
Due to potential difference V speed acquired by proton is v0,
⇒qV=21mv02=K
∵Radius,R=qBmv0=qB2mK
∵K=qV
R=qB2meV
sinα=Rd=2meVdqB=Bd2mVq
A proton (mass m) accelerated by a potential difference V flies through a uniform transverse magnetic field B. The field occupies a region of space by width d. If α be the angle of deviation of proton from the initial direction of motion (see figure), the value of sinα will be:

Held on 10 Apr 2015 · Verified 6 Jul 2026.
2BmVqd
Bd2mVq
dB2mVq
qV Bd2m
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