electromagnetism
When a magnetic field is applied perpendicular to a current carrying wire, a Lorentz force
$$\mathrm{\vec{F}=i(\vec{L}\times\vec{B})}$$
acts on the conducting wire. Of course the applied magnetic field interacts with the magnetic field generated by the current carrying wire.
My question is, does the Lorentz force affect the flow of electrons in the conducting wire? Does it slow them down? Does it affect the resistance of the wire? Does the wire heat up, even slightly?
First, just as a reminder, the force of a particle due to a magnetic field is $$\vec{F_B}=q\vec{v}\times\vec{B}$$. As you mentioned the force due to the magnetic field is always perpendicular to the velocity but it is also perpendicular to the magnetic field so the differential work is $$dw_B=\vec{F_B}\cdot d\vec{s}=q(\vec{v}\times\vec{B})\cdot \vec{v}dt=0 $$ so we have verified that the work is indeed zero due to the magnetic field. $\vec{F_B}$ cannot change the speed of the particle so it cannot change it's kinetic energy, but it can change the direction of $\vec{v}$. The total force on the particle is $$\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$$ so the total work on a particle is $$dw=q\vec{E}\cdot\vec{v}dt$$ which is non zero.
Now lets consider the wire which is really just a group of moving charges. positive particles inside the wire are pushed upward in the wire due to the magnetic force and negative particles are pushed downward (where the up-down directions are orthogonal to the wire velocity v and magnetic field B). The separation of charges induces an E field which produces a force on the respective particles in the opposite direction. In equilibrium the two forces balance and we have $$qvB=qE$$ and the emf across length l is $$\mathcal{E}=El=Bvl$$. When you are talking about an EMF in the wire you should consider the wire in a magnetic field closed by a loop outside the field (see http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html#c4 , In your problem the circuit is not closed so the wire goes nowhere) The induced emf can be seen as the work done per unit charge so it is really the E field that changes the velocity of the particles (changes their kinetic energy)perpendicular to the B field.
You have been tricked by the way this has been drawn. Rotating the wire whilst keeping it perpendicular to the magnetic field does not change the magnitude of the force. Only when you change the angle between the wire and the field, i.e. tilt the wire so that it lines up with the field, does the magnitude reduce.
To prove this, we can look at the origin of this force. It arises directly from the Lorentz force on the electrons in the wire, and is given for each electron by $\textbf F = q(\textbf E + \textbf v \times \textbf B)$. The magnetic contribution to this force is a cross product of the velocity (which is essentially the current) and the field direction: $\textbf v \times \textbf B = vB\sin\theta$. Here $\textbf v$ and $\textbf B$ are perpendicular so the force on each electron is exactly equal to $vB$, which of course translates to $BIL$ on the wire.
Just in case this is still not clear to you, I made a 3D diagram of the situation in the question. The red lines represent the uniform magnetic field, the yellow line is the wire and the green arrow is the force.
As you can see the magnitude of the force does not change as the wire is rotated perpendicularly. However, if we were to rotate in the other direction, the cross product of $\textbf v \times \textbf B$ would have an affect on the magnitude of the force. This can be seen below.
I hope this was useful. OpenSCAD source code:
$fn=30;
for (x=[-10:5:10]) for (y=[-10:5:10])
translate([x, y, 0])
color("red")
translate([0, 0, -10])
cylinder(d=0.5, h=20);
theta = 360*$t;
alpha = 90;//*$t;
f = 10*sin(alpha); //[BIL]sin(theta)
color("green")
rotate(theta)
rotate([90, 0, 0]) {
cylinder(d=1, h=f);
translate([0, 0, f])
cylinder(d1=3, d2=0, h=2);
}
color("yellow")
rotate(theta)
rotate([0, alpha, 0])
translate([0, 0, -10])
cylinder(d=1, h=20);
(gif created with convert -resize 40% -delay 5 -loop 0 frame* gif1.gif)
Best Answer
Let's say the wire is in the $\overrightarrow x$ direction, while the applied magnetic field is in the $\overrightarrow z$ direction.
The Lorentz force law $\overrightarrow F=I\overrightarrow v\times \overrightarrow B $ tells us that the the force will be in the $\overrightarrow y$ direction.
The electrons in the wire will incur a net force in the direction perpendicular to their motion and the applied magnetic field. So if you do further calculations, you will see that the electrons will slow down a bit. If your wire is now a conducting plane, you'll have also a voltage difference across the conductor in the same direction of the force, see Hall effect
The resistance of the wire is an intrinsic parameter of the material of which it's made, so it cannot be modified by the magnetic field. As for the temperature, I think you'll need a strong alternating magnetic field for a non-negligible increase.