conductorselectromagnetismforcesmagnetic fieldsvectors
I don't quite understand why the force is perpendicular, can somebody explain with a magnetic field diagram as to how the perpendicular direction is produced as a result of the interaction of the two fields (one of the wire and the other we apply).
I know the force is proportional to the cross product of the length and the field and hence the perpendicular direction, but how do the individual fields interact to produce the force.
First of all, the force F that you wrote is not due to the interaction between the two magnetic fields. I hope it is clear to you that any current put in the magnetic field will give rise to this force.
Yes, the wire does produce a magnetic field, which is circularly around it with the wire being the axis passing perpendicular to the plane of the circle and if the current is high enough,one would feel force due to this field on the magnetic poles. If you are reading from a book, it has probably neglected the field of wire in this instance.
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
Why is the force perpendicular? It may be helpful to reverse the question. What would happened if the force where parallel or antiparallel to the velocity of the particle? If the force where parallel or antiparallel to the particle's velocity the particle would accelerate along its direction of travel. Either increasing or decreasing in velocity.
Applying the constraint of conservation of energy to the system. If the velocity of the charged particle where to increase. Where would the energy for the increase in velocity come from? If the velocity of the particle decreased. Where would the energy lost by the particle go? There are only two elements to the system, the magnetic field, and the charged particle. If the energy in the particle increases then the field's energy must decrease. If the energy of the particle decreases then the field's energy must increase. How would the energy be transferred?
The force being perpendicular to the magnetic field and the motion of the particle observes conservation of energy. While the velocity of the particle changes the energy of the particle is conserved.