A cyclotron is used to accelerate protons. If the operating magnetic field is 1.0 T and the radius of the cyclotron 'dees' is 60 cm, the kinetic energy of the accelerated protons in MeV will be :
[use mₚ=1.6 × 10⁻²⁷ kg, e=1.6 × 10⁻¹⁹ C ]

Solution

Physics sir · 4 steps

The dee radius caps the orbit, so the largest radius fixes the speed and the kinetic energy.

1. Largest orbit. The proton leaves the cyclotron when its circle reaches the edge of the dees. In the field, qvB = mv²/R, so the exit speed is v = qBR/m.

   K = 1/2mv² = q²B²R²/2m

2. Put in numbers. Use R = 0.6 m and B = 1 T. Then K = ((1.6 × 10⁻¹⁹)² × 1² × 0.36)/(2 × 1.6 × 10⁻²⁷) = 2.88 × 10⁻¹² J.
3. Convert to MeV. Divide by 1.6 × 10⁻¹⁹ J eV⁻¹: K = 1.8 × 10⁷ eV.
4. So. The protons leave with K = 18 MeV, which is option (b).

The idea to keep. In a cyclotron the final energy q²B²R²/2m depends only on the field and the dee radius, not on the voltage across the gap. Remember to change cm to m.

Answer from the NTA final answer key. Solution by PrepWiser.