Calculating accelerations in Induced Electric Fields

In summary: Good. The proton is in the region inside the magnetic field. Let's use the electric field equation to find the force on the proton: \begin{align*} F &= q(\vec vxB) \\ &= q(\vec vxB) - q(\vec vyB) \\ &= m*a*(\vec vx-\vec vy) \\ &= m*a*(\vec vx-\vec y) \\ &= -m*a*B \\ &= 0 \\ \end{align*}So the proton experiences a force of zero due to the magnetic field.
  • #1
Ignitia
21
5

Homework Statement


Over a region of radius R, there is a spatially uniform magnetic field B →. (See below.) At t =0, B=1.0T, after which it decreases at a constant rate to zero in 30 s.

(b) Assume that R=10.0cm. How much work is done by the electric field on a proton that is carried once clock wise around a circular path of radius 5.0 cm?

cff5f780a549fca32486e67d6d45f6095be630d0.jpg

Homework Equations


W = F*d
F=q(vxB)
F=m*a

d=2πr r=5cm

The Attempt at a Solution


Since velocity and field are not given, have to go with F=m*a, but I don't see how [a] can be calculated.

with W = F*d, then W=(m*a)*(2πr)

Is there something I'm missing?
 

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  • #2
Ignitia said:
Is there something I'm missing?
You are missing the fact that the Lorentz force (##q\vec v\times \vec B##) does no work on the proton because it is always perpendicular to ##\vec v##. Besides the problem asks you to find the work done by the electric field. Where do you think that comes from?
 
  • #3
kuruman said:
You are missing the fact that the Lorentz force (##q\vec v\times \vec B##) does no work on the proton because it is always perpendicular to ##\vec v##. Besides the problem asks you to find the work done by the electric field. Where do you think that comes from?

Okay, in part (a) I had calculated the Electric Field both inside and outside the magnetic field. (for reference, inside was was dB/dt * r/2 < answer checks out) If I'm reading this right, I have to take that equation, calculate the field at t = 0 and t = 30, find the difference, and multiply that by the charge of the proton?
 
  • #4
Ignitia said:
Okay, in part (a) I had calculated the Electric Field both inside and outside the magnetic field.
Good. The proton is in the region inside the magnetic field.
Ignitia said:
... I have to take that equation, calculate the field at t = 0 and t = 30, find the difference, and multiply that by the charge of the proton?
Is that consistent with the definition of work ##W = \int \vec F \cdot d\vec s## ? I think you should find the force on the proton (remember there is an electric field at its location) and then do the line integral.
 

FAQ: Calculating accelerations in Induced Electric Fields

What is an induced electric field?

An induced electric field is a type of electric field that is created when a conductor, such as a wire, is placed in a changing magnetic field. This changing magnetic field causes electrons in the conductor to move, creating an electric field.

How do you calculate the magnitude of an induced electric field?

The magnitude of an induced electric field can be calculated using the formula E = -NΔΦ/Δt, where E is the magnitude of the electric field, N is the number of turns in the conductor, and ΔΦ/Δt is the change in magnetic flux over time.

What is the direction of an induced electric field?

The direction of an induced electric field is determined by the right-hand rule. If you point your thumb in the direction of the changing magnetic field, the direction of your fingers will indicate the direction of the induced electric field.

How does the strength of the induced electric field change with distance?

The strength of the induced electric field decreases with distance from the conductor. This is because the magnetic field that is inducing the electric field also decreases with distance.

How can I use the concept of induced electric fields in practical applications?

The concept of induced electric fields is used in many practical applications, such as generators, transformers, and electric motors. It is also important in understanding electromagnetic induction and how electricity is generated and distributed in power grids.

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