# How does a magnetic field perform no work?

• mahrap
In summary, the magnetic field does induce a "force" on a moving charge, but this force does not change the kinetic energy of the charge. Work is defined as transferring energy via a force, and in this case, no work is done because there is no change in kinetic energy.
mahrap
1. I understand that in order for a force to qualify as doing work it must displace an object. The magnetic field does indeed induce a "force" on a moving charge and displaces the moving charge toward a different trajectory. However, it does not change the kinetic energy of the moving charge. So for something to qualify to do work must it also change the kinetic energy of the object? I guess what I really need is a better definition of work and how it relates to force. Please guide me in any way you can. Thanks for all the help.

2. F = q (v x B)

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If the force is always perpendicular to velocity, then no work is done, a change in direction, but not a change in speed.

You should be able to find answers to your questions in your textbook. You probably paid a lot of good money for it, so you should make use of it! What is the definition of work? What does the work-energy theorem tell you?

I like to think of work as transferring energy via a force. In your particular case, if a charged particle is moving parallel to magnetic field lines then no work would be done because, as you said, there would be no change in energy of the particle and consequently no work done.

mahrap said:
The magnetic field does indeed induce a "force" on a moving charge and displaces the moving charge toward a different trajectory.

But consider the total, instantaneous displacement of the charge. It's moving, don't forget. So you can find this infinitesimal displacement, $\vec {ds}$ by by finding the distance traveled over the infinitesimal time, $dt$. Don't forget, $\vec {ds}$ is a vector and it has a direction.

Now ask yourself, knowing what you know about magnetic fields, what is the that vector's direction with respect to the direction of the magnetic force, $\vec f$?

[Edit: bonus hint: how does the charge's instantaneous velocity $\vec v$ relate to $\vec {ds}$ and $dt$?]

I guess what I really need is a better definition of work and how it relates to force. Please guide me in any way you can. Thanks for all the help.

Work is defined as

$$W = \vec f \cdot \vec {s}$$
or in the case of a varying force, or force with a varying direction,
$$W = \int_{s=a}^b \vec f \cdot \vec {ds}$$

Notice that in both cases, there is a dot product involved. (It's not as simple as just multiplying the magnitudes together.) What does that dot product tell you about the work done by magnetic fields, given what you've just found about the charge's displacement and the direction of magnetic force?

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rpthomps said:
I like to think of work as transferring energy via a force. In your particular case, if a charged particle is moving parallel to magnetic field lines then no work would be done because, as you said, there would be no change in energy of the particle and consequently no work done.
The relative direction of the mag field compared to the direction of the particle velocity is immaterial. A mag field produces no work on a moving charge, period. Reason? (v x B)*v = 0 always. Algebraic identity. * means dot-product.

rude man said:
The relative direction of the mag field compared to the direction of the particle velocity is immaterial. A mag field produces no work on a moving charge, period. Reason? (v x B)*v = 0 always. Algebraic identity. * means dot-product.

Can you parse what you said a little more? First, why is this equation (v x B)*v = 0 relevant? And secondly, if no work is done by the mag field then where does the energy come from?

qv x B is force. v is rate of change of displacement. So (qv x B)*v is power. Algebraically, this is always zero since v is always orthogonal to v x B.

The energy of the particle is imparted to it by some other force, usually an electric field. Never by a B field.

rpthomps
rude man said:
qv x B is force. v is rate of change of displacement. So (qv x B)*v is power. Algebraically, this is always zero since v is always orthogonal to v x B.

The energy of the particle is imparted to it by some other force, usually an electric field. Never by a B field.

rpthomps said:
10-4.

## 1. How does a magnetic field perform no work?

A magnetic field performs no work because it does not transfer energy to a charged particle that moves through it. The force exerted by a magnetic field on a charged particle is always perpendicular to the direction of motion, which means that the work done by this force is zero.

## 2. Why is the work done by a magnetic field zero?

The work done by a magnetic field is zero because the force exerted by the field on a charged particle depends on the particle's velocity and direction of motion, but not on the distance it travels. Since work is defined as the product of force and distance, and the distance in this case is zero, the work done by the magnetic field is also zero.

## 3. How does a magnetic field affect a charged particle?

A magnetic field affects a charged particle by exerting a force on it. This force is always perpendicular to the particle's velocity and can either increase or decrease its speed, but it does not change the particle's kinetic energy. The direction of the force is determined by the particle's charge and the direction of the magnetic field.

## 4. Can a magnetic field do work on a non-moving charged particle?

No, a magnetic field cannot do work on a non-moving charged particle. The force exerted by the magnetic field is always perpendicular to the particle's motion, and since the particle is not moving, there is no distance over which the force can act. Therefore, the work done by the magnetic field is zero.

## 5. What is the relationship between magnetic fields and work?

The relationship between magnetic fields and work is that magnetic fields do not perform any work on charged particles that move through them. This is because the force exerted by a magnetic field is always perpendicular to the direction of motion, resulting in zero work being done. However, a magnetic field can still affect the motion of a charged particle by changing its velocity or direction, without transferring any energy to it.

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