Sun and Earth Magnetic Fields

Magnetic Fields of Sun and Earth

The magnetic moment of Earth is $8*10^{22} A/m^2$ which is a big value. Also the Magnetic moment of Sun is big, but the distance Earth-Sun is big and because of this on the Earh the Sun magnetic field value is small, some $5 nT$. Let us say that Earth’s and Sun’s magnetic fields are lined-up and at one moment (after 11 years on average) the magnetic moment of Sun flipps over. Please find the change in energy when the Earth Magnetic anti-aligned to the Sun magnetic field. If Earth axis is tilted at 23.5 degree please find the new change in Energy. Also please find the torque that the Sun field exerts on the Earth Magnetic moment.

Continue reading on magnetic dipoles.

Sun and Earth Magnetic Fields

(By Werner Heil (see “other version” below). [Public domain], via Wikimedia Commons)


The magnetic energy of a magnetic dipole – magnetic field system is defined as the scalar product between the two of them. Thus
$U =-m*B$
where $m$ and $B$ are vectors and $*$ is the scalar (or the dot) product.

Hence, when both the dipole and the external field are aligned the energy has its minimum value of:
$U0 = -8*10^{22}*5*10^{-9}*cos(0) = -4*10^{14} J$

When both the dipole and the field are opposing we write the equation as:
$U1 =-8*10^{22}*5*10^{-9}*cos(180) =+4*10^{14} J$
Thus, the difference between these two energies is:
$\Delta(U) = U1-U0 =2*4*10^{14} =+8*10^{14} J$
When the axis of the earth is tilted we have initial energy of:
$U0 =-8*10^{22}*5*10^{-9}*cos(23.5) =-3.668*10^{14} J$
that changes after flipping to a value of:
$U1 =-8*10^{22}*5*10^{-9}*cos(180+23.5) =+3.668*10^{14} J$

The difference in energies (when the axis is tilted) is
$\Delta(U) =U1-U0 =2*3.668*10^{14}=+7.3365*10^{14} J$
and the percentage difference when tilted is thus
$% = 7.3365*10^{14}/8*10^{14} =0.917 =91.7%$

The torque is defined as vector product between the magnetic moment and the external field. We write the vector product as a matrix. In absolute value have:
$|T| = |m \times B| = m*B*sin(23.5) =8*10^{22}*5*10^{-9}*sin(23.5) =1.595*10^{14} Nm$