Question

A velocity selector has a magnetic field of magnitude 0.22 T perpendicular to an electric field of magnitude 0.44 MV/m.What must the speed of a particle be for it to pass through undeflected?

Answer 1

v = 2 x 10⁶ m/s

Step-by-step explanation:

Given,

Magnetic field, B = 0.22 T

E = 0.44 MV/m

Speed of the particle = ?

We know,

F = q v B

And also

Relation between Force and electric field

F = q E

q v B = q E

`v = (E)/(B)`

`v = (0.44* 10^6)/(0.22)`

v = 2 x 10⁶ m/s

The Speed of the particle is equal to v = 2 x 10⁶ m/s