Question

2) The force that a magnetic field exerts on a charged particle is given by F = qö x B. A particle with mass m = 2.0x10 kg and charge q - +2.5x10°C has an initial speed of v = 4/2 x 103 m/s (in the x- y plane). The magnetic field vector and velocity vector are 5 and 0, respectively are displayed on the coordinate axis below. The angle between the vectors is 135 degrees. Use unit vector notation when describing the vectors. Z (0.0, 10). (0-100) (0, 10, 0) (1000) X a) Write the Cartesian expression for the magnetic field vector: B = b) Write the Cartesian expression for the velocity vector: 3 = c) Write the Cartesian expression for the force on the charge: F= d) Find the magnitude of the force acting on the charge using the Cartesian vector approach: c) Calculate the magnitude of the force using: F = quB sin 2C,B)

Answer 1

Dear Student,

Details step by step solutions are as follows...

171 = 0.06 N. N sin135° ey F =q VB sin Lü,) = 2.5x108 x 402 x 103 x 0.64103 2.5 x 4 /2x0.6 x 1024 2.500. 6x4 410-2 - 2.5 x 4 ish N 0.06 N F = 0.06 N
Happy to help you..