Question

For your senior project, you would like to build a cyclotron that will accelerate protons to 10% of the speed of light. The largest vacuum chamber you can find is 50 cm in diameter.

What magnetic field strength will you need?

Answer 1

Concepts and reason

The concepts that are to be used to solve the given problem are the centripetal force in a circular motion, force on a charged particle in a magnetic field, and equilibrium of force.

Initially, derive the expression for strength of the magnetic field by using the centripetal force in a circular motion, force on a charged particle in a magnetic field. Finally, determine the strength of the magnetic field by using the relationship between the radius and magnetic field, mass, and velocity.

Fundamentals

The centripetal force in a circular motion,

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

Here, F is the force, m is the mass, r is the radius, and v is the velocity.

The expression for force on a charged particle in a magnetic field is,

$F = qvB\sin \theta$

Here, F is the force, q is the charge, and B is the magnetic field, and θ is the angle.

The force on the due to magnetic field is,

$F = qvB$

Here, $F$ is the force on object, $q$is the charge, $v$ is the speed, and $B$ is the magnetic field.

The force on a charged particle in a magnetic field,

$F = qvB\sin \theta$

The velocity of the charged particle is perpendicular to the magnetic field. Hence, the value of θ is 90⁰.

Substitute 90⁰ for θ in$F = qvB\sin \theta$.

$\begin{array}{c}\\F = qvB\sin \left( {{{90}^0}} \right)\\\\ = qvB\\\end{array}$

Use the equilibrium of force and equate the centripetal force with magnetic force and solve for the magnetic field strength B

Solve the equations $F = \frac{{m{v^2}}}{r}$ and $F = qvB$ as follows:

$\begin{array}{c}\\\frac{{m{v^2}}}{r} = qvB\\\\B = \frac{{mv}}{{qr}}\\\end{array}$

The strength of the magnetic field,

$B = \frac{{mv}}{{qr}}$

Here, B is the strength of the magnetic field, m is the mass of the proton, v is the velocity of the proton, q is the charge of the proton, r is the radius of the vacuum chamber.

The diameter of the vacuum chamber is half of the radius. Hence, the radius is,

$\begin{array}{c}\\r = \frac{{50{\rm{ cm}}}}{2}\\\\ = \left( {25{\rm{ cm}}} \right)\left( {\frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)\\\\ = 0.{\rm{25 m}}\\\end{array}$

The proton accelerates at 10% of the speed of light, hence, the speed of the proton is,

$\begin{array}{c}\\v = \frac{{3 \times {{10}^8}{\rm{ m/s}}}}{{10}}\\\\ = 3 \times {10^{7{\rm{ }}}}{\rm{m/s}}\\\end{array}$

The strength of the magnetic field is,

$B = \frac{{mv}}{{qr}}$

Substitute $1.67 \times {10^{ - 27}}{\rm{ kg}}$ for m, $3 \times {10^{7{\rm{ }}}}{\rm{m/s}}$ for v, $1.6 \times {10^{{\rm{ - 19 }}}}{\rm{C}}$ for q, and 0.25 m for r in $B = \frac{{mv}}{{qr}}$.

$\begin{array}{c}\\B = \frac{{\left( {1.67 \times {{10}^{ - 27}}{\rm{ kg}}} \right)\left( {3 \times {{10}^{7{\rm{ }}}}{\rm{m/s}}} \right)}}{{\left( {1.6 \times {{10}^{{\rm{ - 19 }}}}{\rm{C}}} \right)\left( {0.25{\rm{ m}}} \right)}}\\\\ = 1.2525{\rm{ T}}\\\end{array}$

Ans:

The strength of the magnetic field is$1.2525{\rm{ T}}$.