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.

r = mv/ q B 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.

write for the charge to mass ratio of the electron 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

cancel the v's where possible qB = mv/r B = (mv)/(qr) 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 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 We need to solve for the velocity! 56,568 m/s 0.0177 m