Question

A charged particle is moving in a uniform magnetic field in a circular path. Radius of circular path is R. When energy of particle is doubled, then new radius will be
A. \[R\sqrt 2 \]
B. \[R\sqrt 3 \]
C. \[2R\]
D. \[3R\]

Answer 1

Hint: Using Lorentz’s force law, when a charged particle is moving in a uniform magnetic field then the magnetic force keeps the charged particle in the circular path. The radius of the circular path is proportional to the speed of the particle.


Formula used:
\[r = \dfrac{{mv}}{{Bq}}\], here r is the radius of the circular path when a charged particle of mass m and charge q enters into a region of magnetic field strength B with speed v.



Complete answer:
If energy of the particle is E then it is due to the motion of the particle, i.e. the kinetic energy.
If the mass of the charged particle is m and the speed of the particle is v then,
\[\dfrac{{m{v^2}}}{2} = E\]
As the mass of the charged particle is constant, so changing the energy the speed of the particle will change.
Let the initial energy of the charged particle is \[{E_1}\]and the speed is \[{v_1}\]
\[\dfrac{{mv_1^2}}{2} = {E_1}\]
Let the final energy of the charged particle is \[{E_2}\]and the speed is \[{v_2}\]
\[\dfrac{{mv_2^2}}{2} = {E_2}\]
On dividing the first and the second expression for the energy,
\[{\left( {\dfrac{{{v_1}}}{{{v_2}}}} \right)^2} = \dfrac{{{E_1}}}{{{E_2}}}\]
\[\dfrac{{{v_1}}}{{{v_2}}} = \sqrt {\dfrac{{{E_1}}}{{{E_2}}}} \]
It is given that energy is doubled, \[{E_2} = 2{E_1}\]
\[\dfrac{{{v_1}}}{{{v_2}}} = \sqrt {\dfrac{1}{2}} \]
From the expression of the radius of the circular path, the radius of the circular path is proportional to the speed of the charged particle,
\[\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{{{v_1}}}{{{v_2}}}\]
\[\dfrac{R}{{{r_2}}} = \dfrac{1}{{\sqrt 2 }}\]
\[{r_2} = R\sqrt 2 \]
Hence, the on doubling the energy of the particle, the radius of the circular path is \[R\sqrt 2 \]
Therefore, the correct option is (a).






Note:We should be careful about the mass of the charged particle. As the proton is a subatomic particle so the relativistic energy may come into picture, but we have to solve the problem in classical mechanics.