CBSE-XI-Physics
34: Magnetic Field
- #4Which of the following particles will describe the smallest circle when projected with the same velocity perpendicular to a magnetic field?
(a) Electron
(b) Proton
(c) He+
(d) Li+digAnsr: aAns : (a) Electron
When a particle moves in a magnetic field, the necessary centripetal force, for the particle to move in a circle, is provided by the magnetic force acting on the particle.
So, equating the Lorentz force with the cetripetal force for a charged particle of charge q describing a circle of radius r, we get:
`` qvB=\hspace{0.17em}\frac{m{v}^{2}}{r}``
`` \Rightarrow r=\frac{mv}{qB}``
Here, m is the mass of the particle and v is its velocity.
The amount of charge of all the given particles is same; hence, for a given charge, `` r\propto m``. And since the electron is the lightest of all the particles, it describes the smallest circle when projected with the same velocity perpendicular to the magnetic field.
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