A proton of mass m and charge +e is moving in a circular orbit of a magnetic field with energy 1MeV. What should be the energy of α-particle (mass = 4 m and charge = +2e), so that it can revolve in the path of the same radius?
Correct answer: A. 1 MeV
- A. 1 MeV
- B. 4 MeV
- C. 2 MeV
- D. 0.5 MeV
Explanation
By using r = √2mK / qB ; r → same, B → same ⇒ K ∝ q²/mHenceKₐ / Kₚ = (qₐ / qₚ)² × mₚ / mₐ = (2 qₚ / qₚ)² × mₚ / 4 mₚ ⇒ Kₐ = Kₚ = 1 meV.Thus, the correct answer is 1 MeV, while other options do not correctly account for the relationships between charge, mass, and energy.
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About Electromagnetism
Magnetic flux density and magnetic flux describe the strength of a magnetic field and the field passing through a surface. A charged particle moving through a magnetic field experiences a force perpendicular to its velocity and may follow circular or helical motion, depending on the angle between velocity and field.
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