A particle of charge q = 2.73×10-12 C and mass m = 3.32×10-17 kg enters a unifor

Question

A particle of charge q = 2.73×10-12 C and mass m = 3.32×10-17 kg enters a uniform magnetic field B of magnitude 9.41×10-5 T with a velocity v of magnitude 8.34 m/s, and moves in a helix while in the field. The angle between v and B is 84.0°.

(a) What is the magnitude vperp of the component of v perpendicular to B? (m/s)
(b) What is the radius of the helix? (meters)
(c) What is the magnitude vpar of the component of v parallel to B? (m/s)
(d) What is the pitch of the helix, the distance between adjacent loops (measured parallel to B)? (meters)

Explanation / Answer

a) V perpendicular component is V (per) = v sin = 8.34 * sin 84 = 8.294 m/s b) V parallel components is V (parallel) = V cos = 8.34 * cos84 = 0.8717 m/s c) Helical pitch is p = v (parallel) * 2m / q    = 0.8717 * 2 * 3.14 * 3.32 x 10^-17 / 2.73 x 10^-12    = 6.657 x 10^-5 m    = 66.57 m

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