A V = 150-V battery is connected across two parallel metal plates of area 28.5 c

Question

A V = 150-V battery is connected across two parallel metal plates of area 28.5 cm2 and separation 8.50 mm . A beam of alpha particles (charge +2e, mass 6.64×10?27 kg) is accelerated from rest through a potential difference of 1.80 kV and enters the region between the plates perpendicular to the electric field, as shown in the figure.

1) What magnitude of magnetic field is needed so that the alpha particles emerge undeflected from between the plates?

P.S ( I have answered .045T, .047T , .048T and they are all incorrect(3 attempts remaining) please help!)

Explanation / Answer

here,

V = 150 V

area , a = 28.5 cm^2

sepration , d = 0.0085 m

electric feild , E = V/d

E = 150 /( 0.0085)

E = 17647.06 N/C

charge , q = 2 e

mass , m = 6.64 * 10^-27 kg

potential difference of accelration ,V' = 1800 V

let the speed be v

using work energy theorm

work done by potential = change in kinetic energy

q * V = 0.5 * m * v^2

2 * 1.6 * 10^-19 * 1800 = 0.5 * 6.64 * 10^-27 * v^2

v = 4.17 * 10^5 m/s

let the magnetic feild be B

for the particle to not deflect

electric force = magnetic force

q*E = q*v*B

17647.06 = 4.17 * 10^5 * B

B = 0.042 T

the magnitude of the magnetic feild is 0.042 T

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