At a certain location in Africa,a compass points $12^{\circ}$ West of the geographic North. The North tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points $60^{\circ}$ above the horizontal. The horizontal component of the earth's field is measured to be $0.16\, G$. The magnitude of the earth's field at the location will be
- A$0.23 \times 10^{-4}\, T$
- B$0.18 \times 10^{-4}\, T$
- C$0.32 \times 10^{-4}\, T$
- D$0.81 \times 10^{-4}\, T$
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Similar Questions
The dip angle in a vertical plane at an angle of $\cos^{-1} \left( \frac{1}{\sqrt{2}} \right)$ from the magnetic meridian is $60^\circ$. Find the actual dip angle at that place.
Medium
View Solution$A$ short bar magnet placed in a horizontal plane has its axis aligned along the north-south direction. Null points are found on the axis of the magnet at $20 \ cm$ from the centre of the magnet. The Earth's magnetic field at the place is $B$ and the angle of dip is $0^{\circ}$. If the total magnetic field on the normal bisector of the magnet at $20 \ cm$ from the centre of the magnet is $0.6 \ G$,then the magnitude of $B$ is: (in $G$)
The angle of dip at a certain place is $30^o$. If the horizontal component of the earth's magnetic field is $H$,the intensity of the total magnetic field is:
Medium
View SolutionAssume 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 $\pm 45^{\circ}$.
Medium
View SolutionAt a certain place,the horizontal component of the Earth's magnetic field is $B_0$ and the angle of dip is $45^o$. The total intensity of the Earth's magnetic field at that place will be:
Medium
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