Question

An electron after being accelerated at a potential difference of ${10^4\,}V$ enters a magnetic field of $0.04\,T$ perpendicular to its direction of motion. Calculate the radius of curvature of its trajectory?

Answer 1

Hint: For a charged particle when the direction of motion is perpendicular to the magnetic field then the motion of the charged particle will be circular and when the direction of motion (velocity) is parallel to the magnetic field then the motion would be in a straight line.

Formula used:
Force for the moving charge under the influence of magnetic field:
$ \Rightarrow \overrightarrow F = q\overrightarrow v \times \overrightarrow B $........ (1)
Where $q$ is the charge of the particle, $\overrightarrow v $ is the velocity with which the particle is moving, $\overrightarrow B $ is the magnetic field
Force when a body moves in circular orbit :
$ \Rightarrow \overrightarrow F = \dfrac{{m{v^2}}}{r}$ .......... (2)
Where $F$ is the force under circular motion, $m$ is mass of the particle, $v$ is the velocity with which the body is moving, $r$ is the radius of the circular orbit.

Complete step by step answer:
In question, it is given that the particle entered the magnetic field of the strength $0.04T$ and the direction of motion is perpendicular to magnetic field then from equation (1)
$ \Rightarrow \overrightarrow F = qvB\sin {90^ \circ }$
We get,
$ \Rightarrow F = qvB$ ............ (3)
If the velocity is perpendicular to magnetic field so the motion becomes circular and the force under circular motion is,
$ \Rightarrow F = \dfrac{{m{v^2}}}{r}$ ........ (4)
From equation (3) and (4)
We get,
$ \Rightarrow qvB = \dfrac{{m{v^2}}}{r}$
$ \Rightarrow qB = \dfrac{{m{v^{}}}}{r}$
$ \Rightarrow r = \dfrac{{mv}}{{qB}}$ ........ (5)
From the formula of momentum $p = mv$
we get,
 $ \Rightarrow r = \dfrac{p}{{qB}}$ .......... (6)

In question, it is given that the electron is accelerated with the potential difference of ${10^4}V$. So the kinetic energy of electron,
$ \Rightarrow \dfrac{{{p^2}}}{{2m}} = eV$ ....... (7)
We get $p = \sqrt {2meV} $ put the value of momentum in equation (6)
$ \Rightarrow r = \dfrac{{\sqrt {2meV} }}{{qB}}$ .......... (8)
Given that $V = {10^4}V$ and $B = 0.04T$ , mass of electron $m = 9.1 \times {10^{ - 31}}kg$ , charge on electron $q = 1.6 \times {10^{ - 19}}C$
Put all these values in equation (8)
We get,
$ \Rightarrow r = \dfrac{{\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 1.6 \times {{10}^{ - 19}} \times {{10}^4}} }}{{1.6 \times {{10}^{ - 19}} \times 0.04}}$
$ \Rightarrow r = \dfrac{{5.396 \times {{10}^{ - 23}}}}{{6.4 \times {{10}^{ - 21}}}}$
$\therefore r = 0.84 \times {10^{ - 2}}$ Or $r = 8.4 \times {10^{ - 3}}m$

Therefore, the radius of curvature of the trajectory is $8.4 \times {10^{ - 3}}m$.

Note: In practical a current carrying wire also produces a magnetic field of its own so the current flowing through a wire is proportional to the magnetic fields. To find the value of a magnetic field across an infinitely long straight wire we can apply Ampere’s Law.The electric current also represents flow of electrons whereas they both move in opposite directions. The direction of current is in the positive terminal to negative and the electron moves from negative to positive direction.