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

Respuesta :

Answer:

Voltage, V = 450 volts

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, [tex]v=5\times 10^5\ m/s[/tex]

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

[tex]v=\dfrac{E}{B}[/tex]

Also, [tex]E=\dfrac{V}{d}[/tex]

[tex]v=\dfrac{V}{Bd}[/tex]

[tex]V=vBd[/tex]

[tex]V=5\times 10^5\times 0.3\times 0.003[/tex]

V = 450 volts

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