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A charged particle of mass m and charge q is accelerated from rest through a potential difference V. It then enters a region containing a uniform magnetic field B directed perpendicular to its velocity. The field occupies a slab of thickness d. Find the angle theta by which the particle's direction is deflected upon exiting the field region.
- sin(theta) = B*d*(q/(2*m*V))^(1/2)
- cos(theta) = B*d*(q/(2*m*V))^(1/2)
- tan(theta) = B*d*(q/(2*m*V))^(1/2)
- cot(theta) = B*d*(q/(2*m*V))^(1/2)
Correct answer: sin(theta) = B*d*(q/(2*m*V))^(1/2)
Solution
The particle moves in a circular arc of radius r = mv/(qB) inside the magnetic field. The thickness d is the perpendicular distance traversed, so sin(theta) = d/r. Substituting r using v = sqrt(2qV/m) gives sin(theta) = Bd*sqrt(q/(2mV)).
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