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 = 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