- #1
eagleswings
- 16
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[SOLVED] magnetic plus gravitation force
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.
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
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 BThe 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