A particle with a charge q = 1.60 x 10^-19 C, mass m = 2.00 x 10^-27 kg, and an

Question

A particle with a charge q = 1.60 x 10^-19 C, mass m = 2.00 x 10^-27 kg, and an inital velocity v = v0(i+2j+2k), where v0=2.00 x 10^5 m/s, enters a region with a uniform magnetic field B = B0j, where B0= 2.50T.

a) Calculate the force on this moving charge due to the magnetic field. The particle moves in a helical(cork-screw) path as it travels in the y-direction.
b) Calculate the radius of the helical path.
c) How much time is required to make a complete revolution about the y-axis?
d) How far has the particle traveled in the y-direction while making one revolution around the y-axis?

Explanation / Answer

the mass of the particle m = 2.0*10^-27kg the initial speed v = 2.0*10^5 ( i + 2j + 2k) the magnetic field B = 2.5j (a) the force on the particle F = q (v*B)    = q (2*10^5i + 4*10^5j + 4*10^5 k) (2.5j)
   = (1.6*10^-19) [ -10*10^5i + 5*10^5 k]    = -16*10^14 i + 8*10^-14 k Magnitude =17.88*10^-14 N (b) The radius of the helical path    r = mv/qB    = (2.0*10^-27)(6*10^5 m/s) / 1.6*10^-19 (2.5)
   = 3*10^-3 m (c) The time taken to complete the revolutation    T = 2pm / qB = 2p(2.0*10^-27) / (1.6*10^-19)(2.5)    = 3.14*10^-8 sec
(d) The distance traveled
y = 2pr = 18.8*10^-3 m

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