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PhysicsMagnetismMagnetic ForceMedium2 minPYQ_2015
PhysicsMediumsingle choice

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:

Question diagram: A proton (mass m ) accelerated by a potential difference V f

Options:

Answer:
B
Solution:



Due to potential difference V speed acquired by proton is v0,

qV=12mv02=K

Radius, R=mv0qB=2mKqB
K=qV
R=2meVqB
sinα=dR=dqB2meV=Bdq2mV

Stream:JEESubject:PhysicsTopic:MagnetismSubtopic:Magnetic Force
2mℹ️ Source: PYQ_2015

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