* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download PWE 19-1: Magnetic Forces on a Proton and an Electron
Survey
Document related concepts
Hydrogen atom wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Electrostatics wikipedia , lookup
Elementary particle wikipedia , lookup
Condensed matter physics wikipedia , lookup
Magnetic field wikipedia , lookup
Superconductivity wikipedia , lookup
Electromagnetism wikipedia , lookup
Magnetic monopole wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Electromagnet wikipedia , lookup
Neutron magnetic moment wikipedia , lookup
Transcript
Example 19-1 Magnetic Forces on a Proton and an Electron At a location near our planet’s equator, the direction of Earth’s magnetic field is horizontal (that is, parallel to the ground) and due north, and the magnitude of the field is 2.5 * 1025 T. Find the direction and magnitude of the magnetic force on a particle moving at 1.0 * 104 m>s if the particle is (a) a proton moving horizontally and due east, (b) an electron moving horizontally and due east, and (c) a proton moving horizontally in a direction 25° east of north. Recall that a proton has charge e = 1.60 * 10219 C and an electron has charge 2e. Set Up In each case we’ll use Equation 19-1 to find the magnitude of the force on the moving proton or electron. The right-hand rule (Figure 19-5) will tell us the direction of the magnetic force. Solve (a) The velocity vector s v of the proton is s, perpendicular to the magnetic field vector B so in Equation 19-1 the angle u = 90° and s both lie in a v and B sin u = 1. The vectors s horizontal plane; the magnetic force vector Fs must be perpendicular to this plane, so Fs is vertical. The right-hand rule tells us that the force points vertically upward. Magnetic force on a moving charged particle: F = q vB sin u B (19-1) North Charge on the proton: q = e = 1.60 * 10 219 F C The proton velocity vector is perpendicular to the magnetic field vector (u = 90°), so the magnitude of the magnetic force on the proton is B q = +e up v North F = q vB sin u = evB sin u = 11.60 * 10-19 C2 11.0 * 104 m>s2 12.5 * 10-5 T2sin 90 = 4.0 * 10220 N (b) The electron has the same magnitude of charge as the proton and the same velocity, so it experiences the same magnitude of magnetic force. But its charge is negative, so the direction of the magnetic force Fs on the electron is opposite to the direction given by the righthand rule. Hence the force Fs points vertically downward. Charge on the electron: q = 2e = 21.60 * 10 219 C B q = –e The electron velocity vector is perpendicular to the magnetic field vector (u = 90°), so the magnitude of the magnetic force on the electron is East up v North F East F = q vB sin u = evB sin u = 11.60 * 10-19 C2 11.0 * 104 m>s2 12.5 * 10-5 T2sin 90 = 4.0 * 10220 N (c) The proton velocity s v and the magnetic s again both lie in a horizontal plane, but field B now the angle between these vectors is u = 25°. So the magnitude of the magnetic force is less than in part (a). The direction of the force is the same as in part (a), however. The proton velocity vector is at an angle u = 25° to the magnetic field vector, so the magnitude of the magnetic force on the proton is F q = +e F = q vB sin u = evB sin u = 11.60 * 10-19 C2 11.0 * 104 m>s2 * 12.5 * 10-5 T2 sin 25 = 1.7 * 10220 N 25° B v up North East This is smaller than in part (a) because sin 25° = 0.42 compared to sin 90° = 1. Reflect This example illustrates how the magnetic force on a moving charged particle depends on the direction in which the particle is moving. Note that the force magnitudes in parts (a), (b), and (c) are very small because a single electron or proton carries very little charge. These particles also have very small mass, however (1.67 * 10227 kg for the proton and 9.11 * 10231 kg for the electron), so the resulting accelerations are tremendous. Can you use Newton’s second law to show that the proton in part (a) and the electron in part (b) have accelerations of 2.4 * 107 m>s 2 and 4.4 * 1010 m>s 2 respectively?