Question

A singly charged positive ion has a mass of 2.50*10^-26 kg. After being accelerated through a potential difference of 250V, the ion enters a magnetic field of 0.500 T, in a direction perpendicular to the field. Calculate the radius of the path of the ion in the field.

Answer 1

Concepts and reason

The concept used to solve this problem is acceleration in a magnetic field.

Initially, the velocity of the charged particle can be calculated by using the relation potential difference, change in energy and the charge of the particle. Later, expression for the radius of the path can be calculated by equating the magnetic force and the centripetal force. Finally, the value of the radius of the path can be calculated by using the relation between mass of the particle, velocity of the particle, charge of the particle and magnetic field.

Fundamentals

The expression for the velocity is,

$(30)=(10) $

Here, is the potential difference, is the change in kinetic energy and is the charge of the particle.

The expression for the change in energy is,

Here, is the mass of the particle and is the velocity of the particle.

The expression for the force at equilibrium is,

Here, is the magnetic force and is the centripetal force.

The expression for the magnetic force is,

Here, is the magnetic field and q is the charge of the electron.

The expression for the centripetal force is,

Here, is the radius of the path.

The expression for the potential difference is,

$(30)=(10) $

Replace for and rewrite the equation for v.

The expression for the velocity of the particle is,

Substitute for , for and for m.

The expression for the force is,

Substitute for and for .

Rewrite the above relation for r.

$r= $

The expression for the radius of the path is,

$r= $

Substitute for , for , for and for.

Ans:

The radius of the path is.