2. An electron with a charge of -1.60times 10^wedge -19C and a mass of 9.11times 10^wedge -31kg is traveling at a speed of 3.0times 10^wedge 7m/s perpendicular to uniform magnetic field of strength 0.2 T. What is the magnitude of the magnetic ......................................................................electron?

To find the magnitude of the magnetic force acting on the electron, we can use the formula:<br /><br />\[ F = q \cdot v \cdot B \cdot \sin(\theta) \]<br /><br />where:<br />- \( F \) is the magnetic force,<br />- \( q \) is the charge of the electron,<br />- \( v \) is the velocity of the electron,<br />- \( B \) is the magnetic field strength,<br />- \( \theta \) is the angle between the velocity and the magnetic field.<br /><br />Since the electron is traveling perpendicular to the magnetic field, the angle \( \theta \) is 90 degrees, and \( \sin(90^\circ) = 1 \).<br /><br />Given:<br />- \( q = -1.60 \times 10^{-19} \, \text{C} \)<br />- \( v = 3.0 \times 10^7 \, \text{m/s} \)<br />- \( B = 0.2 \, \text{T} \)<br /><br />Now, plug these values into the formula:<br /><br />\[ F = (-1.60 \times 10^{-19} \, \text{C}) \cdot (3.0 \times 10^7 \, \text{m/s}) \cdot (0.2 \, \text{T}) \]<br /><br />\[ F = -9.6 \times 10^{-13} \, \text{N} \]<br /><br />The magnitude of the magnetic force is \( 9.6 \times 10^{-13} \, \text{N} \).