Respuesta :
Answer:
Approximately [tex]4.37 \times 10^{-12}\; {\rm N}[/tex].
Explanation:
When a charged particle moves through a magnetic field, the magnetic force on the particle would be:
[tex]F = q\, v\, B\, \sin(\theta)[/tex],
Where:
- [tex]q[/tex] is the magnitude of the electrostatic charge on the particle,
- [tex]v[/tex] is the speed of the particle,
- [tex]B[/tex] is the magnitude of the magnetic fi7eld, and
- [tex]\theta[/tex] is the angle between the velocity of the particle and the magnetic field.
In this question, it is given that the velocity of the electron is perpendicular to the magnetic field. Hence, [tex]\theta = 90^{\circ}[/tex]. Both [tex]v[/tex] and [tex]B[/tex] are given.
The magnitude of the electrostatic charge on an electron is known as the elementary charge, [tex]e[/tex]. Look up the value of this physical constant:
[tex]q = e \approx 1.602 \times 10^{-19}\; {\rm C}[/tex].
The force on the electron in this question would be:
[tex]\begin{aligned}F &= q\, v\, B\, \sin(\theta) \\ &\approx (1.602 \times 10^{-19}) \, (7.24\times 10^{6})\, (3.77)\, \sin(90^{\circ})\; {\rm N} \\ &\approx 4.37 \times 10^{-12}\; {\rm N}\end{aligned}[/tex].
Final answer:
An electron moving perpendicular to a magnetic field of 3.77 T at a velocity of 7.24 x 10^6 m/s will experience a force of 4.370 x 10^-12 N.
Explanation:
The force experienced by the electron moving perpendicular to a magnetic field can be determined using the equation F = qvB, where F is the force on the electron, q is the charge of the electron (-1.6 x 10-19 C), v is the velocity of the electron, and B is the magnetic field strength. In this case:
F = (1.6 x 10-19 C) * (7.24 x 106 m/s) * (3.77 T)
F = 4.370 x 10-12 N
The electron will experience a force of 4.370 x 10-12 newtons.
