User:
Intensidad del campo magnético de un alambre
Equation
Una alambre por el que circula corriente genera un campo magnético circular en torno de este.
Por ello con el campo magnético se calcula mediante:
| $ H_w = \displaystyle\frac{ I }{ 2\pi r }$ |
ID:(12167, 0)
Charges on a wire
Equation
When considering a segment $dl$ of a wire with a certain cross-sectional area $S$ and length, it results in a volume of wire. Multiplying this volume by the charge density $c$ gives us the number of charges contained within it. Finally, by multiplying it by the unit charge $q$, we obtain the total charge present in the segment.
 $ \Delta Q = q c S dl $ |
ID:(12172, 0)
Current in a wire
Equation
The current is defined by the equation:
| $ I =\displaystyle\frac{ dQ }{ dt }$ |
and the charges within a segment of wire are represented by:
| $ \Delta Q = q c S dl $ |
The ratio of the length of the segment to the corresponding time interval gives us the velocity:
$v =\displaystyle\frac{dl}{dt}$
Therefore, the current in the wire is equal to:
| $ I = q c S v $ |
ID:(12173, 0)
Lorentz force on the wire
Equation
If we consider a segment of a wire, the charge in that segment is given by:
| $ \Delta Q = q c S dl $ |
Furthermore, as charges move, they generate a current equal to:
| $ I = q c S v $ |
If we take these charges into account in the Lorentz force, expressed as:
| $ \vec{F} = q \vec{v} \times \vec{B} $ |
All the charges in a segment $d\vec{l}$ contribute to the force as follows:
$d\vec{F}=\Delta Q\vec{v}\times\vec{B}= q S c v d\vec{l}\times \vec{B} = I d\vec{l}\times \vec{B}$
Therefore, we can conclude that with:
| $ d\vec{F} = I d\vec{l} \times \vec{B}$ |
ID:(12170, 0)
Force on a wire
Equation
If a wire carrying a current $I_1$ generates a magnetic field given by:
| $ H_w = \displaystyle\frac{ I }{ 2\pi r }$ |
This field generates a magnetic flux density represented by:
| $ \vec{B} = \mu_0 \mu_r \vec{H}$ |
Which, in turn, produces a force per segment in a wire with a current $I_2$, defined as:
| $ d\vec{F} = I d\vec{l} \times \vec{B}$ |
With this, the force per segment can be expressed as:
| $ \displaystyle\frac{ dF }{ dl } = \mu_0 \mu_r \displaystyle\frac{ I_1 I_2 }{2 \pi r }$ |
ID:(12169, 0)
Parallel currents
Image
When two currents are allowed to flow in a parallel manner, we observe an attractive force between the wires.
It's worth recalling that currents consist of electrons in motion, and electrons naturally repel each other due to their negative charges. However, when these charges are in motion, this repulsive force turns into an attractive force, resulting in the observed attraction between the negatively charged conductors.
ID:(11772, 0)
Opposite parallel currents
Image
When two currents are allowed to flow in a parallel but opposite direction, we observe a repulsive force between the wires.
Comparing this experiment to the one where the flow is parallel but in the same direction, the key difference lies in the presence of relative velocity in the latter case.
ID:(11773, 0)