Chapter 4: Q45P (page 1015)

Solar Magnetic Energy. Magnetic fields within a sunspot can be as strong as0.4T . (By comparison, the earth’s magnetic field is about 1/10,000 as strong.) Sunspots can be as large as 25,000 km in radius. The material in a sunspot has a density of about $3 \times 10^{- 4} kg / m^{3}$ Assume $μ$ for the sunspot material is $μ_{0}$ . If 100% of the magnetic-field energy stored in a sunspot could be used to eject the sunspot’s material away from the sun’s surface, at what speed would that material be ejected? Compare to the sun’s escape speed, which is about $6 \times 10^{5} kg / m^{3}$ .

Short Answer

Expert verified

The speed at which the material is ejected is $20.6 \times 10^{3} m / s$

Step by step solution

Concept of the energy stored due to the magnetic field

The energy stored due to the magnetic field converts to kinetic energyis given as $K = U_{B}$ The kinetic energy is role="math" localid="1664185693794" $\frac{1}{2} {mv}^{2}$ where the mass m is pV .While the energy storedrole="math" localid="1664185781642" $U_{B} = \frac{B^{2}}{2 μ_{0}} V$ So, we get the speed of escape byrole="math" localid="1664185914696" $\frac{1}{2} {(pV)}^{2} = \frac{B^{2}}{2 μ_{0}} V$ which can be written as $V = \sqrt{\frac{B^{2}}{2 {pμ}_{0}}}$

Calculate the speed at which the material is ejected

The magnetic field $B = 0.4 T$ , the radius is $r = 25000 k m$ and the density is p= $3 \times 10^{- 4} kg / m^{3}$ Substitute the values in the equation $V = \sqrt{\frac{B^{2}}{2 {pμ}_{0}}}$ we have,