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A cyclotron of dee radius R accelerates ions of charge q, mass m in a magnetic field B. The maximum kinetic energy is
AqBR
Bq²B²R²/(2m)
CqBR²/(2m)
Dq²B²/(2mR)
Answer & Solution
Correct answer: B. q²B²R²/(2m)
1. At the edge of the dees, the orbit radius equals R: qvB = mv²/R gives v_max = qBR/m.
2. Kinetic energy K = ½ m v_max².
3. Substitute v_max: K = ½ m (qBR/m)².
4. Simplify: K = q²B²R²/(2m). This is the maximum extracted KE.
_Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", §4.4.2_
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