Problem # 4 . (a) An electron (m- 9.1 x 10 kg: e - 1.602 x 10 19 C) moves in a c

Question

Problem # 4 . (a) An electron (m- 9.1 x 10 kg: e - 1.602 x 10 19 C) moves in a circular path perpendicular to a uniform magnetic field of magnitude 2.0 x 103 T. The speed of the electron is 1.5 x 107 m/s. Determine the radius of the circular path and the time required for one revolution. Ans: 4.27 X 102 m; 1.79 x 10 s - (b) In a velocity selector, the electric field of magnitude 2.5 x 107 V/m is perpendicular to a magnetic field of magnitude 15.0 T. Calculate the velocity of positively charge ions in the beam that would not be deflected. Ans: 167 x 10s m/s n junited

Explanation / Answer

a] for electron in magnetic field r = mv/qB = 9.1e-31*1.5e7/[1.6e-19*2e-3] = 0.0427 m = 4.27*10^-2 m

Time T = 2pir/v = 2pi*0.04265/1.5e7 = 1.79*10^-8 s

b] electrostatic force = mgnetic force

qE = qvB

v = E/B = 2.5e7/15 = 1.67*10^6 m/s answer

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