Question

A uniform electric field of magnitude 106 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.48 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.) m/s

Answer 1

`220833.33\ \text{m/s}`

Step-by-step explanation:

`E` = Electric field =
`106\ \text{kV/m}`

`B` = Magnetic field =
`0.48\ \text{T}`

Velocity is given by

`v=(E)/(B)`

`\Rightarrow v=(106* 10^3)/(0.48)`

`\Rightarrow v=220833.33\ \text{m/s}`

The velocity of the particle is
`220833.33\ \text{m/s}`.