Question

с Homework – Unanswered A doubly charged helium atom whose mass is 6.6x10^-27 kg is accelerated by a voltage V=2700.0 V and enters the area of magnetic field perpendicular to its velocity B=0.20 T. What is its period of revolution? Numeric Answer: Unanswered • 5 attempts left Submit D Homework. Unanswered An electron enters uniform magnetic field B=0.30 T at angle 45 degrees to B with speed v=1.5*10^6 m/s. Find pitch (distance between loops) of the electron's helical path. Numeric Answer: Unanswered . 5 attempts left Submit E Homework. Unanswered A wire with current of linear density of mass 10.0 g/m levitates in magnetic field perpendicular to direction of current B=0.10 T. Find current in the wire, give answer in A. Numeric Answer:

Answer 1

* Given doubly charged helium atom q=20 m = 6.6x102 kg V: 2700 V B=0.21 for cyclotron motion mu2 = qum 7 = q Bg 9 222 air - abr atm - t = an so ta 11 27 22x6.6410 2x1.681019 xo.2 => t : -7 6.478 10'S This is the period f revolution * Given B = 0.37 O= 45° v = 1.5*108 mis

The belical Loop expression for pitch of the P: 2110 vcoso 9 B - p= 27 x 9.1 x 10 31x1.5 x 10to cosus 1.6x1019 x 0.3 :) IP: 1.26*10-4 Given here 10g/m : 102 kg im B = OIT Egravity I magnetic I LO ng I = 2 9/ 10 102x9.8 = I= 0.1 I= 0.98 A