Question

An electron moves at speed \(8.0 \times 10^{5} \mathrm{m} / \mathrm{s}\) in a plane perpendicular to a cyclotron's magnetic field. The magnitude of the magnetic force on the electron is \(1.0 \times 10^{-13} \mathrm{N}\) What is the magnitude of the magnetic field?

Short answer

Answer: The magnitude of the magnetic field is approximately \(7.81 \times 10^{-5}\,T\).

Step-by-step solution

Write down the given variables

We are given these variables: - Speed of the electron: \(v = 8.0 \times 10^{5}\,\mathrm{m/s}\) - Magnetic force on the electron: \(F = 1.0 \times 10^{-13}\,\mathrm{N}\) - Charge of the electron: \(q = -1.6 \times 10^{-19}\,\mathrm{C}\) - Angle between the velocity vector and the magnetic field vector is \(θ = 90^{\circ} ⟹ sin(θ) = 1\)

Solve for the magnetic field strength

Using the magnetic force formula \(F = qvBsinθ\), we can solve for the magnetic field strength \(B\). Since \(sin(θ) =1\), the equation simplifies to \(F=qvB\). Rearrange the formula to isolate the magnetic field strength: \(B= \cfrac{F}{q \times v}\) Now, plug in the given values: \(B = \cfrac{1.0 \times 10^{-13}\,\mathrm{N}}{(-1.6 \times 10^{-19}\,\mathrm{C) \times (8.0 \times 10^{5}\,\mathrm{m/s})}}\)

Calculate the magnetic field strength

Now, calculate the strength of the magnetic field: \(B= \cfrac{1.0\,\times\,10^{-13}\,\mathrm{N}}{((-1.6\,\times\, 10^{-19}\mathrm{C) (8.0\,\times\, 10^{5}\, \mathrm{m/s})}} \approx \boxed{7.81\,\times\,10^{-5}\,T}\) So, the magnitude of the magnetic field is approximately \(7.81 \times 10^{-5}\,T\).