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Centripetal force on charges in magnetic fields Which way does a particle get pushed if the the magnetic field is is always perpendicular to the direction of travel ? • No mater which way the charged particle turns the force on it is always perpendicular to its motion. Circular motion -the force is always perpendicular to the direction of travel -negatively charged particle -magnetic field into the -magnetic force is perpendicular to the velocity, and so velocity changes in direction but not magnitude. Uniform circular motion results. electron moving at right angles to a uniform magnetic field. • Circular motion • F = mv2 / r • F force (Newton, N) • m mass ( Kg) • v velocity (m/s) • B magnetic field • q charge • Force on a charge particle in a magnetic field • F = qvB sinθ • v perpendicular to B: • F = qvB • qvB = mv2 / r • r = mv/ q B • r = mv/ q B • Gives you the radius of a charge particles path in a magnetic field, given its mass and velocity. Relationship between radius and magnetic field, mass and velocity r = mv/ q B • Magnetic field B: The stronger the magnetic field, the stronger the force– and therefore the smaller the radius of the charge • Velocity v: the more speed a charged particles has, the harder it is for the magnetic field to corral ( circle) the particle, and so it travels in a circle with a bigger radius. • Mass m: the more mass the charged particle has, the harder it’ll be to bend its path, sot the more mass, the bigger the radius of the circle travels in. • If the velocity of the electron is due to its having been accelerated through a potential difference of magnitude V (volts), then the kinetic energy of the electron is • ½ mv2 = qV • write for the charge to mass ratio of the electron q/m = 2V / B2r2 Example • Alpha particles of charge q = +2e and mass m = 6.6 x10-27 kg are emitted from a radioactive source at a speed of 1.6 x 10 7 m/s. What magnetic field strength would be required to bend these these in a circular path of radius r = .25 m? • e = 1.6 × 10-19 Alpha particles of charge q = +2e and mass m = 6.6 x10-27 kg are emitted from a radioactive source at a speed of 1.6 x 10 7 m/s. What magnetic field strength would be required to bend these these in a circular path of radius r = .25 m? e = 1.6 × 10-19 • Set the force on the particle due to the magnetic field equal to centripetal force necessary to keep the particle moving in a circle. • qvb = mv2/r • cancel the v's where possible • qB = mv/r • B = (mv)/(qr) • = (6.6 x10-27 * 1.6 x107)/(2e * .25) • = 1.32 T Example • A singly charged positive ion has a mass of 2.5 x 10-26 kg. After being accelerated through a potential difference of 250 V, the ion enters a magnetic field of 0.5 T, in a direction perpendicular to the field. Calculate the radius of the path of the ion in the field A charged positive ion has a mass of 2.5 x 10-26 kg. After being accelerated through a potential difference of 250 V, the ion enters a magnetic field of 0.5 T, in a direction perpendicular to the field. Calculate the radius of the path of the ion in the field q = 1.6x10 -19 C m = 2.5x10 DV = 250V B = 0.5T -26 kg DV = W DK = = q q 1 2mv q 2 2DVq 2(250)(1.6x10 -19 ) 56,568 m/s v= = = -26 m 2.5x10 r =? FB = Fc We need to solve for the velocity! mv 2 qvB = r mv r= qB (2.5x10 -26 )(56, 568) 0.0177 m r= = (1.6x10-19 )(0.5)