An unknown particle moves in a straight line through crossed electric and magnetic fields with e = 1.5 kv/m and b = 0.034 t. if the electric field is turned off, the particle moves in a circular path of radius r = 2.7 cm. what might the particle be

The mass of the particle depends on the type of particle, and the radius of

the path of the particle depend on its mass.

The particle might be an alpha particle.

Reasons:

The given parameters are;

Electric potential of the field, E = 1.5 kV/m

Magnetic field strength, B = 0.034 T

Radius of the circular path, r = 2.7 cm

Required:

Finding the type of particle in the question

Solution:

When moving in a straight line, we have;

B·q·v = F = q·E

Therefore;

E = B·v

When the particle moves in a circular path, we have;

Therefore;

Which gives;

q = n × e

Therefore;

Where;

n = The number of electron charge

Which gives the mass in kilograms as follows;

The particle is larger than a subatomic particle

Therefore;

For an alpha particle, we have;

m ≈ 6.645 × 10⁻²⁷ kg

n = 2

n = 2

Therefore;

Therefore, the particle might be an alpha particle.

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Alpha particle

Explanation:

Initially, the particle moves in a straight line. This means that the electric force and the magnetic force are equal:

where q is the charge, E is the electric field, v is the speed and B is the magnetic field.

In this problem,

E = 1.5 kV/m = 1500 V/m

B = 0.034 T

Solving the equation for v we find the speed of the particle

Now the electric field is turned off, so the particle starts moving in a circular motion, where the magnetic force acts as centripetal force:

where

r = 2.7 cm = 0.027 m is the radius of the circular path

Solving the problem for q/m, we find charge-to-mass ratio of the particle

And this corresponds to the q/m ratio of an alpha particle, which has:

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