What is the magnitude of the magnetic field?

+2 votes

The circuit in the figure below 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. The wire at the bottom has a mass of 10.0 g and is 5.00 cm long. The springs stretch 0.500 cm under the weight of the wire, and the circuit has a total resistance of 12.0 image. When a magnetic field is turned on, directed out of the page, the springs stretch an additional 0.380 cm. What is the magnitude of the magnetic field?
T

image

 

asked Feb 19, 2012 in Physics by Joey33 ~Expert~ (1,216 points)
    

1 Answer

+4 votes
 
Best answer

 

Let Δx1 be the elongation due to the weight of the wire and let Δx2
be the additional elongation of the springs when the magnetic field
is turned on.
And so
 
Magnetic Force  = 2kΔx2  where k is the force constant of the spring.
k = (mg) / (2Δx1 ) (*where m = mass, g = gravity = 9.8 m/s)
 
therefore, plugging k into magnetic force we get:
 
Magnetic Force  = 2((mg) / (2Δx1 )) Δx2 
 
Magnetic Force  = ((mg) Δx2 ) / (Δx1 )
 
The magnitude of magnetic force is equivalent to
 
Magnetic Force = current*length*magnetic field = I*L*B  (*where I=current, L=length, B=magnetic field)
 
Setting the two equations for magnetic force equal we can solve for the magnitude of the magnetic field:
 
 ((mg) Δx2 ) / (Δx1 ) = I*L*B
and so solving for B:
 
B =  ((mg) Δx2 )  /  (I*L*(Δx1 ))
 
we are given
 
m = 10 g = 10*10^-3 kg
 
g = 9.8 m/s
 
Δx1 = 0.5 cm = 0.5*10^-2 m
 
Δx2 = 0.38 cm = 0.38*10^-2 m
 
I = voltage / resistance = 24V /12.0 image = 2 A
 
L = 5 cm = 5*10^-2 m
 
plugging into the formula we get:
 
B =  ((10*10^-3 kg)(9.8 m/s)) (0.38*10^-2 m) )  /  ((2 A)*(5*10^-2 m)*(0.5*10^-2 m ))
 
B = 0.7448 T
answered Feb 25, 2012 by awesome ~Expert~ (1,479 points)
selected Feb 26, 2012 by Joey33

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