Explain the angle of dip.
(N/A) The angle of dip (or magnetic inclination) is the angle that the total magnetic field vector $\vec{B}_{E}$ of the Earth makes with the surface of the Earth (horizontal) at a given point.
If a magnetic needle is perfectly balanced about a horizontal axis so that it is free to rotate only in the vertical plane of the magnetic meridian,it will not remain horizontal but will tilt at an angle with the horizontal. This angle is known as the angle of dip $(I)$.
At the magnetic poles,the needle points vertically downward or upward,so the angle of dip is $90^{\circ}$. At the magnetic equator,the needle remains horizontal,so the angle of dip is $0^{\circ}$.
Figure $(b)$ shows the magnetic meridian plane at a point $P$ on the surface of the Earth,where $\vec{B}_{E}$ is the total magnetic field,$H_{E}$ is the horizontal component,and $Z_{E}$ is the vertical component of the Earth's magnetic field.
If a magnetic needle is perfectly balanced about a horizontal axis so that it is free to rotate only in the vertical plane of the magnetic meridian,it will not remain horizontal but will tilt at an angle with the horizontal. This angle is known as the angle of dip $(I)$.
At the magnetic poles,the needle points vertically downward or upward,so the angle of dip is $90^{\circ}$. At the magnetic equator,the needle remains horizontal,so the angle of dip is $0^{\circ}$.
Figure $(b)$ shows the magnetic meridian plane at a point $P$ on the surface of the Earth,where $\vec{B}_{E}$ is the total magnetic field,$H_{E}$ is the horizontal component,and $Z_{E}$ is the vertical component of the Earth's magnetic field.
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$A$ 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 direction of the null points is on the equatorial line of a bar magnet,when the north pole of the magnet is pointing
Easy
View SolutionThe relation between $Z_E, H_E$,and $B_E$ is . . . . . . .
If the magnetic declination at a place is known to be $15^{\circ} \, E$,and a compass needle points as shown in the figure,then the geographic north is represented by which numbered direction?
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 $|\vec{B}|$ 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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