Magnetic plus gravitation force

Click For Summary
SUMMARY

The discussion centers on calculating the magnitude of the magnetic field in a circuit involving springs and a mass under the influence of gravity and magnetism. The circuit features a 24V battery and a 12-ohm resistance, with a 10-gram mass causing the springs to stretch. The user initially misapplied the gravitational constant instead of the acceleration due to gravity (9.8 m/s²), leading to a significant error in their calculations. The correct magnetic field strength is determined to be 0.588 Tesla after correcting the gravitational force used in the calculations.

PREREQUISITES
  • Understanding of Hooke's Law (F = -kx)
  • Basic knowledge of electric circuits (Ohm's Law: I = V/R)
  • Familiarity with magnetic force equations (F sub B = IL x B)
  • Knowledge of gravitational force (F = mg)
NEXT STEPS
  • Review the application of Hooke's Law in mechanical systems
  • Study the principles of electromagnetism, focusing on magnetic fields and forces
  • Learn about the differences between gravitational constants and acceleration due to gravity
  • Explore circuit analysis techniques involving resistors and power sources
USEFUL FOR

Physics students, electrical engineers, and anyone interested in understanding the interaction between magnetic and gravitational forces in mechanical systems.

eagleswings
Messages
15
Reaction score
1
[SOLVED] magnetic plus gravitation force

Homework Statement



A circuit consists of wires at the top and bottom and identical metal springs in the left and right sides. the upper portion of the circuit is fixed and has a 24 v battery and 12 ohm resistance. the wire at the bottom has a mass of 10 grams and is 5 cm long. the springs stretch 0.5 cm under the weight of the wire. when a magnetic field is turned on, directed out of the page, the springs stretch an additional .3 cm. what is the magnitude of the magnetic field.

Homework Equations

F = -kx, I = v/r = 2 Amps, F = ma = mg; F sub B = IL x B



The Attempt at a Solution


solving for K before the magnetic field is turned on: there are two upward forces from the springs, and one downward force from gravity and mass
-2kx = - mg
substitute -2 (-.005 m)(k) = (-.01kg)(6.673 x 10 -11)
k = 6.673 x 10 -12
springs are in equilibrium between 2 kx forces and mg force

now magnetic field is turned on and you have 2 -kx's and F sub B, where springs stretch another 0.3 cm or .003 meters
-2 (6.673 x 10 -12)(.003) = F sub b = IL x B = (2 A) (5 cm) x B
so B = (4.00 x 10 - 14)/(2)(.05) = 4.0 x 10 - 13 tesla

but the book answer is .588 Tesla - i am only off by a factor of a trillion or so

this should be easy but i am missing a big factor

 
Physics news on Phys.org
magnetic plus gravitation force Picture

i've attached a picture of the problem. i can't figure out how i could be so far off? what element of the problem am i missing?
 

Attachments

  • Slide1.JPG
    Slide1.JPG
    17.9 KB · Views: 755
Magnetic plus gravitation solved

well i was using the Big G gravitational constant instead of little g acceleration = 9.8 meters/sec
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
7
Views
1K
  • · Replies 31 ·
2
Replies
31
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K