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 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:
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In magnetic fields:
K=(q²B²r²)/(2m)
Lighter particles need more kinetic energy for the same orbit radius.
Updated On: Mar 19, 2026
- \(25\,\text{keV}\)
- \(50\,\text{keV}\)
- \(200\,\text{keV}\)
- 100keV
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Verified By Collegedunia
The Correct Option is D
Solution and Explanation
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
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