A 150 V battery is connected across two parallel metal plates of area 28.5 cm2 and separation 0.00820 m. A beam of alpha particles (charge +2e, mass 6.64Ã10â27 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?

B = 4.45mT in the +^k direction

Explanation:

In order to calculate the required magnitude of the magnetic force, to achieve that the beam of particles emerge undeflected of the parallel plates, the electric force between the plates and the magnetic field in that region must be equal.

           (1)

q: charge of the particles beam = +2e = 2*1.6*10^-19C

v: speed of the particles = ?

B: magnitude of the magnetic field = ?

E: electric field between the plates = V/d

V: potential difference between the parallel plates = 150V

d: distance of separation of the plates = 0.00820m

If you assume that the below plate is negative, the electric force on the particles has a direction upward (+^j). Then, the direction of the magnetic force must be downwards (-^j).  

To obtain a downward magnetic force, it is necessary that the magnetic field point out of the page. In fact, if the direction of motion of the particles is toward east (+^i) and the magnetic field points out of the page (+^k), you have:

^i X ^k = -^j

Furthermore, it is necessary to calculate the sped of the particles. It is made by using the information about the charge, the potential difference that accelerates the particles and the kinetic energy.

(2)

You replace the expression (2) into the equation (1) and solve for B:

The required magnitude of the magnetic field is 4.45mT and has a direction out of the page +^k

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg