Answer :
Answer:
Approximately [tex]0.0075\; {\rm m}[/tex], assuming that the electron was initially at rest, and that the relativistic effects are negligible.
Explanation:
Approach this question in the following steps:
- Find the velocity of the electron right before entering the magnetic field.
- Given the velocity of the electron and the magnetic field strength, find the resultant force on the electron.
- Find the radius of the path of the electron using the formula for centripetal motions.
Let [tex]q[/tex] denote the magnitude of the electrostatic charge on this electron, and let [tex]V[/tex] denote the potential difference. In this question:
- [tex]q = 1.602 \times 10^{-19}\; {\rm C}[/tex] (elementary charge).
- [tex]V = 2\; {\rm kV} = 2\times 10^{3}\; {\rm V}[/tex].
The energy gained from the potential difference would be [tex]E = q\, V[/tex].
Let [tex]m[/tex] denote the mass of the electron, and let [tex]v[/tex] denote the velocity of the electron after the acceleration:
- [tex]m = 9.109 \times 10^{-31}\; {\rm kg}[/tex] is the electron rest mass.
- [tex]v[/tex] needs to be found.
Assuming that the relativistic effects of motion are negligible, the kinetic energy of the electron would be [tex](1/2)\, m\, v^{2}[/tex]. If the energy gained from the [tex]2\; {\rm kV}[/tex] potential difference is entirely converted into the kinetic energy of the electron:
[tex]\displaystyle \frac{1}{2}\, m\, v^{2} = q\, V[/tex].
[tex]\displaystyle v = \sqrt{\frac{2\,q\, V}{m}}[/tex].
When a point charge [tex]q[/tex] moves through a magnetic field of strength [tex]B[/tex] at a velocity [tex]v[/tex] perpendicular to the field, magnitude of the magnetic force on the point charge would be:
[tex]F = q\, v\, B[/tex].
This force would be perpendicular to both the magnetic field [tex]B[/tex] and (most importantly) to the direction of motion of the moving charge.
Note that in this question:
- [tex]q = 1.602 \times 10^{-19}\; {\rm C}[/tex]
- [tex]\displaystyle v = \sqrt{\frac{2\,q\, V}{m}}[/tex].
- [tex]B = 0.02\; {\rm T}[/tex].
Assuming that all other forces on the electron are negligible while in this magnetic field, the resultant force (net force) on the electron would also be [tex]q\, v\, B[/tex], perpendicular to the direction of motion of the electron:
[tex]\displaystyle (\text{net force}) = q\, v\, B[/tex].
Because the net force on the electron is perpendicular to the direction of motion, this electron would be in a centripetal motion. The mass [tex]m[/tex] of the electron, tangential velocity [tex]v[/tex] of the motion, and radius [tex]r[/tex] of the circular path should satisfy:
[tex]\displaystyle (\text{net force}) = \frac{m\, v^{2}}{r}[/tex].
(Assuming that relativistic effects are negligible, the mass of the electron would be the same as that of a stationary electron.)
Note that the "tangential velocity" [tex]v[/tex] in this expression is the same as the initial velocity of the electron: [tex]v = \sqrt{2\, q\, V / m}[/tex].
Rearrange this expression and solve for the radius [tex]r[/tex] of the circular path:
[tex]\begin{aligned} r &= \frac{m\, v^{2}}{(\text{net force})} = \frac{m\, v^{2}}{q\, v\, B} = \frac{m\, v}{q\, B} \\ &= \frac{\displaystyle m\, \sqrt{\frac{2\, q\, V}{m}}}{q\, B} = \sqrt{\frac{2\, q\, V\, m^{2}}{m\, q^{2}\, B^{2}}} = \sqrt{\frac{2\, V\, m}{q\, B^{2}}} \\ &\approx \sqrt{\frac{2\, (2 \times 10^{3}\; {\rm V})\, (9.109 \times 10^{-31}\; {\rm kg})}{(1.602 \times 10^{-19}\; {\rm C})\, (0.02\; {\rm T})^{2}}} \\ &\approx 0.0075\; {\rm m}\end{aligned}[/tex].
In other words, the radius of the circular path would be approximately [tex]0.0075\; {\rm m}[/tex] under the assumptions.
