Electrostatics

JEE Main level, single correct, medium

113 of 311

Question 113 of 311

A particle of mass mm and charge qq has an initial velocity v⃗=v0j^\vec{v}={v}_{0}\hat{j} . If an electric field E⃗=E0i^\vec{E}={E}_{0}\hat{i} and magnetic field B⃗=B0i^\vec{B}={B}_{0}\hat{i} act on the particle, its speed will double after a time

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Answer: (C) 3mv0qE0\dfrac{\sqrt{3}\text{m}{v}_{0}}{{\text{qE}}_{0}}

Acceleration produced by electric field,
a=qE0mi^a=\dfrac{q{E}_{0}}{m}\hat{i}
After time tt, velocity of the particle,
v=u+atv=u+at
⇒v=v0j^+(qE0tm)i^[∵u=v=v0j^]\Rightarrow v={v}_{0}\hat{j}+\left(\dfrac{q{E}_{0}t}{m}\right)\hat{i}\left[∵u=v={v}_{0}\hat{j}\right]
So, speed of the particle,
∣v∣=v02+(qE0tm)2\left|v\right|=\sqrt{{v}_{0}^{2}+{\left(\dfrac{q{E}_{0}t}{m}\right)}^{2}}
Now, given ∣v∣=2v0\left|v\right|=2{v}_{0}after time tt, so
2v0=v02+(qE0tm)22{v}_{0}=\sqrt{{v}_{0}^{2}+{\left(\dfrac{q{E}_{0}t}{m}\right)}^{2}}
⇒4v02=v02+(qE0tm)2\Rightarrow 4{v}_{0}^{2}={v}_{0}^{2}+{\left(\dfrac{q{E}_{0}t}{m}\right)}^{2}
⇒3v02=(qE0tm)2\Rightarrow 3{v}_{0}^{2}={\left(\dfrac{q{E}_{0}t}{m}\right)}^{2}
So, time interval, t=3mv0qE0t=\dfrac{\sqrt{3}m{v}_{0}}{q{E}_{0}}
Here, we must note that no change in magnitude of velocity is caused by a perpendicular magnetic field. So, we are not taking effect of magnetic field while calculating change in speed.

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