Finding Magnetic Field from a Box

Click For Summary
SUMMARY

The discussion focuses on calculating the magnetic field magnitude from a uniform magnetic field within a cubic volume of space measuring 50 mm on each side, with a stored magnetic energy of 19 J. The relevant equation used is the energy density formula, U = 1/2 * B^2 / μ, where U represents magnetic energy. The correct approach involves dividing the energy by the volume of the cube, calculated as (side length)^3, leading to a magnetic field magnitude of approximately 2.51E-5 T.

PREREQUISITES
  • Understanding of magnetic energy and its relation to magnetic fields
  • Familiarity with the formula for energy density in magnetic fields
  • Knowledge of cubic volume calculations
  • Basic grasp of magnetic permeability (μ)
NEXT STEPS
  • Review the derivation of the energy density formula for magnetic fields
  • Study the concept of magnetic permeability (μ) and its significance
  • Learn how to calculate the volume of geometric shapes, specifically cubes
  • Explore practical applications of magnetic fields in physics
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone involved in solving problems related to magnetic fields and energy calculations.

Shinwasha
Messages
22
Reaction score
0

Homework Statement

[/B]
A uniform magnetic field exists in a cubic volume of space with a
50-mm side length.
If the magnetic energy stored in this volume is
19J , what is the magnetic field magnitude?
(This is all the information I got. Been a bad week for since it just seem information for my homework is missing)

Homework Equations


This is where things are a little shaky. 0Al

UB=1/2*B^2/μ

The Attempt at a Solution


Taking the length I used it to find the area of the cube. Which is what I take A to be, which I find weird considering I needed l as well.
Setting it up I got

(19*2*4piE-17)/(0.015)(5E-4) = B^2

B = 2.51E-5 T
 
Physics news on Phys.org
I believe the equation you started with is the energy density, not just the Energy (check the units)

[itex]\frac{U}{Volume of cube} = \frac{1}{2} \frac{B^2}{\mu}[/itex]
 
actually, it looks like you did divide by the volume in your calculation, but the volume of the cube should just be (side length)^3.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 15 ·
Replies
15
Views
3K
Replies
4
Views
2K
Replies
13
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
3
Views
1K