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A particle having mass m and charge q is first accelerated through a potential difference V and then enters a zone of uniform magnetic field B acting perpendicular to its velocity. If the magnetic field extends over a width d, what is the angle through which the particle’s direction changes on emerging from the field?

1. sin−1[ Bd (q/2mV)1/2 ]
2. cos−1[ Bd (q/2mV)1/2 ]
3. tan−1[ Bd (q/2mV)1/2 ]
4. zero

Correct answer: sin−1[ Bd (q/2mV)1/2 ]

Solution

From qV=(1/2)mv^2, v=sqrt(2qV/m); radius r=mv/(qB). Deflection angle theta has sin(theta)=d/r = Bd*(q/(2mV))^(1/2), so theta = sin^-1[Bd(q/2mV)^(1/2)].

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