
The radius of circular path in given magnetic field is
r=d=qB0mv
so v=mqB0d
JEE Main 2020 — Physics Electromagnetism
A particle of charge q and mass m is moving with a velocity −vi^(v=0) towards a large screen placed in the Y−Z plane at distance d. If there is magnetic field B=B0k^, the minimum value of v for which the particle will not hit the screen is :
Held on 6 Sept 2020 · Verified 6 Jul 2026.
3mqdB0
m2qdB0
mqdB0
2mqdB0
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