R=qBmV
=qB2m(K.E)
mV=2mK.E
=1.6×10−19×1.5×10−32×9.1×10−31×100×1.6×10−19
=2.248cm≈2.25cm
sinθ=2.2482=0.89
PA=R(1−cosθ)=1.22cm
AB=6tanθ=11.71cm
d=PA+AB=12.93cm
An electron, moving along the x- axis with an initial energy of 100 eV, enters a region of magnetic field B=(1.5× 10^-3 T) k at S (see figure). The…
JEE Main 2019 — Physics Electromagnetism
2019mcqhard
An electron, moving along the x− axis with an initial energy of 100eV, enters a region of magnetic field B=(1.5×10−3T)k^ at S (see figure). The field extends between x=0 and x=2cm. The electron is detected at the point Q on a screen placed 8cm away from the point S. The distance d between P and Q (on the screen) is:
(electron’s charge 1.6×10−19C , mass of electron =9.1×10−31kg )
Official previous-year question
Held on 12 Apr 2019 · Verified 6 Jul 2026.
Options
- A
1.22cm
- B
12.87cm
- C
11.65cm
- D
2.25cm
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