An electron \( \left(q=-1.6 \times 10^{-19} \mathrm{C}\right) \) enters a constant 2-T magnetic field at an angle of \( 45^{\circ} \) to the field with a speed of \( 4 \times 10^{5} \mathrm{~m} / \mathrm{s} \). Find the magnitude of the force on the electron. The magnitude of the force is \( \square \times 10 \square \mathrm{~N} \). (Use scientific notation. Round to two decimal places as needed.)

To find the magnitude of the force on the electron, we can use the formula for magnetic force: \[ F = |q| v B \sin(\theta) \] where: - \( F \) is the magnetic force, - \( |q| \) is the magnitude of the charge of the electron (\( 1.6 \times 10^{-19} \mathrm{C} \)), - \( v \) is the speed of the electron (\( 4 \times 10^{5} \mathrm{m/s} \)), - \( B \) is the magnetic field strength (\( 2 \mathrm{T} \)), - \( \theta \) is the angle (in this case, \( 45^{\circ} \)). First, we need to find \( \sin(45^{\circ}) \): \[ \sin(45^{\circ}) = \frac{\sqrt{2}}{2} \approx 0.7071 \] Now, substituting the values into the formula: \[ F = (1.6 \times 10^{-19} \mathrm{C}) (4 \times 10^{5} \mathrm{m/s}) (2 \mathrm{T}) \left(0.7071\right) \] \[ F = (1.6 \times 4 \times 2 \times 0.7071) \times 10^{-19} \times 10^{5} \] Calculating inside the parentheses: \[ F \approx (12.5664) \times 10^{-14} \mathrm{N} \] Now, converting this to a scientific notation to match the required format: \[ F \approx 1.26 \times 10^{-13} \mathrm{N} \] Thus, rounding to two decimal places, the magnitude of the force on the electron is \[ 1.26 \times 10^{-13} \mathrm{N} \] The magnitude of the force is \( 1.26 \times 10^{-13} \mathrm{~N} \).