Assume the dipole model for Earth's magnetic field $B$,which is given by:
$B_v = \text{vertical component of magnetic field} = \frac{\mu_0}{4\pi} \frac{2m \cos \theta}{r^3}$
$B_H = \text{horizontal component of magnetic field} = \frac{\mu_0}{4\pi} \frac{m \sin \theta}{r^3}$
where $\theta = 90^{\circ} - \text{latitude}$ as measured from the magnetic equator.
$(a)$ Find the loci of points for which the dip angle is $\pm 45^{\circ}$.
$B_v = \text{vertical component of magnetic field} = \frac{\mu_0}{4\pi} \frac{2m \cos \theta}{r^3}$
$B_H = \text{horizontal component of magnetic field} = \frac{\mu_0}{4\pi} \frac{m \sin \theta}{r^3}$
where $\theta = 90^{\circ} - \text{latitude}$ as measured from the magnetic equator.
$(a)$ Find the loci of points for which the dip angle is $\pm 45^{\circ}$.
(A) The dip angle $\delta$ is defined by the relation $\tan \delta = \frac{B_v}{B_H}$.
Given the expressions for the vertical component $B_v$ and the horizontal component $B_H$:
$B_v = \frac{\mu_0}{4\pi} \frac{2m \cos \theta}{r^3}$
$B_H = \frac{\mu_0}{4\pi} \frac{m \sin \theta}{r^3}$
Substituting these into the expression for $\tan \delta$:
$\tan \delta = \frac{\frac{\mu_0}{4\pi} \frac{2m \cos \theta}{r^3}}{\frac{\mu_0}{4\pi} \frac{m \sin \theta}{r^3}} = \frac{2 \cos \theta}{\sin \theta} = 2 \cot \theta$
We are looking for the loci of points where the dip angle $\delta = \pm 45^{\circ}$.
Since $\tan(\pm 45^{\circ}) = \pm 1$,we have:
$\pm 1 = 2 \cot \theta$
$\cot \theta = \pm 0.5$
$\tan \theta = \pm 2$
Since $\theta = 90^{\circ} - \lambda$ (where $\lambda$ is the latitude),the locus of points is defined by the condition $\tan(90^{\circ} - \lambda) = \pm 2$,which simplifies to $\cot \lambda = \pm 2$ or $\tan \lambda = \pm 0.5$.
Given the expressions for the vertical component $B_v$ and the horizontal component $B_H$:
$B_v = \frac{\mu_0}{4\pi} \frac{2m \cos \theta}{r^3}$
$B_H = \frac{\mu_0}{4\pi} \frac{m \sin \theta}{r^3}$
Substituting these into the expression for $\tan \delta$:
$\tan \delta = \frac{\frac{\mu_0}{4\pi} \frac{2m \cos \theta}{r^3}}{\frac{\mu_0}{4\pi} \frac{m \sin \theta}{r^3}} = \frac{2 \cos \theta}{\sin \theta} = 2 \cot \theta$
We are looking for the loci of points where the dip angle $\delta = \pm 45^{\circ}$.
Since $\tan(\pm 45^{\circ}) = \pm 1$,we have:
$\pm 1 = 2 \cot \theta$
$\cot \theta = \pm 0.5$
$\tan \theta = \pm 2$
Since $\theta = 90^{\circ} - \lambda$ (where $\lambda$ is the latitude),the locus of points is defined by the condition $\tan(90^{\circ} - \lambda) = \pm 2$,which simplifies to $\cot \lambda = \pm 2$ or $\tan \lambda = \pm 0.5$.
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What is the most accepted scientific explanation for the origin of the Earth's magnetic field?
Easy
View SolutionAt two different places, the angles of dip are respectively $30^{\circ}$ and $45^{\circ}$. At these two places, the ratio of the horizontal component of Earth's magnetic field is:
Answer the following questions regarding earth's magnetism:
$(a)$ $A$ vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth's magnetic field.
$(b)$ The angle of dip at a location in southern India is about $18^o$. Would you expect a greater or smaller dip angle in Britain?
$(c)$ If you made a map of magnetic field lines at Melbourne in Australia,would the lines seem to go into the ground or come out of the ground?
$(d)$ In which direction would a compass free to move in the vertical plane point to,if located right on the geomagnetic north or south pole?
$(e)$ The earth's field,it is claimed,roughly approximates the field due to a dipole of magnetic moment $8 \times 10^{22} \, J \, T^{-1}$ located at its centre. Check the order of magnitude of this number in some way.
$(f)$ Geologists claim that besides the main magnetic $N-S$ poles,there are several local poles on the earth's surface oriented in different directions. How is such a thing possible at all?
$(a)$ $A$ vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth's magnetic field.
$(b)$ The angle of dip at a location in southern India is about $18^o$. Would you expect a greater or smaller dip angle in Britain?
$(c)$ If you made a map of magnetic field lines at Melbourne in Australia,would the lines seem to go into the ground or come out of the ground?
$(d)$ In which direction would a compass free to move in the vertical plane point to,if located right on the geomagnetic north or south pole?
$(e)$ The earth's field,it is claimed,roughly approximates the field due to a dipole of magnetic moment $8 \times 10^{22} \, J \, T^{-1}$ located at its centre. Check the order of magnitude of this number in some way.
$(f)$ Geologists claim that besides the main magnetic $N-S$ poles,there are several local poles on the earth's surface oriented in different directions. How is such a thing possible at all?
Medium
View SolutionIf $B_V$ and $B_H$ are respectively the vertical and horizontal components of the earth's magnetic field at a place where the angle of dip is $60^{\circ}$, then the total magnetic field at that place is
Assume the dipole model for Earth's magnetic field $B$,which is given by:
$B_{V} = \text{vertical component of magnetic field} = \frac{\mu_{0}}{4\pi} \frac{2m \cos \theta}{r^{3}}$
$B_{H} = \text{horizontal component of magnetic field} = \frac{\mu_{0}}{4\pi} \frac{m \sin \theta}{r^{3}}$
where $\theta = 90^{\circ} - \text{latitude}$ as measured from the magnetic equator.
$(a)$ Find the loci of points for which $|\vec{B}|$ is minimum.
$B_{V} = \text{vertical component of magnetic field} = \frac{\mu_{0}}{4\pi} \frac{2m \cos \theta}{r^{3}}$
$B_{H} = \text{horizontal component of magnetic field} = \frac{\mu_{0}}{4\pi} \frac{m \sin \theta}{r^{3}}$
where $\theta = 90^{\circ} - \text{latitude}$ as measured from the magnetic equator.
$(a)$ Find the loci of points for which $|\vec{B}|$ is minimum.
Difficult
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