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A uniform electric field of magnitude 108 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.80 T perpendicular to the electric field also exists in this region. A beam of positively charged particles travels into the region. Determine the speed of the particles at which they will not be deflected by the crossed electric and magnetic fields. (Assume the beam of particles travels perpendicularly to both fields.)

1 Answer

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Answer:

speed of particle are 135000 m/s

Step-by-step explanation:

given data

electric field = 108 kv/m

magnetic field = 0.80 T

to find out

speed of the particles

solution

we know that when there is no deflection

so that force by electric filed and force by magnetic field is equal ans opposite so that we say

F(E) = F(B)

so

qE = qvB

here

v = E/B

V = 108000 / 0.80

speed of particle are 135000 m/s

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