At time t=0, an electron with kinetic energy 12KeVmoves through x=0in the positive direction of an xaxis that is parallel to the horizontal component of Earth’s magnetic field $\stackrel{\rightharpoonup }{B}$. The field’s vertical component is downward and has magnitude $\mathbf{55}\mathbf{.}\mathbf{0}\mathit{\mu }\mathit{T}$. (a) What is the magnitude of the electron’s acceleration due to$\stackrel{\rightharpoonup }{B}$? (b) What is the electron’s distance from the xaxis when the electron reaches coordinate x=20cm?

1. Kinetic energy of the electron, K=12keV
2. The electron is in positive direction of x-axis parallel to the horizontal component of Earth’s magnetic field $\stackrel{\rightharpoonup }{B}$.
3. Vertical component of the magnetic field is downward.
4. Magnitude of the magnetic field, $|B⃗|=55\times {10}^{-6}T$
5. The given coordinate of the position of the electron, x=20cm.

The electron moving in the Earth’s magnetic field gets accelerated by the magnetic force acting on the electron. Since the electron is moving parallel to the horizontal component of the Earth’s magnetic field, the magnetic force on the electron is due to the vertical component of the magnetic field only.

Formulae:

The applied force on a body Newton’s second law, F=ma …(i)

The Lorentz force on a particle in motion, F=qvBsin$\theta$$\theta$ …(ii)

The kinetic energy of a body in motion, K=1/2mv² …(iii)

The centripetal acceleration of the body in circular motion, $a=\frac{v^2}{r}$ …(iv)

The sine angle of a triangle, $\sin \theta =\frac{Perpendicular}{Base}$ …(v)

Using the given data in equation (iii), the speed of the electron can be given as:

$$\begin{gathered} v=\sqrt{\frac{2K}{m_e}} \\ =\sqrt{\frac{2(12\times {10}^{3}eV)(1.6\times {10}^{-19}J/eV)}{9.11\times {10}^{-31}kg}} \\ =6.49\times {10}^{7}m/s \end{gathered}$$

Now, the acceleration of the electron can be calculated using equation (i) and (ii) with the given data as follows: (As the electron velocity contributes to the vertical component of the magnetic field only)

$$\begin{gathered} m_e{a}_e=qvB \\ a=\frac{qvB}{m} \\ =\frac{(1.6\times {10}^{19}C)(6.49\times {10}^{7}m/s)(55\times {10}^{-6}T)}{(9.11\times {10}^{-31}kg)} \\ =6.27\times {10}^{14}m/s^2 \\ =6.3\times {10}^{14}m/s^2 \end{gathered}$$

Hence, the value of electron’s acceleration is role="math" localid="1662717403899" $6.3\times {10}^{14}m/s^2$.

We ignore any vertical deflection of the beam that might arise due to the horizontal component of Earth’s field. Then, the path of the electron is a circular arc.

Thus, the radius of the arc can be given using equation (iv) can be given as follows:

$$\begin{gathered} R=\frac{v^2}{a} \\ =\frac{(6.49\times {10}^{7}m/s^2)}{6.3\times {10}^{14}m/s^2} \\ =6.68m \end{gathered}$$

Now, using equation (v), we get that

Thus, the height of the electron above ground (let be the x-axis) from the above figure can be given as: (for given x=d=0.20m)

$$\begin{gathered} h=R(1-\cos \theta ) \\ =R(1-\sqrt{1-{\sin }^2\theta }) \\ =R\left(1-\sqrt{1-{(d/R)}^2}\right) \\ =\left(6.68m\right)\left(1-\sqrt{1-{(0.2m/6.68m)}^2}\right) \\ =0.0030m \end{gathered}$$

Hence, the value of the electron distance is 0.0030m.