Question

Magnetic fields within a sunspot can be as strong as 0.4 T. (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 \(3 \times 10^{-4} \, \mathrm{kg/m}^3\). Assume \(\mu\) for the sunspot material is \(\mu_0\). 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 \(6 \times 10^5 \, \mathrm{m/s}\). (\(Hint:\) Calculate the kinetic energy the magnetic field could supply to 1 m\(^3\) of sunspot material.)

Short answer

The ejection speed is approximately 20,600 m/s, much less than the sun's escape speed.

Step-by-step solution

Calculate the Magnetic Energy Density

The energy density (energy per unit volume) of a magnetic field is given by the formula:\[ u = \frac{B^2}{2\mu_0} \]where \( B = 0.4 \, \mathrm{T} \) is the magnetic field and \( \mu_0 = 4\pi \times 10^{-7} \, \mathrm{T\cdot m/A} \) is the permeability of free space. Substituting the values, we find:\[ u = \frac{(0.4)^2}{2 \times 4\pi \times 10^{-7}} = \frac{0.16}{8\pi \times 10^{-7}} = \frac{0.16}{2.5133 \times 10^{-6}} \approx 6.37 \times 10^4 \, \mathrm{J/m^3} \]

Determine Mass of Sunspot Material per Cubic Meter

Given the density of the sunspot material is \( 3 \times 10^{-4} \, \mathrm{kg/m^3} \), we know that each cubic meter of this material has a mass of \( 3 \times 10^{-4} \, \mathrm{kg} \).

Calculate the Kinetic Energy for 1 m^3 of Material

The magnetic energy per cubic meter \( u \) can be converted entirely into kinetic energy. Thus, the kinetic energy \( KE \) per cubic meter is:\[ KE = u = 6.37 \times 10^4 \, \mathrm{J/m^3} \]

Calculate Ejection Speed per 1 m^3

Kinetic energy is also given by the formula:\[ KE = \frac{1}{2} mv^2 \]where \( m = 3 \times 10^{-4} \, \mathrm{kg} \) is the mass of 1 m^3 of material and \( v \) is the ejection speed. Solving for \( v \), we have:\[ 6.37 \times 10^4 = \frac{1}{2} (3 \times 10^{-4}) v^2 \]\[ v^2 = \frac{6.37 \times 10^4 \times 2}{3 \times 10^{-4}} = \frac{1.274 \times 10^5}{3 \times 10^{-4}} \approx 4.247 \times 10^8 \]\[ v = \sqrt{4.247 \times 10^8} \approx 2.06 \times 10^4 \, \mathrm{m/s} \]

Compare with Sun's Escape Velocity

The ejection speed of the material, \( 2.06 \times 10^4 \, \mathrm{m/s} \), is significantly lower than the sun's escape speed of \( 6 \times 10^5 \, \mathrm{m/s} \).

Key concepts

Magnetic Energy Density

Magnetic energy density describes how much electromagnetic energy is stored within a certain volume of space due to a magnetic field. In our scenario, a sunspot on the sun has a very strong magnetic field, much stronger than Earth's, at 0.4 Tesla (T). To calculate the energy density, we use the formula: \[ u = \frac{B^2}{2\mu_0} \] where \( B \) is the magnetic field strength and \( \mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A} \) is the magnetic permeability of free space.

The solution shows us how to substitute the sunspot's magnetic field into the formula, resulting in an energy density of about \( 6.37 \times 10^4 \, \text{J/m}^3 \). This indicates a substantial amount of energy is crammed into each cubic meter of a sunspot just from its magnetic vibrations.

Kinetic Energy Calculation

Kinetic energy is typically associated with the energy of motion. In this context, we are examining how the stored magnetic energy within a sunspot can potentially be converted into kinetic energy. With a given magnetic energy density, we know this energy per cubic meter of sunspot material translates directly into kinetic energy if it were entirely used to eject the material.

For each cubic meter of sunspot material, possessing a density of \( 3 \times 10^{-4} \, \text{kg/m}^3 \), we equate the energy density to kinetic energy using the formula \( KE = u \), where \( u \) is the magnetic energy density previously determined to be \( 6.37 \times 10^4 \, \text{J/m}^3 \). This conversion provides us with a straightforward understanding that each cubic meter of sunspot material carries \( 6.37 \times 10^4 \) joules of kinetic energy when magnetically ejected.

Sun's Escape Speed

The sun's escape speed is the minimum speed a particle or object must reach to break free from the sun’s gravitational pull without further propulsion. Calculated from the sun's mass and radius, this speed is immensely high due to the sun's strong gravitational field, approximately \( 6 \times 10^5 \, \text{m/s} \).

This figure is crucial because it determines the difficulty for any material, like that in a sunspot, to leave the sun. The escape speed acts as a benchmark in the calculations, showing how fast materials need to be ejected if they are to overcome the sun’s gravity and move into space.

Ejection Velocity of Sunspot Material

The ejection velocity of sunspot material is determined by converting magnetic energy into kinetic energy. Given a mass of \( 3 \times 10^{-4} \, \text{kg} \) per cubic meter, we utilize the kinetic energy formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity we are trying to find.

Rearranging the formula and inserting the known values for kinetic energy and mass, we solve for \( v \), resulting in an ejection velocity of approximately \( 2.06 \times 10^4 \, \text{m/s} \).

Comparing this with the sun’s escape speed, about \( 6 \times 10^5 \, \text{m/s} \), it becomes clear that the ejection speed is much lower, incapable of overcoming the sun’s gravitational hold. This highlights the force required to actually propulse sunspot material away from the sun, far exceeding what the magnetic energy alone can achieve.