A beam of electrons moves at right angles to a magnetic field of 4.5 × 10-2 tesla. If the electrons have a velocity of 6.5 × 106 meters/second, what is the force acting on the electrons? The value of q = -1.6 × 10-19 coulombs.
(A)-2.9 × 106 N
(B)-3.9 × 10-14 N
(C)-4.9 × 10-14 N
(D)-6.5 × 10-13 N

Respuesta :

Hello! 

F = Bqv sin theta F=Force B=magnetic flux density q=charge v=velocity theta=angle the moving electrons make with the magnetic field.

^^^You will find the force using that formula^^^

In Short, your Answer would MOST LIKELY have to be "B".

"-3.9 × 10-14 N"

I Hope my answer has come to your Help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead! :)

Explanation :

It is given that,

Magnetic field, [tex]B=4.5\times 10^{-2}\ T[/tex]

Velocity of electrons, [tex]v=6.5\times 10^6\ m/s[/tex]

Charge, [tex]q=-1.6\times 10^{-19}\ C[/tex]

Magnetic force [tex]F=q(v\times B)[/tex]

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

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

So, approximately answer can be option (b) " [tex]-3.9\times 10^{-14}\ N[/tex] "

Hence, this is the required solution.                      

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