A uniform magnetic field B is acting from south to north and is of magnitude 1.5 Wb/m2. If a proton having mass =1.7×10−27 kg and charge =1.6×10−19C moves in this field vertically downwards with energy 5 Me V, then the force acting on it will be
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
7.4×1012 N
b
7.4×10−12 N
c
7.4×1019 N
d
7.4×10−19 N
answer is B.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
F =qvB and K=12mv2⇒F=qB2Km=1.6×10−19×1.52×5×106×1.6×10−191.7×10−27=7.344×10−12 N