Question

An electron after being accelerated through a potential difference of $100\,V$ enters a uniform magnetic field of $0.004\,T$ perpendicular to its direction of motion.Calculate the radius of the path described by the electron.

Answer 1

We shall first equate two expressions for the kinetic energy of a particle $K = eV$ where $K$ is the kinetic energy, $e$ is the electronic charge and $V$ is the potential difference applied and $K = \dfrac{{m{v^2}}}{2}$ where $m$ is the mass and $v$ is the velocity of the particle. From this we will obtain a relation between the potential difference and the velocity of the particle. Then we shall equate the forces acting at any equilibrium position to derive an expression for the radius of the circular path. By making proper substitutions, we shall get the answer.

Complete step by step answer:
The kinetic energy of a charged particle can be given as $K = eV$. Also, the kinetic energy is given as $K = \dfrac{{m{v^2}}}{2}$. Equating both the expressions for the kinetic energy, we get,
$\dfrac{{m{v^2}}}{2} = eV$
$\Rightarrow v = \sqrt {\dfrac{{2eV}}{m}}$
Since the electron enters a magnetic field, it experiences Lorentz force which makes its trace a circular path. Also, since the path is circular, a centripetal force also acts towards the center of the circular path.

At any position during the motion of electrons, the two forces must balance.Hence,
$evB = \dfrac{{m{v^2}}}{r}$
$\Rightarrow eB = \dfrac{{mv}}{r}$
Substituting the value of velocity as calculated before, we get
$eB = \dfrac{{m\sqrt {\dfrac{{2eV}}{m}} }}{r}$
Rearranging the terms we get,
$r = \dfrac{{\sqrt {\dfrac{{2Vm}}{e}} }}{B}$
Now substituting the given values,
$V = 100\,V$ , $B = 0.004\,T$ , $m = 9.1 \times {10^{ - 31\,}}\,kg$ , $e = 1.6 \times {10^{ - 19}}\,C \\\$
$\Rightarrow r = \dfrac{{\sqrt {\dfrac{{2 \times 100 \times 9.1 \times {{10}^{ - 31}}}}{{1.6 \times {{10}^{ - 19}}}}} }}{{0.004}} \\\$
Further solving this we get,
$\therefore r = 8.44\,mm$

Hence, the radius of the circular path is $r = 8.44\,mm$.

Note: The Lorentz force acts perpendicular to both the magnetic field and the velocity of the charged particle. The actual expression is $F = q(v \times B)$ which gives both the direction and magnitude of the force. But in this question, we had to give the magnitude of the radius and hence we neglected the direction of the force. In any case whatsoever, the direction of the force acts perpendicular to both the velocity and the magnetic field.