1. 

Now that you have found out how much energy was released by the Solar flare, you will estimate how strong the magnetic field was inside the coronal loop. To do this you will use the following equation that relates the amount of magnetic energy contained within a region of space where a magnetic field is present:
E = B^{2}/(2μ_{0}) x V
where B is the magnetic field strength, E is energy, V is volume of the coronal loop, and μ_{0} = (4π x 10^{7} kg)(m)/Coulomb^{2} which is called the permeability of free space. Algebraically solve the equation for B. Show your work.
E = B^{2}/(2μ_{0}) x V
B^{2}/(2μ_{0}) = E/V
B^{2} = 2μ_{0}(E/V)
B = [2μ_{0}(E/V)]^{1/2} or B = SQRT[2μ_{0}(E/V)]

2. 

Now you need to estimate the volume of the coronal loop. To make the estimate simple, assume that the loop is a cylinder that has been bent into a semicircle. The volume of a cylinder is V = πR^{2}l, where R is the radius of the cylinder and l is its length. Because, you are assuming that the shape of a coronal loop is a semicircle, then the length of the cylinder is half of the circumference of a full circle. The diameter of the full circle would be the distance, d, between the footprints of the coronal loops. So the length of the cylinder is:
l = π/2 x d
Examine the 9th flare image in the series (23:12:50.50 UT) and find the distance, d, between the footprints in kilometers. You can do this by either measuring the distance with a ruler or using the coordinates of the footprints. Show all work and be mindful of units.
The coordinates of the footprints are: (840, 372) & (864, 348) arcseconds, so using the formula for the distance between two points gives
d = SQRT[(864 – 840)^{2} + (348372)2] = SQRT(1152 arcsec) = 34 arcsec
d = 34 arcsec x 725 km/arcsec = 2.5 x 10^{4} km
For students who measure the distance with a ruler they should find that the distance is approximately 2.5 cm. The scale of that image is 13.3 arcsec/cm.
2.5 cm x 13.3 arcsec/cm x 725 km/arcsec = 2.4 x 10^{4} km

3. 

Now examine the same image and use a ruler to measure the diameter of the coronal loop, use the contours as a guide. The radius of the cylinder, R, will be half of that diameter. Calculate the radius in kilometers. Show all work and be mindful of units.
Using the contours near the top footprint they should find that the width of the loop is about 1 cm.
1 cm x 13.3 arcsec/cm x 725 km/arcsec = 9.6 x 10^{3} km
R = ½ x 9.6 x 10^{3} km = 4.8 x 10^{3} km

4. 

Now calculate the volume, V, of the cylinder in cubic kilometers. Show all work and be mindful of units.
V = πR^{2}l and l = π/2 x d
V = πR^{2}(π/2 x d) = (π^{2}/2) x R^{2}d
V = (π^{2}/2) x (4.8 x 10^{3} km)^{2} x 2.5 x 10^{4} km = 2.8 x 10^{12} km^{3}

5. 

Return to your equation for the magnetic field strength in part 1. We are hoping to estimate the strength of the magnetic field that produced the solar flare and CME from this equation. If you use the amount of energy that the NASA scientists found as the total energy released by the flare in Worksheet 4.6 in this equation, then that will mean that all of the energy contained in the coronal loop would have been transformed from magnetic energy into heat, kinetic, and light energy. Since the loop still exists after the flare, clearly not all of the energy within it was used. So, using this method will actually calculate the minimum magnetic strength of the coronal loop. Set E equal to 2 x 10^{23} Joules and calculate B in units of Gauss. Show all work and be mindful of units. Note on units: 1 Gauss = 10^{4} Tesla, and 1 Tesla = 1 kg/((Coulomb)(seconds)).
B = [2μ_{0}(E/V)]^{1/2}
V = 2.8 x 10^{12} km3 x (10^{3} m/km)^{3} = 2.8 x 10^{18} m^{3}
E = 2 x10^{23} Joules = (2 x 10^{23} kg)(m^{2})/s^{2}
B = [(2 x 4π x 10^{7} kg)(m)/Coulomb^{2} x (2 x 10^{23} kg)(m^{2})/s2 /(2.8 x 10^{18} m^{3})]^{1/2}
B = 4.2 x 10^{1} kg/((Coulomb)(s)) = 4.2 x 10^{1} Tesla x 10^{4} Gauss/Tesla
B = 4.2 x 10^{3} Gauss
