- #1

eagleswings

- 16

- 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