Question

The magnetic field in a cyclotron is \(0.50 \mathrm{T}\). What must be the minimum radius of the dees if the maximum proton speed desired is $1.0 \times 10^{7} \mathrm{m} / \mathrm{s} ?$

Short answer

Answer: The minimum radius of the dees in the cyclotron must be approximately 0.104 m.

Step-by-step solution

Identify the relevant formula

The Lorentz force (F) acting on a charged particle of charge q moving with a velocity v in a magnetic field B is given by: F = q * (v x B) Where "x" denotes the cross product between velocity and magnetic field vectors. For a cyclotron, the charged particle moves in a circular path due to the magnetic field, so F also represents the centripetal force. Centripetal force is given by: F = (m * v²) / R Where m is the mass of the particle, and R is the radius of the circular path.

Equate the two expressions for force

As the Lorentz force and the centripetal force both represent the force acting on the proton in a cyclotron, we can equate the two expressions: q * (v x B) = (m * v²) / R

Simplify the equation

In a cyclotron, the magnetic field B is perpendicular to the velocity v. Therefore, the cross product simplifies to: v x B = v * B Now, we can substitute this back into the equation: q * (v * B) = (m * v²) / R

Solve for the radius R

We can rearrange the equation to find the minimum radius of the dees (R): R = (m * v) / (q * B) For a proton, the mass (m) is 1.67 × 10^(-27) kg, the charge (q) is 1.6 × 10^(-19) C, the maximum proton speed (v) is 1.0 × 10^(7) m/s, and the magnetic field (B) is 0.50 T. Using these values, we can now calculate R.

Calculate the minimum radius

Substitute the given values into the equation for R: R = (1.67 × 10^(-27) kg * 1.0 × 10^(7) m/s) / (1.6 × 10^(-19) C * 0.50 T) R ≈ 0.104 m The minimum radius of the dees in the cyclotron must be approximately 0.104 m.