The angle of dip at a place on the earth gives:

  • A
    The horizontal component of the earth's magnetic field
  • B
    The location of the geographic meridian
  • C
    The vertical component of the earth's field
  • D
    The direction of the earth's magnetic field

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Similar Questions

The horizontal component of the earth's magnetic field at any place is $0.36 \times 10^{-4} \; Wb/m^2$. If the angle of dip at that place is $60^{\circ}$,then the value of the vertical component of the earth's magnetic field will be ........ $\times 10^{-4} \; Wb/m^2$.

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?

$A$ magnetic compass needle oscillates $30$ times per minute at a place where the angle of dip is $45^{\circ}$,and $40$ times per minute where the angle of dip is $30^{\circ}$. If $B_1$ and $B_2$ are respectively the total magnetic field due to the earth at the two places,then the ratio $B_1/B_2$ is best given by:

$A$ small bar magnet $A$ oscillates in a horizontal plane with a period $T$ at a place where the angle of dip is $60^o$. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian,its period will be

At a certain place,the vertical component of the Earth's magnetic field is $\sqrt{3}$ times the horizontal component of the Earth's magnetic field. If a magnetic needle is suspended freely in the air,then it will incline:

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