FiniteStraightWire

class plasmapy.formulary.magnetostatics.FiniteStraightWire(p1: Quantity, p2: Quantity, current: Quantity)[source]

Bases: Wire

Finite length straight wire class.

The p1 to p2 direction is the positive current direction.

Parameters:

Methods Summary

 Calculate magnetic field generated by this wire at position p. Convert this Wire into a GeneralWire.

Methods Documentation

magnetic_field(p) [source]

Calculate magnetic field generated by this wire at position p.

Parameters:

p (astropy.units.Quantity) – Three-dimensional position vector

Returns:

B – Magnetic field at the specified position

Return type:

astropy.units.Quantity

Notes

The magnetic field generated by a straight, finite wire with constant electric current can be found at a point in 3D space using the Biot–Savart law.

Let the point where the magnetic field will be calculated be represented by the point $$p_0$$ (or p) and the wire’s beginning and end as $$p_1$$ and $$p_2$$, respectively (with corresponding position vectors $$\vec{p}_0$$, $$\vec{p}_1$$, and $$\vec{p}_2$$, respectively). Further, the vector from points $$p_i$$ to $$p_j$$ can be written as $$\vec{p}_{ij} = \vec{p}_j - \vec{p}_i$$.

Next, consider the triangle with the points $$p_0$$, $$p_1$$, and $$p_2$$ as vertices. The vector from the vertex $$p_0$$ to the perpendicular foot opposite the vertex $$p_0$$, which will be used to find the unit vector in the direction of the magnetic field, can be expressed as

$\vec{p}_f = \vec{p}_1 + \vec{p}_{12} \frac{\vec{p}_{10} \cdot \vec{p}_{12}} {|\vec{p}_{12}|^2}.$

The magnetic field $$\vec{B}$$ generated by the wire with current $$I$$ can be found at the point $$p_0$$ using the Biot–Savart law which in this case simplifies to

$\vec{B} = \frac{\mu_0 I}{4π} (\cos θ_1 - \cos θ_2) \hat{B}$

where $$\mu_0$$ is the permeability of free space, $$\theta_1$$ ($$\theta_2$$) is the angle between $$\vec{p}_{10}$$ ($$\vec{p}_{20}$$) and $$\vec{p}_{12}$$ with

$\cos\theta_1 = \frac{\vec{p}_{10} \cdot \vec{p}_{12}} {|\vec{p}_{10}| |\vec{p}_{12}|}, \quad \cos\theta_2 = \frac{\vec{p}_{20} \cdot \vec{p}_{12}} {|\vec{p}_{20}| |\vec{p}_{12}|},$

and

$\hat{B} = \frac{\vec{p}_{12} \times \vec{p}_{f0}} {|\vec{p}_{12} \times \vec{p}_{f0}|}$

is the unit vector in the direction of the magnetic field at the point $$p_0$$.

to_GeneralWire()[source]

Convert this Wire into a GeneralWire.