In this article, we shall study the force acting on poles and torque action on bar magnet suspended in a uniform magnetic field.

Force Between Two Magnetic Poles (Inverse Square law):

If two poles of strengths m₁ and m₂ separated by a distance ‘r’ from each other in a vacuum, then the force between the two magnetic poles is given by

Problems on Inverse Square Law

Example – 01:

The force between two magnetic poles in the air is 9.604 mN. If one pole is ten times stronger than other, calculate the pole strength of each magnet. The distance between the poles of two magnets is 0.1 m.

Given: Force = F = 9.604 mN = 9.604 x 10^-3 N, distance between poles = r = 0.1 m, Relation between pole strength m₂ = 10 m₁, μ_o/4π = 10^-7 Wb/Am.

To Find: pole strengths = ?

Solution:

m₂ = 10 m₁ = 10 x 9.8 = 98 Am

Ans: The pole strengths of magnets are 9.8 Am and 98 Am.

Example – 02:

What is the force of repulsion between two magnetic poles of strengths 1.6 Am and 7.2 Am separated by a distance 0.06 m in a vacuum?

Given: pole strengths m₁ = 1.6 Am and m₂ = 7.2 Am, distance between poles = r = 0.06 m, μ_o/4π = 10^-7 Wb/Am.

To Find: The force between poles = F = ?

Solution:

Ans: Force Between Poles is 3.2 x 10^-4 N

Force Experienced by a Pole of Magnet in Uniform Magnetic Field:

If we try to break the bar magnet at the centre and separate the poles, new poles are formed at the broken ends. It means it is not possible for us to separate the poles and study them individually hence the magnetic circuit is studied in analogy with the electrical circuit. Analogous to the quantity of electrical field intensity we have similar quantity magnetic field intensity. It is also referred as magnetic induction. The electrical field intensity at a point is defined as the force experienced by a unit charge kept at that point.

By analogy, we can define magnetic induction at a point as the force experienced by a pole of unit strength kept at that point. Let the strength of the magnetic field be B.

Thus the N-pole is acted upon by a force of magnitude mB and the south pole is acted upon by a force of magnitude mB, in the opposite direction to that on the north pole.

Torque Acting on Bar Magnet in a Uniform Magnetic Field:

Let us consider a bar magnet of pole strength, ‘m’ and magnetic length ‘2l’ suspended in a magnetic field of induction B such that it is free to rotate about a transverse axis passing through its centre. Let θ be the angle between the axis of the bar magnet and the direction of the magnetic field.

The magnetic dipole moment of a bar magnet is given by

M = m . 2l …………… (1)

Now, each pole of the bar magnet is acted upon by a force whose magnitude is given by

F = mB …………… (2)

The force acting on the north pole is equal to the force acting on the south pole but they act in the opposite direction. Similarly, the lines of action of these two forces are different, Hence they form what is called a couple.

The magnitude of the couple or moment of force is given by

Torque ( τ ) = Force (F) . Perpendicular distance between the forces

∴ τ  =    F  . SP

∴ τ  =    F  . 2l sin θ          ……………….   (3)

From equations (2) and (3)

τ =    m B . 2l sin θ

∴ τ  =    (m . 2l ) . B  sin θ   ………………..(4)

From equations (1) and (4)

τ =    M B sin θ

This is an expression for the torque acting on a magnet kept in the magnetic field. The direction of the torque is perpendicular to the plane passing through .i.e. perpendicular to the plane of the paper.

Tangent Law:

Statement:

If a bar magnet is free to rotate about an axis at right angles to two mutually perpendicular uniform magnetic fields of inductions B₁ and B₂, then it comes to rest in a direction making an angle q with the direction of , such that

B₂ = B₁ . tan θ

Proof:

It is found that when a bar magnet is suspended in two cross magnetic fields, it comes to rest with its axis along the direction of the resultant of the two magnetic fields. Consider a bar magnet suspended in two cross magnetic fields B₁ and B₂ .

Under the action of these two magnetic fields the needle rotates through angle θ and come to rest along the resultant of these two magnetic fields.

In the equilibrium condition, the needle is acted upon by two torques. one due to the magnetic field of induction given by m B₂ .2l. cosθ and second due to the horizontal component of the earth given by m B₁ .2l. sin θ. Under the action of these two torques, the needle remains in equilibrium. Hence the two couples should be equal.

∴  m B₂ . 2l . cos θ  = m B₁  . 2l. sin θ

∴    B₂ . cos θ   = B₁  . sin θ

B₂ = B₁ . (sin θ / cos θ)

B₂ = B₁ . tan θ

This relation is called as the tangent law.

Numerical Problems on Torque Acting on Magnet:

Example – 03:

A torque of moment 25 x 10^-2 Nm acts on a magnet suspended in a uniform magnetic field of induction 0.5 Wb/m² when making an angle of 30° with the field. Find the magnetic dipole moment of the magnet.

Given: Torque = τ = 25 x 10^-2 Nm, Magnetic induction = B = 0.5 Wb/m², angle with field = θ = 30°.

To find: Magnetic moment = M = ?

Solution:

τ = MB sin θ

∴ M = τ/ Bsin θ

∴ M = 25 x 10^-2/ (0.5 x sin 30°) = 25 x 10^-2/ (0.5 x 0.5)

∴ M = 25 x 10^-2/ (0.25) = 1 Am²

Ans: The magnetic dipole moment of the magnet is 1 Am²

Example – 04:

A magnet of magnetic dipole moment 2 Am² is deflected through 30° from the direction of a magnetic field of induction 2 Wb/m². Find the magnitude of the torque or couple.

Given: Magnetic moment = M = 2 Am², Magnetic induction = B = 2 Wb/m², angle with field = θ = 30°.

To find: Torque = τ =?

Solution:

τ = MB sin θ

∴ τ = 2 x 2 x sin 30° = 2 x 2 x 0.5 = 2 Nm

Ans: The torque acting on the magnet is 2 Nm

Example – 05:

A bar magnet of dipole moment 7.5 Am² experiences a torque of 1.5 x 10^-4 Nm, when placed inclined at 30° in a uniform magnetic field. Find the magnetic induction of the field.

Given: Magnetic moment = M = 7.5 Am², , angle with field = θ = 30°, Torque = τ = 1.5 x 10^-4 Nm

To find: Magnetic induction = B =?

Solution:

τ = MB sin θ

∴ B = τ/ Msin θ

∴ B = 1.5 x 10^-4/ (7.5 x sin 30°) = 1.5 x 10^-4/ (7.5 x 0.5)

∴ B = 4 x 10^-5 Wb/m²

Ans: The magnetic induction is  is 4 x 10^-5 Wb/m²  or 4 x 10^-5 T

Example – 06:

A magnet of dipole moment 0.05 Am² is suspended so to move freely in a horizontal plane. Find the couple required to hold it at right angles to the Earth’s horizontal magnetic induction of 0.32 x 10^-4 Wb/m².

Given: Magnetic moment = M = 0.05 Am², Magnetic induction = B = 0.32 x 10^-4 Wb/m², angle with field = θ = 90°.

To find: Torque = τ =?

Solution:

τ = MB sin θ

∴ τ = 0.05 x 0.32 x 10^-4 x sin 90°

∴ τ = 0.05 x 0.32 x 10^-4 x 1 = 1.6 x 10^-6 Nm

Ans: The torque acting on magnet is 1.6 x 10^-6 Nm

Example – 07:

Calculate the dipole moment of magnet which when placed at right angles to the earth’s horizontal magnetic induction 2 x 10^-5 Wb/m² experiences a couple of 2 x 10^-5 Nm.

Given: Torque = τ = 2 x 10^-5 Nm, Magnetic induction = B = 2 x 10^-5  Wb/m², angle with field = θ = 90°.

To find: Magnetic moment = M = ?

Solution:

τ = MB sin θ

∴ M = τ/ Bsin θ

∴ M = 2 x 10^-5/ (2 x 10^-5 x sin 90°) = 2 x 10^-5/ (2 x 10^-5 x 1)

∴ M = 1 Am²

Ans: The magnetic moment of the magnet is 1 Am²

Example – 08:

A magnet of moment 0.6 Am² makes an angle of 30° with the magnetic meridian. Calculate the torque that tends to bring  the magnet back to the meridian given that the earth’s horizontal field is 3.2 x 10^-5 Wb/m².

Given: Magnetic moment = M = 0.6 Am², Magnetic induction = B = 3.2 x 10^-5 Wb/m², angle with field = θ = 90°.

To find: Torque = τ =?

Solution:

τ = MB sin θ

∴ τ = 0.6 x 3.2 x 10^-5 x sin 30°

∴ τ = 0.6 x 3.2 x 10^-5 x 0.5 = 9.6 x 10^-6 Nm

Ans: The torque required is 9.6 x 10^-6 Nm

Example – 09:

A torque of 5 x 10^-3  Nm is required to hold a freely suspended horizontal bar magnet with the axis at an angle of 60° to uniform horizontal field of induction 3 x 10^-3 Wb/m². Find the magnetic moment of the magnet.

Given: Torque = τ = 5 x 10^-3 Nm, Magnetic induction = B = 3 x 10^-3  Wb/m², angle with field = θ = 60°.

To find: Magnetic moment = M = ?

Solution:

τ = MB sin θ

∴ M = τ/ Bsin θ

∴ M = 5 x 10^-3/ (3 x 10^-3 x sin 60°) 

∴ M = 5 x 10^-3/ (3 x 10^-3 x 0.8660)

∴ M = 1.925 Am²

Ans: The magnetic moment of the magnet is 1.925 Am²

Example – 10:

A short bar magnet placed with its axis at 30° to a uniform magnetic field of 0.2 T experiences a torque of 0.06 Nm. Calculate the magnetic moment of the magnet and find out what orientation of the magnet corresponds to stable equilibrium in magnetic field.

Given: Torque = τ = 0.06Nm, Magnetic induction =0.2 T, angle with field = θ = 30°.

To find: Magnetic moment = M = ? θ = ? for stable equilibrium.

Solution:

τ = MB sin θ

∴ M = τ/ Bsin θ

∴ M = 0.06/ (0.2 x sin 30°) = 0.06/ (0.2 x 0.5)

∴ M = 0.6 Am²

For stable equilibrium torque acting on magnet is zero.

τ = MB sin θ

∴  0 = MB sin θ

∴  sin θ = 0

∴  θ = 0°

Ans: The magnetic moment of the magnet is 0.6 Am² and for stable equilibrium θ = 0°

i.e. the axis of magnet should be parallel to the magnetic field.

Example – 11:

A bar magnet of moment  0.9 JT^-1 experiences a torque of 0.063 Nm, when placed inclined at 30° in a uniform magnetic field. Find the magnetic induction of the field.

Given: Magnetic moment = M = 0.9 JT^-1, angle with field = θ = 30°, Torque = τ = 0.063 Nm

To find: Magnetic induction = B =?

Solution:

τ = MB sin θ

∴ B = τ/ Msin θ

∴ B = 0.063/ (0.9 x sin 30°) = 0.063/ (0.9 x 0.5)

∴ B = 0.14 T

Ans: The magnetic induction is  is 0.14 T.

Example – 12:

A magnet of dipole moment 6.4 Am² and length 5 x 10^-3 m makes an angle of 60° with the magnetic field of 0.4 T. calculate the torque acting on it. Also find the pole strength of the magnet.

Given: Magnetic moment = M = 6.4 Am², Magnetic induction = B = 0.4 T, angle with field = θ = 60°, magnetic length = 5 x 10^-3 m

To find: Torque = τ =? pole strength = m = ?

Solution:

τ = MB sin θ

∴ τ = 6.4 x 0.4x sin 60° = 6.4 x 0.4x 0.8860 = 2.2 Nm

M = m x magnetic length

∴  m = M/magnetic length = 6.4/5 x 10^-3 = 1.28 x 10³ Am

Ans: The torque required is 1.28 x 10³ Nm

Example – 13:

Two magnets of magnetic moments M and √3M are joined together to form a cross. The combination is suspended freely in a uniform magnetic field. In the equilibrium position the magnet of magnetic moment M makes an angle θ with the field. Determine θ.

Solution:

The magnet of magnetic moment M makes an angle θ with the field.

As the two magnets are forming cross, the magnet of magnetic moment √3M makes an angle (90° – θ) with the field.

Let B be the magnetic strength of the external field.

For equilibrium the torque acting on the two magnets must be equal.

τ₁ = τ₂

∴  MB sin θ =  √3MB sin (90° – θ)

∴  sin θ =  √3 cos θ

∴  tan θ =  √3

∴  θ =  60°

Ans: Hence the value of θ is  60°

Science > Physics > Magnetism > Torque Acting on Bar Magnet in Uniform Magnetic Field

The post Torque Acting on Bar Magnet in Uniform Magnetic Field appeared first on The Fact Factor.

In this article, we shall study the terminology of a bar magnet and the concept of the magnetic dipole moment of a magnet.

Bar Magnet:

A bar magnet is a rectangular parallelepiped body which exhibits magnetic properties. When a bar magnet is suspended in the air such that it is free to rotate about the transverse axis passing through its centre, then it is found that the bar magnet always aligns itself in the north-south direction. The end of the magnet which is pointing towards the geographical north is called north-seeking pole or simply north pole, while the end of the magnet pointing towards the geographical south is called south seeking pole or simply south pole. From the behaviour of a bar magnet, we can say that earth itself is behaving like a magnet. The magnetic north pole of the earth is at geographical south pole while the magnetic south pole of the earth is at the geographical north pole.

 A Magnetic Monopole Does Not Exist:

A bar magnet is said to have two poles located at the two ends of the magnet. If we try to break the bar magnet at the centre and separate the poles, new poles are formed at the broken ends. Thus two new magnets are formed each having two opposite poles at their ends. If the magnet is broken down into very small pieces further each piece will be a magnet with two poles. It means it is not possible for us to separate the poles. Thus we can say that a magnetic monopole does not exist.

Fictitious Poles:

It is not possible to separate the two poles (the south pole and the north pole) by breaking the magnet into two parts. Similarly, it is not possible to locate the position of the poles of the magnet. Hence a magnetic dipole is supposed to be made up of two fictitious or imaginary poles. The position of these fictitious poles is found using a compass needle. And it is found that they are not located exactly at the ends of the magnet but slightly inside.

• Geometric Length of magnet: The length of the edge parallel to the magnetic axis is called the geometric length of the bar magnet.
• Magnetic Length of Magnet: The distance between the poles of a bar magnet is called magnetic length.

Magnetic length of bar magnet × 1.2 = Geometric length of the bar magnet.

• Axis of Magnet: The line joining the poles of the bar magnet to called an axis of the magnet.

• Equator of Magnet: The perpendicular drawn to the magnetic axis through the centre of magnet is called the equator of the magnet.

Need of Analogy of Magnetic Circuit with Electric Circuit:

If we try to break the bar magnet at the centre and separate the poles, new poles are formed at the broken ends. It means it is not possible for us to separate the poles and study them individually hence the magnetic circuit is studied in analogy with the electrical circuit. Formulae & concepts are derived from an electrical circuit and by analogy formula for a magnetic circuit is written.

Magnetic Induction at a Point in Magnetic Field:

The magnetic intensity or magnetic induction at any point in the magnetic field is equal to the number of tubes of force passing through the unit area of a small surface element drawn at that point.

B = ∅ / A

Where B = magnetic intensity or intensity of the magnetic field

or magnetic induction.

∅ = Magnetic flux.  Its unit as weber (Wb)

A = Area through which magnetic flux pass.

Unit of Magnetic induction B is Wb/m² or tesla (T)

 Magnetic Dipole:

A magnetic dipole can be defined as two equal and opposite magnetic poles separated by a finite distance. A magnetic dipole consists of two equal and opposite magnetic charges having pole strength +m & -m separated by finite distance ‘2l’.

Magnetic Dipole Moment:

The magnetic dipole moment is defined as the product of the pole strength and the magnetic length of a magnet.

M = m × 2l

The magnetic dipole moment is a vector quantity. Its direction is from -m to + m i.e. from the south pole to the north pole. SI unit of Magnetic dipole moment is ampere. metre²  i.e. Am²

A current-carrying coil behaves like a magnet.

Magnetic dipole moment of current-carrying coil = NiA

Where N = Number of turns of coil

I= Current through the coil

A = Area of the coil

Numerical Problems on Bar Magnet:

Example – 01:

A bar magnet of geometric length 18 cm has pole strength 100 Am. Find the magnetic dipole moment of the bar magnet.

Given: Geometric length of bar magnet = 18 cm = 0.18 m, Pole strength = m = 100 Am

To Find: Magnetic dipole moment = M = ?

Solution:

Magnetic length = (5/6) x Geometric length = (5/6) x 0.18 = 0.15 cm

M = m x magnetic length

∴  M = 100 x 0.15 = 15Am²

Ans: The magnetic dipole moment of bar magnet is 15 Am².

Example – 02:

A bar magnet of magnetic length 0.1 m has pole strength of 10 Am. What is the dipole moment of this magnet?

Given: Magnetic length of bar magnet = 0.1 m, Pole strength = m = 10 Am

To Find: Magnetic dipole moment = M = ?

Solution:

M = m x magnetic length

∴  M = 10 x 0.1 = 1Am²

Ans: The magnetic dipole moment of bar magnet is 1 Am².

Example – 03:

A bar magnet of the magnetic moment 5 Am² has poles 0.2 m apart. Calculate the pole strength.

Given: Magnetic moment = 5 Am², magnetic length of bar magnet = 0.2 m,

To Find: pole strength = m = ?

Solution:

M = m x magnetic length

∴ m = M/ magnetic length = 5/0.2 = 25 Am

Ans: The pole strength of bar magnet is 25 Am.

Example – 04:

A bar magnet has pole strength 48 Am which are 0.25 m apart. What is the magnetic moment of the magnet?

Given: Magnetic length of bar magnet = 0.25 m, Pole strength = m = 48 Am

To Find: Magnetic dipole moment = M = ?

Solution:

M = m x magnetic length

∴  M = 48 x 0.25 = 12Am²

Ans: The magnetic dipole moment of bar magnet is 12 Am².

Problems on Magnetic Dipole Moment of Current-Carrying Coil:

Example – 05:

A circular coil of 20 turns and radius 10 cm has a magnetic dipole moment of 3.142 Am². What is the current flowing through the coil.

Given: Number of turns = n = 20, radius of coil = 10 cm = 0. 1 m, magnetic dipole moment = M = 3.142 Am².

To Find: Current through the coil = i = ?

Solution:

M = n I A

∴   i = M/nA = M/ (n x π r² )

∴   i = 3.142/ (20 x 3.142 x (0.1)² ) = 5 A

Ans: The current through the coil is 5 A.

Example – 06:

A circular coil of 50 turns and 10 cm radius carries a current of 2 A. Find the magnetic moment of the coil.

Given: Number of turns = n = 50, radius of coil = 10 cm = 0. 1 m, Current through coil = i = 2 A.

To Find: Magnetic dipole moment = M = ?

Solution:

M = n i A = n i (π r²)

∴  M =  50 x 2 x 3.142 x (0.1)² = 3.142 Am²

Ans: The magnetic dipole moment is 3.142 Am².

Example – 07:

A circular coil has 1000 turns each of area 2 m². If a current of 3 mA flows through the coil, find the magnetic moment of the coil.

Given: Number of turns = n = 1000, Area of coil = A = 2 m², Current through coil = i = 3 mA = 3 x 10^-3 A

To Find: Magnetic dipole moment = M = ?

Solution:

M = n i A

∴  M =  1000 x 3 x 10^-3 x 2 = 6Am²

Ans: The magnetic dipole moment is 6 Am².

Example – 08:

A rectangular coil of length 8 cm and breadth 5 cm has 200 turns of insulated wire. Find the magnetic dipole moment of the coil. When a current of 2 A flows through it.

Given: Number of turns = n = 200, Area of coil = A = 8 cm x 5 cm = 40 cm² = 40 x 10^-4 m², Current through coil = i = 2 A

To Find: Magnetic dipole moment = M = ?

Solution:

M = n i A

∴  M =  200 x 40 x 10^-4 x 2 = 1.6Am²

Ans: The magnetic dipole moment is 1.6 Am².

Example – 09:

A steel wire of length ‘l‘ has a magnetic moment M. If the wire is bent in a semicircular arc, what would its moment be?

Solution:

Let m be the pole strength

Now, M = m x magnetic length

∴  m = M/l

Curved length of semicircle = half circumference

∴   l = π r

∴ r = l /π

Now new magnetic dipole moment

M = m x new magnetic length

∴ M = (M/l)x 2r = (M/l)x 2 x (l /π) = 2M/π

Ans: New magnetic moment is 2M/π

Previous Subject: Types of Magnetic Materials

Science > Physics > Magnetism > Magnetic Dipole Moment of a Magnet

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In this article, we shall see the properties of magnet and experiments to verify them

Types of Material:

Magnetic Materials: Materials which are attracted by magnets are called magnetic material. e.g.  Iron, cobalt, nickel

Nonmagnetic Materials: Materials which are not attracted by magnets are called non-magnetic material. e.g. glass, plastic, rubber.

Experimental Verification of Properties of Magnet

Experiment to Find Magnetic Material:

Bring a magnet near the material to be categorized. If the material is getting attracted towards the magnet, then the material is categorized as magnetic material. If the material is not getting attracted towards the magnet, then the material is categorized as nonmagnetic material.

Experiment to Locate Poles of a Magnet:

Suspend a bar magnet with twistless thread to a wooden stand such that it is capable of rotating about a transverse axis passing through its centre. Thus the magnet is horizontal. Let the magnet comes to rest. When it comes to rest, the end pointing towards the geographical north is called the north pole and the end pointing towards the geographical south is called the south pole.

Experiment to Show That the Strength of Magnet is Located at the Poles:

Take some iron filings in a dish. Place a bar magnet in it. We observe that the iron filings stick to the magnet but cluster around the poles rather than the middle portion of the magnet. This shows that the strength of the magnet is located at the poles.

Experiment to Show That the Like Poles Repel and Unlike Poles Attract:

Suspend a bar magnet with twistless thread to a wooden stand such that it is capable of rotating about a transverse axis passing through its centre. Thus the magnet is horizontal. Let the magnet comes to rest. When it comes to rest, the end pointing towards the geographical north is called the north pole and the end pointing towards the geographical south is called the south pole.

Now bring north pole of another magnet near the north pole of the suspended magnet. We observe that the north pole of the suspended magnet moves away from the north pole of the other magnet. This phenomenon is called the magnetic repulsion.

Now take away the other magnet and allow the suspended magnet to come to rest. Now bring south pole of another magnet near the north pole of the suspended magnet. We observe that the north pole of the suspended magnet moves towards the south pole of the other magnet. This phenomenon is called magnetic attraction.

Thus we can conclude that like poles repel and unlike poles attract. In this experiment, if you interchange the poles, the result will be the same.

Note:

If an iron rod is suspended in case of the magnet, then in both the cases the rod will get attracted towards the other magnet. Hence attraction is not a sure test of magnetism.

Experiment to Show That the Poles of Magnet Cannot Be Separated:

Take a thin bar magnet which can be cut by scissors. Mark its north pole and south pole. Cut this magnet into two halves at the centre. Put these pieces in iron filings. Iron filings get attracted to both pieces at the ends.

Suspend these pieces with twistless thread to a wooden stand such that it is capable of rotating about a transverse axis passing through its centre. Thus the magnet is horizontal. We observe that both the pieces come to rest in the north-south direction.

This shows that no matter how small you cut the magnet, each piece will have both the north pole and south pole. Thus the two poles of a magnet cannot be separated from each other.

A horizontally suspended magnet always comes to rest in the north south direction:

The earth itself is a giant magnet. Its magnetic North pole is near geographical South pole. Its magnetic South pole is near geographical North pole.

Now, unlike poles of magnet always attract each other and like poles of magnet always repel each other. Thus the north pole of the suspended magnet gets attracted towards the magnetic south pole of the earth (geographical north). Similarly, the south pole of the suspended magnet gets attracted towards the magnetic north pole of the earth (geographical south).

Thus the magnet when suspended in the air such that it is free to rotate about a transverse axis passing through its centre, it always comes to rest in the north-south direction.

If a bar magnet is suspended vertically, it does not hang in the north-south direction:

When a bar magnet is suspended vertically it is acted upon by two forces. Magnetic force due to earth’s magnetic field and the gravitational force due to earth’s gravitational field.

The magnetic force tries to align the magnet in the north-south direction, while gravitational force tries to move the magnet downward. The gravitational force acting on the magnet is much stronger than the magnetic force acting on the magnet. Hence, a bar magnet when suspended vertically, it does not hang in the north-south direction.

Repulsion rather than attraction is the test for identifying magnet:

Unlike poles of the magnet always attract each other and like poles of the magnet always repel each other. Thus there are two phenomena involved: attraction and repulsion. If there is neither attraction nor repulsion due to bringing a magnet near the material, then the material is nonmagnetic. If it is getting attracted then it may be magnetic or a magnet.

If the material is brought near a north pole of a magnet and is getting attracted. Then there are two possibilities Firstly, the material is magnetic and itself is not a magnet and is getting attracted towards the magnet and secondly, the material is a magnet and its south pole is getting attracted towards the north pole of the magnet.

Now, if a material is brought near a north pole of a magnet and is getting repelled. Then it means that the material is a magnet and its north pole is brought near the north pole of the magnet. Thus, repulsion rather than attraction is the test for identifying magnet.

These properties of magnet are used in different practical applications.

Previous Topic: Magnets

Next Topic: Magnetization, Demagnetization, and Induced Magnetization

Science > Physics > Magnetism > Properties of Magnet

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