What Magnetic Field Strength Keeps an Electron on a Straight Path?

In summary, an electron accelerated through a potential difference of 1.7 kV and enters a gap between two parallel plates with a separation of 20.0 mm and a potential difference of 110 V. In order for the electron to travel in a straight line, a uniform magnetic field is needed. Using the equations F=qvB and F=qE, the velocity of the electron can be found by solving for qU=\frac{1}{2}mv^2. After finding the velocity, the equation vB=E can be used to solve for the magnetic field, where E is the electric field found by dividing the potential difference by the distance between the plates.
  • #1
22steve
12
0

Homework Statement



In the figure, an electron accelerated from rest through potential difference 1.7 kV enters the gap between two parallel plates having separation 20.0 mm and potential difference 110 V. The lower plate is at the lower potential. Neglect fringing and assume that the electron's velocity vector is perpendicular to the electric field vector between the plates. What uniform magnetic field allows the electron to travel in a straight line in the gap?


Homework Equations


F=qvB
q=charge
v=velocity
B=magnetic field




The Attempt at a Solution



I honestly don't know where to even begin, I know I've got to be missing a formula because distance has got to be used somewhere in there, I'm just interested in getting an idea how to start this problem.
 
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  • #2
If the electron travels in a straight line it doesn't accelerate upwards/downwards, which essentially means that the sum of all forces acting in the direction of the y-axis is zero. The magnetic force has to be equal, but opposite to that of the electric force.
 
  • #3
Ok, so that means that F=-F, and i found another formula qvB=(mv^2)/r, but I'm sure that I need to find the velocity first and then use that and substitute and since i have two equations i would be able to solve for F which would allow me to find B, but I'm not sure how to find the velocity by only knowing the potential differences.
 
  • #4
Forget qvB=(mv^2)/r. This equation describes an electron (or any other particle with a charge q) moving in a circular path, this is not true in your assignment.

If a particle with the charge q is accelerated with the voltage U it will gain kinetic energy according to

[tex]qU=\frac{1}{2}mv^2[/tex]

From this equation you can solve the speed of the electron.
 
  • #5
Ok so, I found the velocity to be 24,449 m/s and then by combining the equations qvB=F and F=qE I got the equation vB=E so 24449 m/s x B = E and E=V(volts) / d

so I plugged in numbers and 1.27e-10 but wrong... Is my math wrong?
 

1. What is a uniform magnetic field?

A uniform magnetic field is a region in space where the strength and direction of the magnetic field are constant. This means that the magnetic field lines are evenly spaced and parallel to each other.

2. How is a uniform magnetic field created?

A uniform magnetic field can be created by passing an electric current through a straight wire or by using a bar magnet. It can also be created using specialized equipment such as an electromagnet.

3. What are the properties of a uniform magnetic field?

A uniform magnetic field has the following properties:

  • Constant strength and direction
  • Uniformly spaced and parallel magnetic field lines
  • Produces a force on a charged particle moving through it
  • Can be shielded by certain materials, such as iron
  • Can be used to create a compass

4. What are some applications of a uniform magnetic field?

A uniform magnetic field has various applications, including:

  • Particle accelerators
  • Magnetic resonance imaging (MRI) machines
  • Cathode ray tubes in TVs and computer monitors
  • Magnetic levitation trains
  • Electric motors and generators

5. How can a uniform magnetic field be measured?

A uniform magnetic field can be measured using a magnetometer, which is a device that can detect and measure the strength and direction of a magnetic field. Other methods include using a compass or a hall effect sensor.

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