Question

Magnetic fields within a sunspot can be as strong as 0.4T. (By comparison, the earth's magnetic field is about 1/10,000 as strong.) Large sunspots can be as much as 25,000km in radius. The material in a sunspot has a density of about 3×10^−4kg/m3.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? (Hint: Calcualte the kinetic energy the magnetic field could supply to 1m^3 of sunspot material.)

Answer 1

Answer:

The speed of ejection is $2.06\times 10^{4}\ m/s$

Solution:

As per the question:

Magnetic field density, B = 0.4 T

Density of the material in the sunspot, $\rho = 3\times 10^{4}\ kg/m^{3}$

Now,

To calculate the speed of ejection of the material, v:

The magnetic field energy density is given by:

$U_{B} = \frac{B^{2}}{2\mu_{o}}$

This energy density equals the kinetic energy supplied by the field.

Thus

$KE = U_{B}$

$\frac{1}{2}mv^{2} = \frac{B^{2}}{2\mu_{o}}$

where

m = mass of the sunspot in $1\ m^{3}$ = $3\times 10^{- 4}\ kg/m^{3}$

$v = \frac{B}{\sqrt{\mu_{o}m}}$

$v = \frac{0.4}{\sqrt{4\pi \times 10^{- 7}\times 3\times 10^{- 4}}} = 2.06\times 10^{4}\ m/s$