Question
Download Solution PDFA particle of charge e and mass m moves with a velocity v in a magnetic field B applied perpendicular to the motion of the particle. The radius r of its path in the field is _______
This question was previously asked in
Agniveer Navy SSR: 25th May 2025 Shift 2 Memory-Based Paper
Answer (Detailed Solution Below)
Option 1 : \(\frac{mv}{Be}\)
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Agniveer Navy SSR Full Test - 01
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Detailed Solution
Download Solution PDFCONCEPT:
- When moving through a magnetic field, the charged particle experiences a force.
- When the direction of the velocity of the charged particle is perpendicular to the magnetic field:
- Magnetic force is always perpendicular to velocity and the field by the Right-Hand Rule.
- And the particle starts to follow a curved path.
- The particle continuously follows this curved path until it forms a complete circle.
- This magnetic force works as the centripetal force.
- Centripetal force (FC) = Magnetic force (FB)
⇒ qvB = mv2/R
⇒ R = mv/qB
where q is the charge on the particle, v is the velocity of it, m is the mass of the particle, B is the magnetic field in space where it circles, and R is the radius of the circle in which it moves.
EXPLANATION:
Given that particle has charge e; mass = m; and moves with a velocity v in a magnetic field B. So
- Centripetal force (FC) = Magnetic force (FB)
⇒ qvB = mv2/R
\(\Rightarrow R=\frac{mv}{qB}\)
\(\Rightarrow r=\frac{mv}{Be}\)
So the correct answer is option 1.