Question

An electron emitted by a heated cathode and accelerated through a potential difference of $2.0\,{\text{kV}}$, enters a region with uniform magnetic field of $0.15\,{\text{T}}$. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of $30^\circ$ with the velocity.

Answer 1

Use the equations for the kinetic energies of the electron. Also use the equations for the magnetic force on the electron and centripetal force on the electron. For the velocity of an electron perpendicular to the magnetic field, the electron follows a circular path.

Formula used:
The kinetic energy $K$ of an electron is
$K = eV$ …… (1)
Here, $e$ is the charge on the electron and $V$ is the potential difference.
The kinetic energy $K$ of an electron is
$K = \dfrac{1}{2}m{v^2}$ …… (2)
Here, $m$ is the mass of the electron and $v$ is the velocity of the electron.
The magnetic force ${F_B}$ on an electron is
${F_B} = evB$ …… (3)
Here, $e$ is the charge on the electron, $v$ is the velocity of the electron and $B$ is the magnetic field.
The centripetal force ${F_C}$ on an electron in the circular motion is
${F_C} = \dfrac{{m{v^2}}}{r}$ …… (4)
Here, $m$ is the mass of the electron, $v$ is the velocity of the electron and $r$ is the radius of the circular path.

Complete step by step answer:
An electron emitted by a heated cathode and accelerated through a potential difference $V$ of $2.0\,{\text{kV}}$, enters a region with uniform magnetic field $B$ of $0.15\,{\text{T}}$.
$B = 0.15\,{\text{T}}$
And
$V = 2.0\,{\text{kV}}$
$\Rightarrow V = 2.0 \times {10^3}\,{\text{V}}$

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

The kinetic energy is gained by the electron when it moves in the magnetic field of potential difference $V$.

Equate equation (1) and (2) for the kinetic energies of the electron.
$eV = \dfrac{1}{2}m{v^2}$
$\Rightarrow v = \sqrt {\dfrac{{2eV}}{m}}$

Substitute $1.6 \times {10^{ - 19}}\,{\text{C}}$ for $e$, $2.0 \times {10^3}\,{\text{V}}$ for $V$ and $9.1 \times {10^{ - 31}}\,{\text{kg}}$ for $m$ in the above equation.
$v = \sqrt {\dfrac{{2\left( {1.6 \times {{10}^{ - 19}}\,{\text{C}}} \right)\left( {2.0 \times {{10}^3}\,{\text{V}}} \right)}}{{9.1 \times {{10}^{ - 31}}\,{\text{kg}}}}}$
$\Rightarrow v = 2.652 \times {10^7}\,{\text{m/s}}$

Hence, the velocity of the electron is $2.652 \times {10^7}\,{\text{m/s}}$.

(a)When the electron moves in the magnetic field with its velocity transverse to the magnetic field, the force ${F_B}$ acting on the electron is perpendicular to the direction of the magnetic field and the direction of motion of the electron.

This perpendicular force causes the electron to move in a circular path providing the required centripetal force ${F_C}$.

Hence, the forces ${F_B}$ and ${F_C}$ are equal.
${F_B} = {F_C}$

Substitute $evB$ for ${F_B}$ and $\dfrac{{m{v^2}}}{r}$ for ${F_C}$ in the above equation.
$evB = \dfrac{{m{v^2}}}{r}$
$\Rightarrow r = \dfrac{{mv}}{{eB}}$ …… (5)

Substitute $9.1 \times {10^{ - 31}}\,{\text{kg}}$ for $m$, $2.652 \times {10^7}\,{\text{m/s}}$ for $v$, $1.6 \times {10^{ - 19}}\,{\text{C}}$ for $e$ and $0.15\,{\text{T}}$ for $B$ in the above equation.
$r = \dfrac{{\left( {9.1 \times {{10}^{ - 31}}\,{\text{kg}}} \right)\left( {2.652 \times {{10}^7}\,{\text{m/s}}} \right)}}{{\left( {1.6 \times {{10}^{ - 19}}\,{\text{C}}} \right)\left( {0.15\,{\text{T}}} \right)}}$
$r = 100.5 \times {10^{ - 5}}\,{\text{m}}$
$\Rightarrow r = 1\,{\text{mm}}$

Hence, the electron moves in a circular path of radius $1\,{\text{mm}}$ in a direction perpendicular to the magnetic field.

(b)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.

Determine the radius of the helical path of an electron.

Rewrite equation (5) for the radius of the helical path of an electron.
$\Rightarrow r = \dfrac{{mv\sin 30^\circ }}{{eB}}$

Substitute $9.1 \times {10^{ - 31}}\,{\text{kg}}$ for $m$, $2.652 \times {10^7}\,{\text{m/s}}$ for $v$, $1.6 \times {10^{ - 19}}\,{\text{C}}$ for $e$ and $0.15\,{\text{T}}$ for $B$ in the above equation.
$r = \dfrac{{\left( {9.1 \times {{10}^{ - 31}}\,{\text{kg}}} \right)\left( {2.652 \times {{10}^7}\,{\text{m/s}}} \right)\sin 30^\circ }}{{\left( {1.6 \times {{10}^{ - 19}}\,{\text{C}}} \right)\left( {0.15\,{\text{T}}} \right)}}$
$r = 50.2 \times {10^{ - 5}}\,{\text{m}}$
$\Rightarrow r = 0.5\,{\text{mm}}$

Hence, the electron moves in a helical path of radius $0.5\,{\text{mm}}$.

Note:
Since the horizontal and vertical components of velocity follow a straight line and circular path, the resulting path of the electron is helical or spiral which is the combination of the straight line path and circular path.