A 150-V battery is connected across two parallel metal plates of area 28.5 cm2 a

Question

A 150-V battery is connected across two parallel metal plates of area 28.5 cm2 and a separation of 8.20 mm. A beam of alpha particles (charge +2e, mass 6.64E-27 kg) is accelerated from rest through a potential difference of 1.75 kV and enters the region between the plates perpendicular to the electric field. What magnitude and direction of magnetic field are needed so that the alpha particles emerge undeflected from between the plates?

3 Find the magnitude and direction of the magnetic field needed

Explanation / Answer

# Accelerating Voltage = V1=1.75 kV = 1750 V

Charge on alpha= q=3.2*10^-19 C

Mass = m =6.64*10^-27 kg

Kinetic energy = electric potential energy

(1/2)mv^2 = qV

v = sqrt [2qV/m]

The alpha particles will emerge undeflected from between the plates if magnetic force=elctric force

qvB =qE

E = Vo/d

B = E /v

B =(Vo/d) sqrt [m/2qV]

B =0.04968 T

B =4.968*10^-2 T

# As electric field points upward , electric force is upwards. For alpha to remain undeflected,direction of magnetic field should be out of the page.

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