ exists in that region. Then maximum distance of the particle from x-axis is