A proton moving in a uniform magnetic field with V1 = 1.18 × 106 m/s experiences force F₁ = 1.39 × 10-16 N. A second pro
Posted: Sun May 22, 2022 1:13 pm
A proton moving in a uniform magnetic field with V1 = 1.18 × 106 m/s experiences force F₁ = 1.39 × 10-16 N. A second proton with v₂ = 2.21 ×106 m/s experiences → F2: -16% N in the same field. 3.62 x 10 == What is the magnitude of B? Express your answer with the appropriate units. ► View Available Hint(s) 0 μA ? B = Value T Submit X Incorrect; Try Again Part B What is the direction of B? Give your answer as an angle measured ccw from the +x-axis. Express your answer in degrees. Previous Answers