Question
Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 154 V/m and the magnetic field is 3.18×10-2 T. The ions next enter a uniform magnetic field of magnitude 1.75×10-2 T that is oriented perpendicular to their velocity.
a)How fast are the ions moving when they emerge from the velocity selector?
v=____________m/s
b)If the radius of the path of the ions in the second magnetic field is 17.5 cm, what is their mass?
m=__________kg
Explanation / Answer
Given E = 151 V/m B =3.18×102 T (a) Now the velocity of the ion when it emerges from the velocity selector is v = E / B................(1) = (151 V/m) / (3.18×102 T) = 4748.42 m/s (b) magnetic field B =1.75×102 T radius of the path of the ions r = 17.5 cm = 0.175 m magnetic force F = Bvq ................... (1) from Newton's second law of motion , force F = ma here , centripetal acceleration a = v2/r force F = mv2/r ................ (2) compare eq (1) & eq (2) , we get Bvq = mv2/r Bq = mv/r mass m = Bqr/v = (1.75×102 T)(1.6*10-19 C)(0.175 m) /(4748.42 m/s) = 10.32*10-26 kg Given E = 151 V/m B =3.18×102 T (a) Now the velocity of the ion when it emerges from the velocity selector is v = E / B................(1) = (151 V/m) / (3.18×102 T) = 4748.42 m/s (b) magnetic field B =1.75×102 T radius of the path of the ions r = 17.5 cm = 0.175 m magnetic force F = Bvq ................... (1) from Newton's second law of motion , force F = ma here , centripetal acceleration a = v2/r force F = mv2/r ................ (2) compare eq (1) & eq (2) , we get Bvq = mv2/r Bq = mv/r mass m = Bqr/v = (1.75×102 T)(1.6*10-19 C)(0.175 m) /(4748.42 m/s) = 10.32*10-26 kg magnetic force F = Bvq ................... (1) from Newton's second law of motion , force F = ma here , centripetal acceleration a = v2/r force F = mv2/r ................ (2) compare eq (1) & eq (2) , we get Bvq = mv2/r Bq = mv/r mass m = Bqr/v = (1.75×102 T)(1.6*10-19 C)(0.175 m) /(4748.42 m/s) = 10.32*10-26 kg force F = mv2/r ................ (2) compare eq (1) & eq (2) , we get Bvq = mv2/r Bq = mv/r mass m = Bqr/v = (1.75×102 T)(1.6*10-19 C)(0.175 m) /(4748.42 m/s) = 10.32*10-26 kg