5 Electromagnetic Force on a Moving Charge Vector Weberwbs T

5. Electromagnetic Force on a Moving Charge (Vector) (Weber-wbs Trn2 = Nm/A-JA) Magnetic Flux through any closed surface) 6. Magnetic Flux: 6. Gauss\'s Law for Magnetism: Example Problems 1. A small particle m/s f + 4 m/s in a unifo no gravity, (a) what is the fo of the particle at t = 0? (c) take the particle to complete of mass 2 g and charge 10 C s emitted at 1-0 fromthe origin with the velocity 3 m\'s k in a uniform magnetic field of strength 100 T in the z-direction. If there is no electrical field and rcie at t0? (c) What is the radius of the helix about which the particle moves? (d) How long does it (a) what is the force on the particle at t 02 (b) What is the magnitude and direction of the acceleration during this revolution? Given: Equations Answers: m = 2.0x10-3 kg (a) IFs-3.0x10-3 N

Solution

m = 2 x 10^-3 kg; q = 10 x 10^-6 C; v(0) = 3 i + 4k ms^-1 ; B = 100 T; E = 0; g = 0;

(a) Force on the particle at t =0;

using the same formula which are given the problem image, then we got the answer is, F = -3 x 10 ^-3 N j;

Note: using Flemings left hand rule for find direction of the Force.

(b) Acceleration:

F = ma; => a =F /m;; = -1.5 ms^-2 (magnitude) & direction is -j (acceleration direcction is along the force direction)

(c) Radius of the helix about the particle moves:

Lorentz force & Centrifugal force are balanced. So,

q v x B = (mv²) / r; => r = (mv) / (qB); Where v is the perpendicular to the field. So, v = 3 i ms^-1;

So, r = 6 m;

(d) Time period of one revolution: (T)

T= (2 * pie * m) / (qB) ; = 12. 571429 s.

(e) Pitch: (along z direction):

pitch distance = velocity x T; Where velocity along z direction is 4 ms^-1 K;

So, pitch distance = 4 x 12. 571429 = 50. 285716 m.

Note: my network is so slow.