Question

A velocity selector can be used to measure the speed of a charged particle. A beam of particles is directed along the axis of the instrument. The plates are separated by 3 mm and the magnitude of the magnetic field is 0.3 T. What voltage between the plates will allow particles of speed 5.0 x 10^5 m/sa. 1200 V b. 3800 V c. 7500 V d. 190 V e. 380 V

Answer 1

Voltage, V = 450 volts

Step-by-step explanation:

It is given that,

Separation between plates, d = 3 mm = 0.003 m

magnitude of magnetic field, B = 0.3 T

Speed of the particle,
`v=5* 10^5\ m/s`

The relation between the magnetic field, electric field and the velocity of the particle is given by :

`v=(E)/(B)`

Also,
`E=(V)/(d)`

`v=(V)/(Bd)`

`V=vBd`

`V=5* 10^5* 0.3* 0.003`

V = 450 volts

So, the voltage between the plates will be 450 V. Hence, this is the required solution.