Select the correct answer.

A beam of electrons moves at right angles to a magnetic field of [tex]4.5 \times 10^{-2}[/tex] tesla. If the electrons have a velocity of [tex]6.5 \times 10^6[/tex] meters/second, what is the force acting on the electrons? The value of [tex]q = -1.6 \times 10^{-19}[/tex] coulombs.

A. [tex]-2.9 \times 10^6 \, \text{N}[/tex]
B. [tex]-3.9 \times 10^{-14} \, \text{N}[/tex]
C. [tex]-4.9 \times 10^{-14} \, \text{N}[/tex]
D. [tex]-6.5 \times 10^{-13} \, \text{N}[/tex]



Answer :

To find the force acting on the electrons moving in a magnetic field, we can use the formula for the magnetic force on a charged particle moving in a magnetic field:

[tex]\[ F = q \cdot v \cdot B \][/tex]

where:
- [tex]\( F \)[/tex] is the force acting on the electron,
- [tex]\( q \)[/tex] is the charge of the electron,
- [tex]\( v \)[/tex] is the velocity of the electron,
- [tex]\( B \)[/tex] is the magnetic field strength.

Given:
- The magnetic field strength [tex]\( B = 4.5 \times 10^{-2} \text{ T} \)[/tex],
- The velocity of the electrons [tex]\( v = 6.5 \times 10^6 \text{ m/s} \)[/tex],
- The charge of the electron [tex]\( q = -1.6 \times 10^{-19} \text{ C} \)[/tex].

Using the formula:

[tex]\[ F = q \cdot v \cdot B \][/tex]

Substitute the given values into the equation:

[tex]\[ F = (-1.6 \times 10^{-19} \text{ C}) \cdot (6.5 \times 10^6 \text{ m/s}) \cdot (4.5 \times 10^{-2} \text{ T}) \][/tex]

Calculate the product of the numbers (keeping track of the exponents):

[tex]\[ F = (-1.6) \cdot (6.5) \cdot (4.5) \times 10^{-19 + 6 - 2} \][/tex]

[tex]\[ F = -46.8 \times 10^{-15} \][/tex]

Simplify the exponent:

[tex]\[ F = -4.68 \times 10^{-14} \text{ N} \][/tex]

Therefore, the force acting on the electrons is:

[tex]\[ F = -4.68 \times 10^{-14} \text{ N} \][/tex]

The correct answer among the given choices is:

C. [tex]\(-4.9 \times 10^{-14} \text{ N}\)[/tex]