PHYS 102 Problems - Chapter 20 – Set 8 Feb. 2, 2010
... Alpha particles of charge q = +2e and mass m = 6.6 x 10-27 kg are emitted from a radioactive source at a speed of 1.6 x 107 m/s. What magnitude field strength would be required to bend them into a circular path of radius r = 0.25 m? In this scenario, the magnetic force is causing centripetal motion, ...
... Alpha particles of charge q = +2e and mass m = 6.6 x 10-27 kg are emitted from a radioactive source at a speed of 1.6 x 107 m/s. What magnitude field strength would be required to bend them into a circular path of radius r = 0.25 m? In this scenario, the magnetic force is causing centripetal motion, ...
ESS154_200C_Lecture7_W2016
... – Energy of charged particles is usually given in electron volts (eV) – Energy that a particle with the charge of an electron gets in falling through a potential drop of 1 Volt – 1 eV = 1.6x10-19 Joules (J). • Energies in space plasmas go from electron Volts to kiloelectron Volts (1 keV = 103 eV) to ...
... – Energy of charged particles is usually given in electron volts (eV) – Energy that a particle with the charge of an electron gets in falling through a potential drop of 1 Volt – 1 eV = 1.6x10-19 Joules (J). • Energies in space plasmas go from electron Volts to kiloelectron Volts (1 keV = 103 eV) to ...
Week 11 Monday
... a current loop will be equal and opposite (if the field is uniform and the loop is symmetric), but there may be a torque. The magnitude of the torque is given by ...
... a current loop will be equal and opposite (if the field is uniform and the loop is symmetric), but there may be a torque. The magnitude of the torque is given by ...
PHYS 102 Midterm Exam 2 (09.04.2016) Solutions
... perpendicular to the current. This results in a force with only radial and vertical components. By symmetry, we find that the radial force components from segments on opposite sides of the loop cancel. The net force then is purely vertical and upward. Symmetry also shows us that each current element ...
... perpendicular to the current. This results in a force with only radial and vertical components. By symmetry, we find that the radial force components from segments on opposite sides of the loop cancel. The net force then is purely vertical and upward. Symmetry also shows us that each current element ...
A ball of mass M is thrown vertically upward with an initial speed of vo
... b. An electron is released from rest at point B. i. Qualitatively describe the electron's motion in terms of direction, speed, and acceleration. ii. Calculate the electron's speed after it has moved through a potential difference of 10 V. c. Points B and C are separated by a potential difference of ...
... b. An electron is released from rest at point B. i. Qualitatively describe the electron's motion in terms of direction, speed, and acceleration. ii. Calculate the electron's speed after it has moved through a potential difference of 10 V. c. Points B and C are separated by a potential difference of ...
Since we will be studying electromagnetic waves, let`s review some
... We just showed how the electromagnetic (E&M) wave is initially generated by the ac voltage source near the antenna (near field). As the wave moves farther away, it propagates itself by the changing E-field producing a B-field and the changing B-field producing an E-field (radiation field). è E&M wa ...
... We just showed how the electromagnetic (E&M) wave is initially generated by the ac voltage source near the antenna (near field). As the wave moves farther away, it propagates itself by the changing E-field producing a B-field and the changing B-field producing an E-field (radiation field). è E&M wa ...
Tuesday, October 23 rd
... Magnetic forces can only change the direction (of the Velocity) of charged particles. They cannot change the magnitude (of their velocity). Example: J.J. Thomson’s experiment (1897) for q/m 1. Determine v by measuring E/B: 2. Determine q/m by measuring R with E turned off. ...
... Magnetic forces can only change the direction (of the Velocity) of charged particles. They cannot change the magnitude (of their velocity). Example: J.J. Thomson’s experiment (1897) for q/m 1. Determine v by measuring E/B: 2. Determine q/m by measuring R with E turned off. ...
velocity of propagation
... surface of the shield. The dielectric keeps the center conductor supported in the middle of the cable, and also insulates it from the shield. When RF travels through coaxial cable, RF currents propagate on and near the outer surface of the center conductor, and on and near the inside surface of the ...
... surface of the shield. The dielectric keeps the center conductor supported in the middle of the cable, and also insulates it from the shield. When RF travels through coaxial cable, RF currents propagate on and near the outer surface of the center conductor, and on and near the inside surface of the ...
printable version - Gosford Hill School
... Demo 120D Demonstration 'Deflecting electron beams in a magnetic field' Warm 80W 'Getting F = q v B' – link to Chapter 15 Dis 120O 'How a magnetic field deflects an electron beam' Dis 130O 'Force on current: force on moving charge' SoftAct Activity 140S 'Circular motion in a magnetic field' Dis 140O ...
... Demo 120D Demonstration 'Deflecting electron beams in a magnetic field' Warm 80W 'Getting F = q v B' – link to Chapter 15 Dis 120O 'How a magnetic field deflects an electron beam' Dis 130O 'Force on current: force on moving charge' SoftAct Activity 140S 'Circular motion in a magnetic field' Dis 140O ...