3. A charged particle moves through a velocity selector at a constant speed in a straight line. The electric field of the velocity selector is 3.20 × 10° N/C, while the magnetic field is 0.390 T. When the electric field is turned off, the charged particle travels on a circular path whose radius is 4.00 cm. Find the charge-to-mass ratio of the particle. C/kg

The aurora is caused when electrons and protons, moving in the earth’s magnetic field of  ≈5.0×10^−5 T, collide with molecules of the atmosphere and cause them to glow. a.  What is the frequency of the circular orbit for an electron with speed 1.0×10^6 m/s? Assume that the electron moves in a plane perpendicular to the magnetic field. b.  What is the frequency of the circular orbit for a proton with speed 5.0×10^4 m/s?

4. Please refer again to the diagram. This time we have a charge of 3.86 x 10^-6 C moving at velocity 7.00 x 10^5 m/s through a magnetic field of strength 1.28 T and an electric field of strength 8.92 x 10^5 N/C. What is the magnitude of the total force on this charge?     0.02 N     0.01 N     0.03 N     0.03 N

1. A magnetic field has a magnitude of 1.50 × 10-³ T, and an electric field has a magnitude of 4.00 × 10³ N/C. Both fields -3 point in the same direction. A positive 2.5 µC charge moves at a speed of 2.50 × 106 m/s in a perpendicular to both fields. Determine the magnitude of the net force that acts on the charge. N ssfánection that is 0 ssf60 sf60 ss 50 ssf6 €60 ssf60 ss

A charged particle moves through a velocity selector at constant velocity. The velocity selector is configured with "crossed" electric and magnetic fields of magnitude E = 1.50 x 104 N/C and B = 0.4 T. %3D Hint a. What is the velocity of the charged particle? Velocity of the charged particle is x 104 m/s. b. When the electric field is turned off, the charged particle travels in a circular path of radius 6 mm, as it travels through the magnetic field (still at B = 0.4 T). What is the mass-to-charge ratio of the particle? Hint for (b) Mass-to-charge ratio of the part is kg/C. (Use the "E" notation to enter your answer in scientific notation. For example, to enter 3.14 × 10´ 12".) - 12 enter "3.14E-