Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

An electron is moving with a speed of [tex]3.5 \times 10^5 \, \text{m/s}[/tex] when it encounters a magnetic field of [tex]0.701 \, \text{T}[/tex]. The direction of the magnetic field makes an angle of [tex]60.0^\circ[/tex] with respect to the velocity of the electron. What is the magnitude of the magnetic force on the electron?

A. [tex]3.2 \times 10^{-13} \, \text{N}[/tex]
B. [tex]3.4 \times 10^{-14} \, \text{N}[/tex]
C. [tex]4.9 \times 10^{-13} \, \text{N}[/tex]
D. [tex]1.7 \times 10^{-13} \, \text{N}[/tex]
E. [tex]2.9 \times 10^{-14} \, \text{N}[/tex]


Sagot :

To solve the problem of finding the magnitude of the magnetic force on an electron moving through a magnetic field, we will use the formula for the magnetic force acting on a charged particle:

[tex]\[ F = q \cdot v \cdot B \cdot \sin(\theta) \][/tex]

where:
- [tex]\( F \)[/tex] is the magnetic force,
- [tex]\( q \)[/tex] is the charge of the electron,
- [tex]\( v \)[/tex] is the speed of the electron,
- [tex]\( B \)[/tex] is the magnetic field strength,
- [tex]\( \theta \)[/tex] is the angle between the velocity of the electron and the magnetic field.

Given values:
- [tex]\( v = 3.5 \times 10^5 \, \text{m/s} \)[/tex]
- [tex]\( B = 0.701 \, \text{T} \)[/tex]
- [tex]\( \theta = 60.0^\circ \)[/tex]
- [tex]\( q = 1.602 \times 10^{-19} \, \text{C} \)[/tex] (the elementary charge of an electron)

We need to follow these steps to find the magnitude of the magnetic force:

1. Convert the angle to radians:
[tex]\[ \theta_\text{rad} = \frac{60.0 \times \pi}{180} = \frac{\pi}{3} \][/tex]

2. Calculate the sine of the angle:
[tex]\[ \sin\left( \frac{\pi}{3} \right) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \][/tex]

3. Substitute the known values into the magnetic force formula:
[tex]\[ F = (1.602 \times 10^{-19} \, \text{C}) \cdot (3.5 \times 10^5 \, \text{m/s}) \cdot (0.701 \, \text{T}) \cdot \frac{\sqrt{3}}{2} \][/tex]

4. Perform the multiplication and simplification to find [tex]\( F \)[/tex]:
[tex]\[ F = 1.602 \times 10^{-19} \cdot 3.5 \times 10^5 \cdot 0.701 \cdot 0.866 \][/tex]

Putting all constants together:
[tex]\[ 1.602 \times 3.5 \times 0.701 \times 0.866 \approx 3.403918911752562 \times 10^{-14} \, \text{N} \][/tex]

Thus, the magnitude of the magnetic force on the electron is approximately
[tex]\[ 3.4 \times 10^{-14} \, \text{N} \][/tex]

Therefore, the correct answer is:
[tex]\[ 3.4 \times 10^{-14} \, \text{N} \][/tex]
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.