Question

A velocity selector has magnetic field of magnitude 0.28 T perpendicular to the electric field of magnitude 0.46 M V /m.What is the speed of the particle for it to pass through undeflected?

Answer 1

The speed of the particle for it to pass through the velocity selector undeflected is 1.64 × 10^6 m/s.

Step-by-step explanation:

To pass through the velocity selector undeflected, the electric force acting on the particle must equal the magnetic force acting on it. Therefore, we can equate the electric force, qE, to the magnetic force, qvB, where q is the charge of the particle, E is the electric field magnitude, v is the velocity of the particle, and B is the magnetic field magnitude. Setting these two forces equal to each other and solving for the velocity, we have:

qE = qvB

Dividing both sides of the equation by q and rearranging, we get:
v = E/B

Substituting the given values, with E = 0.46 MV/m and B = 0.28 T, we have:
v = (0.46 × 106 V/m) / (0.28 T)

Simplifying, we find:
v = 1.64 × 106 m/s