A uniform electric field of magnitude 110 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.64 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.)

1.7*10^5 ms-1

Explanation:

From

qE = qvB

q = charge on the electron

E = electric field

v = velocity

B = magnetic field

E = vB

v = E/B = 110*10^3/0.6

v = 1.7*10^5 ms-1

Answer: 1.71*10^5 m/s

Explanation:

Given

Magnitude of electric field, E = 110 kV/m

Magnetic field, B = 0.64 T

The forces acting on the charged particle.

Due to electric field = qE, and it's an upward one

Due to magnetic field = qVB, this is a downwards one judging from right hand screw rule

Since the two forces are equal and opposite, then

qE = qVB

E = VB and then,

V = E / B

V = 110*10^3 / 0.64

V = 171.875*10^3 m/s

Therefore, we can say, the speed of the particle at which they will not be deflected is 1.72*10^5 m/s