Question

A deuteron of kinetic energy 50keV is describing a circular orbit of radius 0.5m in a plane perpendicular to a magnetic field B. The kinetic energy of the proton that describes a circular orbit of radius 0.5m in the same plane with the same B is:

• A. \(25\,\text{keV}\)
• B. \(50\,\text{keV}\)
• C. \(200\,\text{keV}\)
• D. 100keV

Hint:

In magnetic fields: K=(q²B²r²)/(2m) Lighter particles need more kinetic energy for the same orbit radius.

Answer 1

The Correct Option is D

Step 1: Radius of circular motion in magnetic field: r=(mv)/(qB)
Step 2: Kinetic energy: K=(1)/(2)mv²=(q²B²r²)/(2m)
Step 3: For same r, B, and q: K ∝ (1)/(m)
Step 4: Mass of deuteron is approximately twice that of proton: Kp = 2Kd = 2×50 = 100keV