One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories are 1.00 cm and 2.40 cm. The trajectories are perpendicular to a uniform magnetic field of magnitude 0.044 0 T. Determine the energy (in keV) of the incident electron.

F.j = (-e)(-v,k) × (B,i + B.k) ev Bzj. Solving for B gives F. Br 8.50x10 16 N = 1.13 T. Therefore B = 1.13 Ti – 0.772 Tk. The magnitude o A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.70 km/s in the +x-direction experiences a force of 2.10x10-16 N in the +y- direction, and an electron moving at 4.70 km/s in the eve (1.60x10 19 C)(4700 m/s) is B = VB? + B = (1.13 T)² + (-0.772 T)² = 1.37 T. -z-direction experiences a force of 8.50x10-16 N in the +y-direction. Part B For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Magnetic force on a proton. What is the direction of the magnetic field? (in the xz-plane) Express your answer in degrees. 0 = 60.59 ° from the -z-direction to the +x-direction Submit Previous Answers Request Answer X Incorrect; Try Again; 9 attempts remaining

The Hall affect is used to measure the carrier density of a thin sheet of electrons. When a current of 80.0 µA flows through the length of the electron sheet, which is 1.2-mm long, 0.27- mm wide and 12-nm thick, a magnetic field perpendicular to the sheet produces a potential difference of 0.53 mV across the width of the sheet. If the carrier density in the electron sheet is 6.74 x 1025 m-3, what is the magnitude of the magnetic field?

When you shine a certain source of EM radiation on a sheet of gold (work function W1 = 5.1 eV), the stopping voltage required to bring all the ejected electrons to a halt is V = 3.5 V. If you were to use this same source of EM radiation on a sheet of potassium (work function W2 = 2.3 eV), what would be the maximum speed of ejected electrons?