EM Orbit

Electricity and Magnetism Level pending

Particle 1 1 has mass m m and charge + q +q , and Particle 2 2 has mass m m and charge − q -q . At a particular time, Particle 1 1 is at position ( x , y ) = ( 1 , 0 ) (x,y) = (1,0) with speed v v in the + y +y direction. At the same time, Particle 2 2 is at position ( x , y ) = ( − 1 , 0 ) (x,y) = (-1,0) with speed v v in the − y -y direction.

If the particles both maintain uniform circular motion around the origin, what is the value of v v ? Be sure to read the "Details and Assumptions" carefully.

Details and Assumptions:
1) Each particle exerts both an electric and a magnetic influence on the other
2) The magnetic flux density calculation is described in this link
3) m = 1 m = 1
4) q = 6 q = 6
5) Permittivity and permeability ϵ 0 = μ 0 = 1 \epsilon_0 = \mu_0 = 1
6) I think this approach is still ultimately wrong, since accelerating charged particles are supposed to lose energy due to emission of electromagnetic radiation. But that will be a subject for another time. For the purposes of this problem, consider electric and magnetic forces but neglect electromagnetic radiation.


The answer is 1.5886.

Steven Chase by Steven Chase , 37, USA

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1 solution

Post a problem in which particle lose energy due to emission of electromagnetic radiation.

Due to a moving charge magnetic field B ⃗ \vec{B} is generated as B ⃗ = μ 0 4 π q v ⃗ × r ⃗ r 3 \textcolor{#20A900}{\vec{B}=\frac{\mu_{0}}{4π} \frac{q\vec{v}×\vec{r}}{r^{3}}} Using this the magnetic field on − q -q comes B − q ⃗ = μ 0 4 π q v r 2 ( − k ⃗ ) \vec{B_{-q}}=\frac{\mu_0}{4π}\frac{qv}{r^{2}}(\vec{-k}) B ⃗ = 3 v 8 π ( − k ⃗ ) \vec{B}=\frac{3v}{8π}(\vec{-k})
Force on a particle moving in magnetic field F ⃗ = q ( v ⃗ × B ⃗ ) \textcolor{#20A900}{\vec{F}=q(\vec{v}×\vec{B})}
In the − q -q , F m F_m (magnetic force) will be in + x +x direction
Of magnitude F m ⃗ = 9 v 2 4 π i ⃗ \vec{F_m}=\frac{9v^{2}}{4π} \vec{i}
And while calculating electric force using coulomb's law F c ⃗ = k q 1 q 2 r ⃗ r 3 \vec{F_c}=\frac{kq_{1}q_{2}\vec{r}}{r^{3}} As the charges are unlike force will be attractive. F e ⃗ = 9 4 π i ⃗ \vec{F_e}=\frac{9}{4π}\vec{i}
F m a g n e t i c ⃗ + F e l e c t r i c ⃗ = F c e n t r i p e t a l ⃗ \vec{F_{magnetic}}+\vec{F_{electric}}=\vec{F_{centripetal}}
9 v 2 4 π + 9 4 π = v 2 1 \frac{9v^{2}}{4π}+\frac{9}{4π}=\frac{v^{2}}{1}
v ⃗ ≈ 1.5886 ( − j ⃗ ) \textcolor{#3D99F6}{\boxed{\vec{v} \approx 1. 5886(\vec{-j})}}


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