How Do You Solve an Electromagnetism Problem Involving a Cylindrical Conductor?

[cw88]

Hello, I'm new on this forum :)

I seriously need help with this problem. It has to be perfect if I want to pass the course, so I will really appreciate any solution to check my results.

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An electrical current of 100[A] runs through a cilindrical conductor (radius 0.01[m], infinite length). The (axis of the) conductor is 2[m] above the earth, and it is held at a constant voltage of -1000[V] with respect to the ground.

1. Write the differential equations that describe the movement of an electron that escapes from the surface of the cable. Use a cartesian coordinate system whose x-axis coincides with the axis of the cable.

2. Solve the differential equations with the initial conditions x(0)=0, y(0)=0.01, and the electron starts at rest. (y(t) refers to the vertical axis)

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That's it. I suppose you start by obtaining the magnetic induction and electric field and then use the Lorentz force on Newton's second law.

Sorry for my english if I made any mistake...

Thanks in advance.

... maybe this question is for the advanced physics forum, sorry for that :S

 

[Irid]

I suppose you start by obtaining the magnetic induction and electric field and then use the Lorentz force on Newton's second law.

That's right, go ahead.

 

[cw88]

Allright. Using Gauss's law and Ampere's circuital law it's easy to obtain the fields (in cylindrical coordinates, with the vertical axis fixed at the conductor's axis and oriented in the current's direction):

$$\vec{B}=\frac{\mu_0 I}{2 \pi r}\hat{\phi}$$

$$\vec{E}=\frac{\lambda}{2 \pi \epsilon_0 r}\hat{\rho}$$

$$r > d$$ in both cases

To obtain $$\lambda$$ (charge per unit of length), we use the potential difference between the conductor's surface and the ground:

$$V(d) - V(D) = \int_\gamma \vec{E} \cdot \vec{dr} = \frac{\lambda}{2 \pi \epsilon_0} \int^D_d \frac{dr}{r} = \frac{\lambda}{2 \pi \epsilon_0} ln(D/d)$$

Since $$V(D) = 0, V(d) = V$$

$$\lambda = \frac{2 \pi V \epsilon_0}{ln(D/d)}$$

Thus, the electric field is:

$$\vec{E} = \frac{V}{r ln(D/d)} \hat{\rho}$$

Is it all right to this point?

Note:

• d: conductor's radius
• D: distance between the conductor's axis and the ground
• V: voltage at which the conductor is held
• I: the current that flows through the conductor

 

[cw88]

Using the stuff above, I've obtained a system of differential equations, but I don't know if it is correct...

I want to use Newton's second law, F=ma, being

$$\vec{F}=q( \vec{E} + \vec{v} \times \vec{B})$$

the Loretz force that acts on the electron (of charge q and mass m).

Velocity in cylindrical coordinates is

$$\vec{v} = \dot{\rho} \hat{\rho} + \rho \dot{\phi} \hat{\phi} + \dot{z}\hat{k}$$

Hence

$$\vec{v} \times \vec{B} = \frac{ \mu_0 I \dot{\rho} }{ 2 \pi \rho } \hat{k} - \frac{ \mu_0 I \dot{z} }{ 2 \pi \rho } \hat{\rho}$$

$$\vec{F} = ( \frac{qV}{ \rho ln(D/d) } - \frac{ q \mu_0 I \dot{z} }{ 2 \pi \rho } ) \hat{\rho} + \frac{ q \mu_0 I \dot{\rho} }{ 2 \pi \rho } \hat{k} = m \vec{a}$$

Writing the acceleration in cylindrical coordinates, we obtain the following equations:

$$1.\qquad \frac{qV}{ \rho ln(D/d) } - \frac{ q \mu_0 I \dot{z} }{ 2 \pi \rho } = m \ddot{\rho}$$

$$2.\qquad 2 \dot{\rho} \dot{\phi} + \rho \ddot{\phi} = \frac{d}{dt} ( \rho^2 \dot{\phi} ) = 0$$

$$3.\qquad \frac{q \mu_0 I \dot{\rho}}{2 \pi \rho} = m\ddot{z}$$

From the second equation, since the electron starts with no velocity at all:

$$\rho^2 \dot{\phi(t)} = \rho^2 \dot{\phi(0)} = 0 \qquad \forall t$$

$$\qquad \Rightarrow \phi(t) = c$$

Being c an arbitrary constant. Here I choose c = 0 for simplicity. This way the transformation to the asked coordinate system is trivial.

I'm really bad at this, so it would be a big achievement to me if it's correct to this point :D

 

[Irid]

I didn't check everything explicitly, but your procedure is fine, so if you haven't made some trivial error, the equations should be correct.

 

[cw88]

thanks a lot, from this point it's not very hard to find the equations... I'm an expert at trivial errors, so wish me luck...

 

[Irid]

Good luck.