Kinetic Energy in B field (is it really conserved)

In summary, kinetic energy in a magnetic field refers to the work required to move a charged particle against the magnetic force. It is conserved as long as there are no external forces acting on the particle. The velocity of the particle affects its kinetic energy by changing the magnitude of the magnetic force, but the direction of the magnetic field does not affect its conservation. Kinetic energy cannot be converted into other forms in a magnetic field, as the magnetic force does not do any work on the particle.
  • #1
hms.tech
247
0
Here is a quote from one of the MIT notes on magnetism :

"Note that F(magnetic force) is perpendicular to v and B , and cannot change the particle’s speed v (and thus the kinetic energy). In other words, magnetic force cannot speed up or slow down a charged particle. "

How is this possible, it kind of conflicts with the idea of acceleration and "change" in velocity.

lets say the velocity of an electron is v in the direction of increasing x axis, while the direction of the B field is along the positive y axis. A force will act on the electron directed towards the negative z axis.

The initial velocity was in the x-y plane only but after the force acts on the charge, the "z" component of its velocity changes.

Would this not change the total Kinetic Energy possessed by the electron?
Clearly the velocity is not the same as before !
Thus the magnetic force HAS changed the velocity.

Can anyone help clarify this ?
 
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  • #2
hms.tech said:
Here is a quote from one of the MIT notes on magnetism :

"Note that F(magnetic force) is perpendicular to v and B , and cannot change the particle’s speed v (and thus the kinetic energy). In other words, magnetic force cannot speed up or slow down a charged particle. "

How is this possible, it kind of conflicts with the idea of acceleration and "change" in velocity.

lets say the velocity of an electron is v in the direction of increasing x axis, while the direction of the B field is along the positive y axis. A force will act on the electron directed towards the negative z axis.

The initial velocity was in the x-y plane only but after the force acts on the charge, the "z" component of its velocity changes.

Would this not change the total Kinetic Energy possessed by the electron?
Clearly the velocity is not the same as before !
Thus the magnetic force HAS changed the velocity.

Can anyone help clarify this ?

Kinetic energy ([itex]K=\frac{1}{2}mv^2[/itex] non-relativistically) depends on the speed of the particle. the direction of the velocity can change, but its magnitude, the speed of the particle cannot.
 
  • #3
gabbagabbahey said:
Kinetic energy ([itex]K=\frac{1}{2}mv^2[/itex] non-relativistically) depends on the speed of the particle. the direction of the velocity can change, but its magnitude, the speed of the particle cannot.

Speed, v , initially = √(x^2+y^2)

speed , v , final (after magnetic force acts on the electron) = √(x^2+y^2+z^2)

Note that the x and y components of velocity remain constant thus the only difference between the final speed and the initial speed is the "z" component of velocity.

As you said the the kinetic energy is same then :

x^2+y^2 = x^2 + y^2 + z^2

thus z=0

Which is clearly another contradiction with the statement that the kinetic energy is conserved.
 
  • #4
hms.tech said:
Note that the x and y components of velocity remain constant thus the only difference between the final speed and the initial speed is the "z" component of velocity.

No. As the z component increases, the x and/or y components decrease in a way that exactly compensates to keep the magnitude of the velocity constant.

It may help to thing about a simpler example. Consider say, a ball on a string being whirled round in a circle at constant speed. Here the string's force always points perpendicular to the ball's current velocity, with the result that the ball's speed is constant.
 
  • #5
hms.tech said:
Speed, v , initially = √(x^2+y^2)

speed , v , final (after magnetic force acts on the electron) = √(x^2+y^2+z^2)

Note that the x and y components of velocity remain constant thus the only difference between the final speed and the initial speed is the "z" component of velocity.

No, initially [itex]\mathbf{v}(t=0)=v_0\mathbf{e}_x[/itex] and the force is [itex]\mathbf{F}(t=0)=v_0|\mathbf{B}|\mathbf{e}_z[/itex].

An infinitesmal time later, the velocity is [itex]\mathbf{v}(t=0+dt) = v_0\mathbf{e}_x +\frac{v_0|\mathbf{B}|}{m}dt\mathbf{e}_z[/itex] and the force is [itex]\mathbf{F}(t=0 +dt)=v_0|\mathbf{B}|\mathbf{e}_z - \frac{v_0|\mathbf{B}|^2}{m}dt\mathbf{e}_y[/itex] (Notice the force has a negative y-component).

But we don't measure at infinitesimal time intervals, only at finite time intervals. To do that, we integrate the acceleration:

[tex]\mathbf{v}(t) = \int_0^{t} \frac{\mathbf{F}(t')}{m} dt' = \int_0^{t} \frac{q}{m}(\mathbf{v}(t') \times \mathbf{B}(t'))dt'[/tex]

The easiest way to see that the speed is constant is to look at [itex]v^2 = \mathbf{v} \cdot \mathbf{v}[/itex] (if the square of the speed is constant, then so is the speed):

[tex]\frac{d}{dt}v^2= \frac{d}{dt}\left[ \mathbf{v} \cdot \mathbf{v}\right] = 2\mathbf{v} \cdot \frac{d\mathbf{v}}{dt} = 2\mathbf{v} \cdot \left(\frac{q}{m} \mathbf{v} \times \mathbf{B} \right) = \frac{2q}{m}\mathbf{B} \cdot (\mathbf{v} \times \mathbf{v}) = 0 [/tex]

where the second last step comes from the triple product identity [itex]\mathbf{A} \cdot ( \mathbf{B} \times \mathbf{C}) = \mathbf{B} \cdot ( \mathbf{C} \times \mathbf{A}) = \mathbf{C} \cdot ( \mathbf{A} \times \mathbf{B})[/itex].

Evidently, [itex]v^2[/itex], and hence [itex]v[/itex] and [itex]K[/itex], is constant.
 
  • #6
I understood the method of Kinetic energy but i didn't really catch the first part of your post no.5

Would you please explain/define what the variables represent .
 
  • #7
I used [itex]v_0[/itex] to represent the initial speed of the particle, [itex]\mathbf{v}(t)[/itex] to represent the velocity vector of the particle at time [itex]t[/itex], [itex]\mathbf{B}[/itex] to represent the external magnetic field, and [itex]\mathbf{F}[/itex] to represent the force on the particle.
 
  • #8
Alright, but u missed one which is troubling me

e (subscript x or y or z)
 
  • #9
gabbagabbahey said:
No, initially [itex]\mathbf{v}(t=0)=v_0\mathbf{e}_x[/itex]

what does e(x) represent
 
  • #10
hms.tech said:
what does e(x) represent

[itex]\mathbf{e}_x[/itex] is just one of many common ways to write the unit vector in the positive [itex]x[/itex]-direction. Other common ways to write it include [itex]\mathbf{i}[/itex], [itex]\mathbf{\hat{x}}[/itex], [itex]\mathbf{\hat{e}}_1[/itex], and [itex]\hat{i}[/itex] .
 

1. How is kinetic energy defined in a magnetic field?

Kinetic energy refers to the energy an object possesses due to its motion. In a magnetic field, this energy is defined as the work required to move a charged particle against the magnetic force.

2. Is kinetic energy conserved in a magnetic field?

Yes, kinetic energy is conserved in a magnetic field as long as there is no work done by external forces. This is because the magnetic force does not do any work on the charged particle, only changing its direction of motion.

3. How does the velocity of a charged particle affect its kinetic energy in a magnetic field?

The velocity of a charged particle affects its kinetic energy in a magnetic field by changing the magnitude of the magnetic force acting on it. As velocity increases, the magnetic force also increases, resulting in a higher kinetic energy.

4. Does the direction of the magnetic field affect the conservation of kinetic energy?

No, the direction of the magnetic field does not affect the conservation of kinetic energy. As long as there are no external forces acting on the charged particle, its kinetic energy will remain constant regardless of the direction of the magnetic field.

5. Can kinetic energy be converted into other forms of energy in a magnetic field?

No, kinetic energy cannot be converted into other forms of energy in a magnetic field. As mentioned earlier, the magnetic force does not do any work on the charged particle, so its kinetic energy remains constant.

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