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PhysicsMediumMCQ2015 · 10 Apr Online

Q21.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 sin α will be: (1) B (2) q 2 √qdmV Bd√ 2mV (3) B q (4) qV d √ 2mV √Bd2m

What This Question Tests

Tests the ability to combine concepts of energy conservation (potential to kinetic) and motion of a charged particle in a uniform magnetic field to determine the angle of deviation.

Concepts Tested

Kinetic energy from potential differenceLorentz force on charged particleRadius of circular path in magnetic fieldTrigonometry

Formulas Used

KE = qV

KE = 1/2 mv²

r = mv / (qB)

sin α = d/r

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