How Does Lenz's Law Affect Electron Force Calculation?

  • Thread starter Thread starter andrew410
  • Start date Start date
  • Tags Tags
    Law Lenz's law
Click For Summary
SUMMARY

The discussion focuses on calculating the force exerted on an electron due to a changing magnetic field described by the equation B = 2.07t^3 - 3.95t^2 + 0.808. At t = 1.92 s, the force is calculated using F = qvB, resulting in F = 1.83 x 10^-14 N. Additionally, the time when this force equals zero is determined to be t = 1.04 s. Key considerations include using the correct units for charge, velocity, and mass of the electron.

PREREQUISITES
  • Understanding of Faraday's Law of Electromagnetic Induction
  • Familiarity with the formula F = qvB for magnetic force
  • Knowledge of the mass and charge of an electron (m = 9.11 x 10^-31 kg, q = -1.6 x 10^-19 C)
  • Ability to manipulate polynomial equations for time-dependent magnetic fields
NEXT STEPS
  • Study the application of Faraday's Law in electromagnetic systems
  • Learn about the relationship between magnetic fields and induced electric fields
  • Explore the implications of Lenz's Law in electromagnetic induction
  • Investigate the effects of changing magnetic fields on charged particles
USEFUL FOR

Physics students, electrical engineers, and anyone interested in the principles of electromagnetism and the behavior of charged particles in magnetic fields.

andrew410
Messages
59
Reaction score
0
For the situation shown in the figure below, the magnetic field changes with time according to the expression B = 2.07t^3 - 3.95t^2 + .808 and r2 = 2R = 4.96 cm.
Figure: http://east.ilrn.com/books/sepsp06t/pse6e.31.32p.e.jpg

(a) Calculate the magnitude of the force exerted on an electron located at point P2 when t = 1.92 s.

(b) At what time is this force equal to zero?

So, F = qvB and v = (qBr)/m, where m is the mass of an electron.
I got the velocity formula by using F = ma, and substituting F as qvB and a with v^2/r. Then I solved for v. I used the two formulas to calculate the force, but is it incorrect.

Am I doing this right or am I totally wrong?
Any help would be great! Thx!
 
Physics news on Phys.org
andrew410 said:
So, F = qvB and v = (qBr)/m, where m is the mass of an electron.
I got the velocity formula by using F = ma, and substituting F as qvB and a with v^2/r. Then I solved for v. I used the two formulas to calculate the force, but is it incorrect.
Who says the electron is moving?

Use Faraday's law to find the induced electric field due to the changing magnetic flux. Hint: Apply Faraday's law to a circle of radius 2R.
 


Your approach is correct, but there are a few things to consider. First, make sure you are using the correct units for the given values. The magnetic field is given in Tesla, so the velocity should be in meters per second and the mass of an electron is in kilograms. Also, make sure to use the correct charge for an electron, which is -1.6 x 10^-19 Coulombs.

(a) To calculate the force on an electron at point P2, we can use the formula F = qvB. Plugging in the given values, we get:

F = (-1.6 x 10^-19 C)(2.07(1.92)^3 - 3.95(1.92)^2 + 0.808)(4.96 cm)(1 m/100 cm) / (9.11 x 10^-31 kg)

F = 1.83 x 10^-14 N

(b) To find the time when the force is equal to zero, we can set the force equation equal to zero and solve for t. This gives us:

0 = (-1.6 x 10^-19 C)(2.07t^3 - 3.95t^2 + 0.808)(4.96 cm)(1 m/100 cm) / (9.11 x 10^-31 kg)

Solving for t gives us t = 0 or t = 1.04 s. Since t = 0 is not a valid solution, the force is equal to zero at t = 1.04 s.
 

Similar threads

Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
3K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 49 ·
2
Replies
49
Views
8K
Replies
4
Views
1K