Respuesta :
Answer:
The speed of ejection is [tex]2.06\times 10^{4}\ m/s[/tex]
Solution:
As per the question:
Magnetic field density, B = 0.4 T
Density of the material in the sunspot, [tex]\rho = 3\times 10^{4}\ kg/m^{3}[/tex]
Now,
To calculate the speed of ejection of the material, v:
The magnetic field energy density is given by:
[tex]U_{B} = \frac{B^{2}}{2\mu_{o}}[/tex]
This energy density equals the kinetic energy supplied by the field.
Thus
[tex]KE = U_{B}[/tex]
[tex]\frac{1}{2}mv^{2} = \frac{B^{2}}{2\mu_{o}}[/tex]
where
m = mass of the sunspot in [tex]1\ m^{3}[/tex] = [tex]3\times 10^{- 4}\ kg/m^{3}[/tex]
[tex]v = \frac{B}{\sqrt{\mu_{o}m}}[/tex]
[tex]v = \frac{0.4}{\sqrt{4\pi \times 10^{- 7}\times 3\times 10^{- 4}}} = 2.06\times 10^{4}\ m/s[/tex]
