Find the magnitude of the magnetic force (FB) of an electron moves through space with a speed of v= 8.0x106 m/s along the x axis. At its location, the magnetic field of the Earth has a magnitude of B= 5 X10-5 T, and is directed at an angle of 90o to the x axis. (Hint: q= -1.6X10-6)

An electron that is moving through a uniform magnetic field has a velocity v = (44 km/s)i + 31 km/s)j when it experiences a force F = (-3.000×10-15 N)i + (4.258×10-15 N)j due to the magnetic field. If Bx = 0, calculate the magnitude of the magnetic field B. (Tesla = T)   1) what is the magnitude of the magnetic field B?   2) what is By?   3) What is Bz?

The earth's magnetic field is about 45 micro Tesla (μT) and points approximately due north. What is the force exerted by the field on a particle with a charge of 3.0 μC that is moving due west at a speed of 250 m/s? What is the direction of the force on the particle (i.e. north, south, east, west, up or down)?

One component of a magnetic field has a magnitude of 0.0383 T and points along the +x axis, while the other component has a magnitude of 0.0735 T and points along the -y axis. A particle carrying a charge of +2.22 × 10-5 C is moving along the +z axis at a speed of 4.97 x 10³ m/s. (a) Find the magnitude of the net magnetic force that acts on the particle. (b) Determine the angle that the net force makes with respect to the +x axis. (a) Number (b) Number i IN Units Units By +q