An ?-particle has a charge of +2e and a mass of 6.64 10-27 kg. It is accelerated

Question

An ?-particle has a charge of +2e and a mass of 6.64 10-27 kg. It is accelerated from rest through a potential difference that has a value of 1.04 106 V and then enters a uniform magnetic field whose magnitude is 2.90 T. The ?-particle moves perpendicular to the magnetic field at all times. (a) What is the speed of the ?-particle? (b) What is the magnitude of the magnetic force on it? (c) What is the radius of its circular path? I have found the speed to be 1.04e7, but I can't figure out b and c, please help!

Explanation / Answer

the charge is q = +2e = +2 * 1.6 * 10^-19 C mass is m = 6.64 * 10^-27 kg potential difference is V = 1.04 * 10^6 volts magnetic field is B = 2.90 T we know that V = 1/4pie_o * q/r where 1/4pie_o = 9 * 10^9 Nm^2/C^2 and r is radius of circular path or r = 1/4pie_o * q/V we know that qvB = mv^2/r where v is speed of the particle or qB = mv/r or v = qBr/m the magnetic force is F = qvB

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