Question

Answer the following questions about force on a moving charge. An electron \(\left(q=-1.6 \times 10^{-19} \mathrm{C}\right)\) enters a constant \(2-\mathrm{T}\) magnetic field at an angle of \(45^{\circ}\) to the field with a speed of \(2 \times 10^{5} \mathrm{m} / \mathrm{s} .\) Find the magnitude of the force on the electron.

Short answer

Question: Calculate the magnitude of the magnetic force acting on an electron moving with a velocity of \(2 \times 10^5\,\text{m/s}\) in a magnetic field of strength \(2\,\text{T}\) at an angle of \(45^{\circ}\) with the field. The charge of the electron is \(q = -1.6 \times 10^{-19}\,\text{C}\). Answer: The magnitude of the magnetic force acting on the electron is approximately \(9.067 \times 10^{-14}\,\text{N}\).

Step-by-step solution

Identifying the given parameters

The given parameters are as follows: Charge of the electron, \(q = -1.6 \times 10^{-19}\,\text{C}\) Magnetic field strength, \(B = 2\,\text{T}\) Velocity of the electron, \(v = 2 \times 10^5\,\text{m/s}\) Angle between the velocity and the magnetic field, \(\theta = 45^{\circ}\)

Convert the angle from degrees to radians

To plug the angle into the formula correctly, we need to convert it to radians. We can do this using the formula: \(\text{Radians} = \frac{\text{Degrees} \times \pi}{180}\) So, \(\theta = \frac{45 \times \pi}{180} = \frac{\pi}{4}\,\text{radians}\).

Calculate the force on the electron

Now, we will use the formula for the magnetic force: \(F = qvB\sin{\theta}\) Substituting the given values: \(F = (-1.6 \times 10^{-19}\,\text{C}) \times (2 \times 10^5\,\text{m/s}) \times (2\,\text{T}) \times \sin{\frac{\pi}{4}}\) Simplify the equation: \(F = (-1.6 \times 10^{-19}) \times (2 \times 10^5) \times (2) \times \frac{\sqrt{2}}{2}\) Calculate the force: \(F = (-1.6 \times 10^{-19}) \times 4 \times 10^5 \times \frac{\sqrt{2}}{2}\) \(F = 6.4 \times 10^{-14} \times \sqrt{2}\) \(F \approx 9.067 \times 10^{-14}\,\text{N}\) Since we are asked to find the magnitude of the force, we can ignore the negative sign. So, the magnitude of the force on the electron is approximately: \(F \approx 9.067 \times 10^{-14}\,\text{N}\)