Question

Consider the mass spectrometer shown schematically in the figure below. The magnitude of the electric field between the plates of the velocity selector is 2.40 103 V/m, and the magnetic field in both the velocity selector and the deflection chamber has a magnitude of 0.040 0 T. Calculate the radius of the path for a singly charged ion having a mass m = 2.04 10-26 kg.

Answer 1

Concepts and reason

The concepts used to solve this problem is the motion of a charged particle in a uniform magnetic field.

First, use the relationship between magnetic field and electric field to calculate the speed of the moving charge.

Then, use the relationship between magnetic force and centripetal to calculate the radius of the path for a single charged ion.

Fundamentals

A moving charge travelling at a speed in a uniform magnetic field experiences a force,

$F = qvB$

Here, F is the magnetic force, q is the charge, and v is the speed of the moving charge.

The force is always perpendicular to the direction of the charge.

This principle is experienced in equipment called as velocity selector.

When a charged particle enters the velocity selector, it experiences a force due to magnetic field and electric field.

The particle will experience a force in the upward direction due to electric field which is expressed as:

$F = qE$

Here, F is the electric force due to electric field and E is the electric field.

The particle will also experience a force in the downward direction due to its motion through magnetic field which is expressed as:

$F = qvB$

In order to pass the charged particles through the space without getting deflected either upwards or downwards, the force in the upward direction must be equal to force in the downward direction.

$\begin{array}{c}\\qvB = qE\\\\v = \frac{E}{B}\\\end{array}$

The force is always perpendicular to the direction of the charge which causes the charged particle to change the direction, but does not affect the speed.

The charged particle will follow a circular path. The force directed towards the center of the circle equals the centripetal force to keep the charge, moving on a circle.

$qvB = \frac{{m{v^2}}}{r}$

Here, m is the mass of the charge ion and r is the radius of the circular path.

Rearrange the expression for the radius in the circular path,

$r = \frac{{mv}}{{qB}}$

The expression for the speed of the charged ion is,

$v = \frac{E}{B}$

Substitute $2.40 \times {10^3}\,{\rm{V/m}}$ for E and $0.0400\,{\rm{T}}$ for B.

$\begin{array}{c}\\v = \frac{{\left( {2.40 \times {{10}^3}\,{\rm{V/m}}} \right)}}{{\left( {0.0400\,{\rm{T}}} \right)}}\\\\ = 6.00 \times {10^4}\,{\rm{m/s}}\\\end{array}$

Expression for radius of the circular path is,

$r = \frac{{mv}}{{qB}}$

Substitute $2.04 \times {10^{ - 26}}\,{\rm{kg}}$ for m, $6.00 \times {10^4}\,{\rm{m/s}}$ for v, $1.6 \times {10^{ - 19}}\,{\rm{C}}$ for charge of the ion, and $0.0400\,{\rm{T}}$ for B.

$\begin{array}{c}\\r = \frac{{\left( {2.04 \times {{10}^{ - 26}}\,{\rm{kg}}} \right)\left( {6.00 \times {{10}^4}\,{\rm{m/s}}} \right)}}{{\left( {1.60 \times {{10}^{ - 19}}\,{\rm{C}}} \right)\left( {0.0400\,{\rm{T}}} \right)}}\\\\ = 0.19125\,{\rm{m}}\\\\ \approx {\rm{0}}{\rm{.191}}\,{\rm{m}}\\\end{array}$

Ans:

The radius of the circular path is $0.191\,{\rm{m}}$ .