In which direction does a free hanging magnet get stabilized? Explain.
(N/A) freely suspended magnet always comes to rest in the $North-South$ direction.
This happens because the Earth acts like a giant magnetic dipole.
The magnetic North pole of the Earth is located near the geographic South pole,and the magnetic South pole of the Earth is located near the geographic North pole.
When a magnet is suspended freely,its North pole is attracted by the Earth's magnetic South pole (near the geographic North),and its South pole is attracted by the Earth's magnetic North pole (near the geographic South).
Thus,the magnet aligns itself along the Earth's magnetic field lines,which run approximately in the $North-South$ direction.
This happens because the Earth acts like a giant magnetic dipole.
The magnetic North pole of the Earth is located near the geographic South pole,and the magnetic South pole of the Earth is located near the geographic North pole.
When a magnet is suspended freely,its North pole is attracted by the Earth's magnetic South pole (near the geographic North),and its South pole is attracted by the Earth's magnetic North pole (near the geographic South).
Thus,the magnet aligns itself along the Earth's magnetic field lines,which run approximately in the $North-South$ direction.
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Similar Questions
If $B_V$ 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
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:
Medium
View SolutionAt which place does the Earth's magnetism become horizontal?
Easy
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
View SolutionThe value of horizontal component of earth's magnetic field at a place is $0.35 \times 10^{-4} \,T$. If the angle of dip is $60^{\circ}$,the value of vertical component of earth's magnetic field is nearly ............. $\times 10^{-4} \,T$.
Easy
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