Question

The magnetic field in a cyclotron is \(0.50 \mathrm{T}\). Find the magnitude of the magnetic force on a proton with speed $1.0 \times 10^{7} \mathrm{m} / \mathrm{s}$ moving in a plane perpendicular to the field.

Short answer

Answer: The magnitude of the magnetic force on the proton is \(8.0 \times 10^{-19} \mathrm{N}\).

Step-by-step solution

Identify the given quantities and the required formula

We are given that: Magnetic field, B = 0.50 T Speed of the proton, v = \(1.0 \times 10^7 \mathrm{m/s}\) Charge of a proton, q = \(1.6 \times 10^{-19} \mathrm{C}\) The angle between the magnetic field and the velocity, θ = 90° (perpendicular) We need to find the magnetic force, F, which can be calculated using the formula: \(F = qvB\sin\theta\)

Calculate the sine of the angle

Since the angle between the magnetic field and the velocity is 90° (perpendicular), the sine function is as follows: \(\sin\theta = \sin 90° = 1\)

Calculate the magnetic force

Now we can plug in the given values and the sine function into the formula for F: \(F = qvB\sin\theta = (1.6 \times 10^{-19}\mathrm{C})(1.0 \times 10^7 \mathrm{m/s})(0.50 \mathrm{T})(1)\) Multiplying the given values, we get: \(F = 8.0 \times 10^{-19} \mathrm{N}\) Therefore, the magnitude of the magnetic force on the proton is \(8.0 \times 10^{-19} \mathrm{N}\).