What does the work on a current carrying wire in a Magnetic Field?

Question

We consider that the force acting on a current carrying wire placed in a uniform magnetic field perpendicular to the length of the wire is given by $IBl$. If the wire moves by a distance $x$ in a direction perpendicular to its length the raise in its kinetic energy is $IBlx$

Now basically the magnetic forces are acting on the electrons moving inside the wire and is always perpendicular to the instantaneous velocity so it cannot perform any work on it and normals are also unable to provide any net work done on the system, so what provides the change in kinetic energy of the system?

Answer 1

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.

Answer 2

Your wire must be part of an electrical circuit (no closed loop = no current). Let us assume that the "return wire" is far away, where there is no magnetic field. So we have a large loop.

Now the motion of the wire in the magnetic field will induce a current that will oppose the magnetic flux change. What happens is that the magnetic field "seen" by the wire becomes less - and since the energy stored in the magnetic field goes as the integral of $B^2$ over the volume, there will be less energy in the magnetic field.

So the "source" of the energy of the wire is the energy that used to be stored in the magnetic field.

Answer 3

so what provides the change in kinetic energy of the system?

In standard EM theory, there are no magnetic poles, only electric ones. Magnetic force on charged particle is always perpendicular to the velocity of the particle, and thus magnetic forces do not work. Generally, if any work is done on the matter, it has to be due to other forces, either mechanical force (like normal or friction force) or electrical force due to electric field.

In your case, since there are no external forces other than magnetic, the working forces have to be internal, i.e. due to the wire itself.

^{There is no problem with internal forces doing net work - this happens daily. For example, consider how person on a bicycle acquires kinetic energy - the force working on the bike is due to biker and the force working on the biker is due to bike while net positive work is done on the system. The energy for this comes from the energy stored in the biker.}

In the case of wire, imagine that wire consists of charged particles (conduction charges) moving inside and along a perfectly reflecting tube (the wire without the conduction charges). Magnetic field acts on the particles and would curve their motion as it happens in vacuum, if the tube did not prevent that. If the wire is held static (by external mechanical force), part of the normal force of the tube cancels the magnetic force and the rest curves the motion of the charges so that they can move along the tube.

When the external mechanical force is removed and the wire is allowed to move, the reaction forces of the charged particles on the tube can do work on the tube. The tube also does work on the charges, while net result is that net work is done on the system.

In this view the work is due to contact mechanical forces, but sometimes it is believed that all mechanical contact forces are ultimately electromagnetic forces on the microscopic level. So if we went deeper into this, the working forces are actually microscopic electric forces - the electric force of the charge works on the lattice and the electric force of the lattice works on the charge.

If the circuit is made of perfect conductor without any source of voltage, the kinetic energy increases at the expense of the magnetic energy in the space around the wire.

If the circuit is made of Ohmic wire and has a source of voltage, kinetic and internal energy of the wire increases also due to energy supplied by the source of voltage.

Answer 4

Newton's 3rd law of motion : Every action has equal and opposite reaction.

When a rod is moving in a magnetic field, An emf is developed in it similarly when current is passed or emf is induced across a stationary wire, It starts moving or a force it acted on it which is responsible for motion or tendency to bring it in motion