# If a magnetic field is increasing, will a charge in that field accelerate?

1. Jun 16, 2005

### michaelw

I dont know what would happen..

Fm = qvBsin(theta)
If B is increasing, will velocity of a charge increase? Or will only its radius decrease? Or both?

Fc = Fm
mv^2/R = qvB
mv/qR = B

anyone know?

*acceleration was a bad choice of wording :) I mean will velocity increase..

2. Jun 16, 2005

### dextercioby

The centripetal acceleration will increase.

Daniel.

3. Jun 16, 2005

### Pengwuino

Accelerating is the correct word. Acceleration is defined as a change in velocity.

4. Jun 16, 2005

### whozum

It will fall out of uniform circular motion.

edit: I retract, it would only happen if the B field was not uniform, but rather distance-dependant.

Last edited: Jun 16, 2005
5. Jun 16, 2005

### michaelw

6. Jun 16, 2005

### dextercioby

$$R=\frac{mv}{qB}$$

So if the initial velocity doesn't change,then the radius will decrease.

Daniel.

7. Jun 16, 2005

### michaelw

but how can you tell?
i thought magnetic fields cannot do work on charges, they cannot change acceleration.. is this true for changing magetic fields? the equation is

$$B=\frac{mv}{qR}$$

so will velocity change? or radius? or both? how can you tell?

8. Jun 16, 2005

### dextercioby

$$m\vec{a}=q\vec{v}\times\vec{B}$$

Okay.Now project that relation on two axisne parallel to the magnetic field and one perpendicular to it.Then see what happens.

Daniel.

9. Jun 16, 2005

### Dr.Brain

As a fact, when an electron enters a magnetic field which is prependicular to the velocity direction of the electron , the electron starts moving in a circle for which necessary centripedal force is provided by the magnetic field. As the magnetic field is increased the charge accelerate in a sense that the magnitude of velocity remains same but now the velocity direction changes more rapidly .As for radius is concerned , it is given by $\frac{mv}{qB}$ , so the radius keeps on decreasing.

So now you can imagine it, as the magnetic field is increased the radius goes on decreasing and the electron moves revolves more rapidly , ...leading to some comnclusion?....What does an accelerating electron do? ..Why is it going inwards towards the centre? ACTUALLY, AN ACCELERATING ELECTRON RADIATES, loses energy .....so what happens?

10. Jun 16, 2005

### OlderDan

There is no question that the magnetic field exerts a force perpendicular to the velocity that does no work on the charge. Increasing the strength of the field will increase the force and reduce the radius, but there is more going on here. What does Faraday's law tell you about the relationship between a time varying magnetic field and the induced electric field?

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/maxeq2.html#c3

Dr. Brain is correct about the radiation of an accelerating charge, but my guess is you have not encountered that yet in your studies and can neglect that effect.

11. Jun 16, 2005

### michaelw

i honestly have no idea..
if the magnetic field is increasing, the charge will produce a magnetic field to oppose that increasing magnetic field (make a magnetic field in the opposite direction).. but i dont understand how velocity will be affected..

12. Jun 16, 2005

### OlderDan

You are on the right track. Faraday's Law says that there is a component of electric field tangent to every point on the circular path of the charge. Lenz's Law helps you figure out which direction the field is pointing, which leads to the statement you have made.

Suppose you had a whole circle full of charges moving in this magnetic field. Their combined charge and rate of movement consititute a current like in a single loop. What happens to the current in a loop when the magnetic flux through that loop is changing? Will the current from these charges increase or decrease? Will they speed up or slow down?

13. Jun 16, 2005

### Gokul43201

Staff Emeritus
This is not a valid comparison to make here. The path of the electrons are confined to the conducting loop in the case of the current. They are not, in the case of the free charge.

14. Jun 16, 2005

### OlderDan

I agree it is not an exact comparison. The radius of the charge trajectory is going to change. I think that has already been established in the discussion. I just think it is helpful in trying to understand the direction of the tangential electric field due to the time varying magnetic field to compare this situation to the case of a current loop. The OP still needs to decide whether these charges are going to speed up or slow down.

Last edited: Jun 16, 2005
15. Jun 17, 2005

### michaelw

argh so that means that they will change the magnitude of their velocity?
There is a question on the mcat im trying to solve
it talks about the aurora borealis/australis and how they are caused by electrons moving in a circle due to earths magnetic field

the period is how long it takes an electron to move around in a circular path once
the pitch is the difference in distance between one circle made and another

the magnetic field is positioned with N being at the left (in the drawing), and the magnetic fields are strongest at the poles, weakest at the equator

so basically the electron follows a spiral path going from N to S, because it has a componemt of velocity both parallel and perpendicular to the magnetic field

"Which of the following most accurately describes the change in pitch and period of the helical path travelled by a negatively charged particle moving through a magnetic field that is gradually strengthening?"
a) Pitch increases, period decreases
b) Pitch/period decreases
c) Pitch decreases, period remains the same
d) Pitch and period decreases

The equation to solve this is B = mv/qR
q and m must remain constant..

now the question becomes will velocity change, or radius, or both, and how

This is what I think..
if parallel velocity increases, pitch increases, period is same
if parallel velocity decreases, pitch decreases, period is same
if perpendicular velocity increases, period increases, pitch stays same
if perpendicular velocity decreases, period decreases, pitch stays same
but a change in radius can also affect pitch/period.. smaller radius means both will decrease

parallel velocity will not change, but how about perpendicular velocity? will it incraese or decrease? and what about the radius? will it decrease?

Last edited: Jun 17, 2005
16. Jun 17, 2005

### OlderDan

By Fardaday's Law, there will be an electric field induced by the time varying magnetic field that has a component aligned with the direction of motion of the charge. So yes, the magnitude of the velocity will change. Gokul43201's comment that this problem is not the same as a current carrying loop is entirely correct, but the same electric field that induces an emf in a coil affects the velocity of the charge in this problem.

There is another form of Faraday's Law that I assumed you have not yet seen. Perhaps I should not assume that. In differential form Faraday's Law does not rely on calculations of "flux". It is

$$\nabla \times E = - \frac{\partial B}{\partial t}$$

The left hand side of this equation is called the "curl" of the electric field. If we take the direction of the magnetic field in this problem to be the z direction, then its derivative with respect to time is also in the z direction and this equation reduces to

$$\frac{\partial E_y}{\partial x} - \frac{\partial Ex}{\partial y} = - \frac{\partial B}{\partial t}$$

If you have seen this before, great. If not, all you need to know about it is that it tells you that there is an electric field generated in the area of your charge because the magnetic field at every point in space is changeing with time. The equation shows that the field has "circulation", which means the Faraday path integral in the integral form of Faraday's Law is not zero. Work will be done on the charge by this electic field.

This is in contrast to your new problem, where you have a charge moving through a magnetic field that varies from one point in space to another, but is independent of time. The moving charge will experience different forces due to its motion through the magnetic field, but there will be no induced electric field that will do work on the charge. The magnetic force will always be perpendicular to the motion, so no work will be done on the charge.