Given that:
$$ \nabla V_m = - \bigl( { {\partial}\over {\partial r} } { {m \cos \theta} \over {4 \pi r^2}} + { {1\over r} } { {\partial}\over {\partial \theta} } { { m\cos \theta}\over { 4 \pi r^2} } \bigr) $$it follows that:
$$ H_r = -{ {\partial}\over {\partial r} } { {m \cos \theta} \over {4 \pi r^2}} = { {2m\cos \theta } \over {4 \pi r^3}} = { {m\cos \theta } \over {2 \pi r^3} } $$and $$ H_{\theta}= { {1\over r} } { {\partial}\over {\partial \theta} } { { m\cos \theta}\over { 4 \pi r^2} } ={{ m\sin \theta} \over {4 \pi r^3}}. $$ Plug in the values for $m, \theta, r$:
$$ H_r = { {8 \times 10^{22} \cos(45)} \over {2 \pi \cdot (6 \times 10^6)^3} } [ {{Am^2}\over{m^3} } ] = 41.7 A/m $$$$ H_{\theta} = { {8 \times 10^{22} \sin(45)} \over {4 \pi \cdot (6 \times 10^6)^3} } [ {{Am^2}\over{m^3} } ] = 20.8 A/m $$Because $B = \mu_o H$, and $\mu_o = 4 \pi \times 10^{-7} [ {{kg m^2} \over {A^2s^2}}\cdot {1\over m} ]$, $B_r = \mu_o H_r = 52 \mu$T and $B_{\theta} = 26 \mu$T.
We can use NumPy functions, radians( ) to convert from degrees to radians (required for trigonometric functions), and sin( ) and cos( ).
In [1]:
# code to calculate H_r and H_theta
import numpy as np # import the numpy package
m=80e21 # magnetic moment of the dipole in Am^2
theta=np.radians(45) # 45 degrees converted to radians
r= 6e6 # approximately the radius of the Earth in meters
H_r=(m*np.cos(theta))/(2.*np.pi * r**3) # calculate H_r
H_theta = (m*np.sin(theta))/(4.*np.pi *r**3) # calculate H_an
print ('H_r= ','%7.1f'%(H_r), 'H_theta= ', '%7.1f'%(H_theta)) # print the results
And now for the second part:
In [2]:
mu_o=(4.*np.pi)*1e-7 # mu_o in Henry's per meter
B_r=mu_o*H_r # get the radial field in units of tesla
B_theta=mu_o*H_theta # same for tangential field
print ('B_r = ','%i'%(B_r*1e6), 'uT') # B_r in microtesla
print ('B_theta =','%i'%(B_theta*1e6),'uT') # B_theta in microtesla
Here is an interactive script:
In [3]:
def convert_units(B,m,H,units):
"""
This function converts input values of B, m, and H in
units of either cgs or SI to SI and cgs respectively.
"""
if units=="cgs": # cgs=> SI
B_out = B*1e-4 # convert gauss to tesla
m_out= m*1e-3 # convert emu to Am^2
H_out= H/(4*np.pi*1e-3) # convert oe to A/m
elif units=="SI": # SI=> cgs
B_out= B*1e4 # convert tesla to gauss
m_out=m*1e3 # convert Am^2 to emu
H_out=H*(4*np.pi*1e-3) # convert A/m to oe
else: # mistake
print ('mistake in your units, try again ')
return B_out,m_out,H_out # returns the output variables
# use these values to input from entry box
#B,m,H,units=3.5e5,2.78e-20,128.,'cgs' # define some values
B=float(input('B: '))
m=float(input('m: '))
H=float(input('H: '))
units=input('units: ')
B_SI,m_SI,H_SI = convert_units(B,m,H,units) # call the function
print ('B_SI: ',B_SI) # print out the converted values
print ('m_SI: ',m_SI)
print ('H_SI: ',H_SI)
B_cgs,m_cgs,H_cgs = convert_units(B_SI,m_SI,H_SI,'SI') # do the reverse
print ('B_cgs: ','%5.3e'%(B_cgs))
print ('m_cgs: ','%8.4e'%(m_cgs))
print ('H_cgs: ',H_cgs)
a) This problem boils down to finding the value for ${\bf m}$ in Equation 1.8 in Chapter 1 that would give rise to a radial field of 10 $\mu$T at a depth of 2890 km (radius of the Earth minus radius of the dipole source). Because $\theta$ is 0$^{\circ}$, we do not have to worry about the tangential component yet, only the radial field, which was given by:
$$ H_r = {1\over {2\pi}}{{m \cos\theta}\over {r^3}}. $$We were given the field strength in units of tesla, so we have to convert to $H$ by the relation $H=B/\mu_o $. Taking $\cos \theta$ as unity here, we solve for $m$ and get:
$$ {m}={{2\pi H_r r^3}} = {{2\pi B_r r^3}\over{\mu_o}} $$Plugging in the values, we get:
In [4]:
mu_o=(4.*np.pi)*1e-7 # mu_o in Henry's per meter
B_r=10e-6 # radial field in microtesla
r=2890*1e3 # radius converted to meters
m=2.*np.pi*B_r*r**3/mu_o
print (m)
print ('m = ', '%7.1f'%(m*1e-21) , ' ZAm^2') # m in ZAm^2
b) To compare 10 $\mu$T with the field produced by an axial dipole of 80 ZAm$^2$, we need the second part of Equation 1.8 in the text:
$$ H_{\theta} = {{m \sin \theta}\over{4\pi r^3}}. $$Realizing the $\theta$ is 90 minus the latitude, we have $\theta=30$ and $$ {H_r} ={ {80 \times 10^{21}\times \cos 30 }\over {2 \pi (6370 \times 10^3)^3 }} $$ and $$ {H_{\theta}} = { {80 \times 10^{21}\times \sin 30 }\over {4 \pi (6370 \times 10^3)^3 }} $$ which are 42.7 and 12.3 Am$^{-1}$ respectively. The total field would be $\sqrt{H_r^2+H_{\theta}^2}$ or 44.4 Am$^{-1}$ . To compare this with the value of 10$\mu$T, we must convert this to units of $B$, so multiplying 44.4 by $\mu_o$ we have 55.8 $\mu$T.
In [5]:
mu_o=(4.*np.pi)*1e-7 # mu_o in Henry's per meter
m_dipole=80e21 # axial dipole of the Earth's magnetic field in ZAm^2
theta=90.-60. # 60 degree latitude expressed as a colatitude
theta=np.radians(theta) # convert to radians
r_earth=6371*1e3 # radius of the Earth in meters
H_r= (m_dipole * np.cos(theta))/(2.*np.pi*r_earth**3)
H_theta= (m_dipole * np.sin(theta))/(4.*np.pi*r_earth**3)
print ('H_r = ','%7.1f'%(H_r) , ' A/m')
print ('H_theta = ','%7.1f'%(H_theta), ' A/m')
print ('total field =','%7.1f'%(np.sqrt(H_r**2+H_theta**2)), ' A/m')
print ('in uT, this is =','%7.1f'%(mu_o*np.sqrt(H_r**2+H_theta**2)*1e6))
Knowing that $B = \mu_o H$, work out the fundamental units of $\mu_o$ in SI units. Fundamental units are the units for time (seconds), distance (meters), mass (kilograms), electrical potential (volts), electrical current (amps) and so on. Units work just like other variables in algebra; you can multiply and divide them. So,
$B$ is in tesla which boil down to [kg A$^{-1}$ s $^{-2}$] in fundamental units
$H $ is in [A/m] which already is in fundamental units.
We know that $\mu_o = {B\over H}$, so the units of $\mu_o$ are the units for $B$ divided by the units for $H$:
i.e.,
$ {\mu_o [} {{ kg \cdot m} \over {A^2 s^2} }]. $