A particle of mass m and charge q has an initial velocity v=v_0 j . If an electric field E=E_0 i and magnetic field B=B_0 i act on the particle, its…
JEE Main 2020 — Physics Electromagnetism
2020mcqmedium
A particle of mass m and charge q has an initial velocity v=v0j^ . If an electric field E=E0i^ and magnetic field B=B0i^ act on the particle, its speed will double after a time
Official previous-year question
Held on 7 Jan 2020 · Verified 6 Jul 2026.
Options
A
qE02mv0
B
qE03mv0
C
qE03mv0
D
qE02mv0
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Solution
Acceleration produced by electric field,
a=mqE0i^
After time t, velocity of the particle,
v=u+at
⇒v=v0j^+(mqE0t)i^[∵u=v=v0j^]
So, speed of the particle,
∣v∣=v02+(mqE0t)2
Now, given ∣v∣=2v0after time t, so
2v0=v02+(mqE0t)2
⇒4v02=v02+(mqE0t)2
⇒3v02=(mqE0t)2
So, time interval, t=qE03mv0
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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