Question

1) An electron emitted by a heated cathode and accelerated through a potential difference of $2.0KV,$ enters a region with uniform magnetic field of $0.15T.$ Dtermine the trijectory of the electron if the field is trannsverse to its initial velocity?

2) An electron emitted by a heated cathode and accelerated through a potential difference of $2.0KV,$ enters a region with uniform magnetic field of $0.15T.$ Dtermine the trijectory of the electron if the field makes an angle of ${30}^{\circ }$ with the initial velocity?

Answer 1

1) Step 1: find the velocity of the electron.
Given, Maggnetic field strength,
$B=0.15T$

Potential difference,

$V=2.0KV=2.0\times {10}^{3}V$

Let the electrons move through the magnetic field with velocity $v$.

Electron possess kinetic energy in electric field.
So,

$eV=\frac{1}{2}m{v}^2$

where, $m=\text{mass of the electron}$

$=9.1\times {10}^{-31}kg$

Then we get,

$\Rightarrow v=\sqrt{\frac{2eV}{m}}$

Substituting the values we get

$v=\sqrt{\frac{2(1.6\times {10}^{-19})(2.0\times {10}^3)}{9.1\times {10}^{-31}}}$

$\Rightarrow v=2.652\times {10}^{7}m/s$

Step 2: Find the trajectory of the electron.

When the electron moves in the magnetic field with its velcity transverse to the magnetic field,magnetic force $F_B$ acts von the electron.This force causes the electron to move in a circular path providing the required centripetal force $F_C.$

Hence, the forces $F_B\text{and}{F}_C$ are equal.

$F_B=F_C$

$evB=\frac{m{v}^2}{r}$

$\Rightarrow r=\frac{mv}{eB}$

Substituting the values, we get,

$r=\frac{(9.1\times {10}^{-31})(2.652\times {10}^7)}{(1.6\times {10}^{-19})\left(0.15\right)}$

$r=100.5\times {10}^{-5}m$

$\to r=1mm$

Final Answer: Circular trajectory of radius $1.0mm\text{normal to}B.$

2) Step 1 : Determine the path followed by the electron.

When the electron moves in the magnetic field making an angle of ${30}^{\circ }$ with the magnetic field, the electron will have two components in horizontal and vertical direction to the magnetic field.

The horizontal component of velocity $v\cos {30}^{\circ }$ parallel to the magnetic field causes the electron to move in a straight-line path.

The vertical component of velocity $v\sin {30}^{\circ }$ perpendicular to the magnetic field causes the electron to move in a circular path.

Therefore, the path followed by the electron is helical.

Step 2: Find the velocity of the electron.

Given, Magnetic field strength,

$B=0.15T$

Potential difference,

$V=2.0\text{kV}=2.0\times {10}^{3}V$

$\theta ={30}^{\circ }$

Let the electrons move through the magnetic field velocity $v$.

Electron posses kinetic energy in electric field.

So, $eV=\frac{1}{2}m{v}^2$

Where, $m=$ mass of the electron

$=9.1\times {10}^{-31}\text{kg}$

ANSWER :

Then we get,

$\Rightarrow v=\sqrt{\frac{2eV}{m}}$

Substituting the values we get

$v=\sqrt{\frac{2(1.6\times {10}^{-19})(2.0\times {10}^3)}{9.1\times {10}^{-31}}}$

$\Rightarrow v=2.652\times {10}^{7}m/s$

The velocity along magnetic field $\left(B\right)$ is $v^{\prime }=v\cos {30}^{\circ }=2.652\times {10}^7\times \sqrt{3}/2$

$v^{\prime }=2.3\times {10}^{7}m/s$

Step 3: Calculate the radius of the helical path for the electron.

For the radius of the helical path of an electron.

$\Rightarrow r=\frac{mv\sin {30}^{\circ }}{eB}$

$r=\frac{(9.1\times {10}^{-31})(2.652\times {10}^7)\sin {30}^{\circ }}{(1.6\times {10}^{-19}C)\left(0.15T\right)}$

$r=50.2\times {10}^{-5}m$

$\Rightarrow r=0.5\text{mm}$

Final answer : $0.5\text{mm},$ Helical path, $v=2.3\times {10}^{7}m/s$ (along magnetic field)