In this article, we shall study to derive an expression for magnetic induction and magnetic potential at any point in a magnetic field created by a bar magnet.

Magnetic Induction at Any Point Due to a Short Bar Magnet:

Consider a short magnetic dipole NS.  Let  be the magnetic moment of the dipole

M = m x 2l ………………(1)

The direction of magnetic induction is along the axis from S-pole to N-pole inside the magnet.

Consider a point ‘P’ near the dipole at distance ‘r’ from its centre O. i.e. OP = r Let ‘ θ’ be the angle between the line joining the point from the centre O and the axis of the dipole (angle between OP and SN). Resolving magnetic moment into two mutually perpendicular components, we have,  the component M Cosθ along OP and M Sinθ perpendicular to OP.

Now, the point P lies on the axis of M Cosθ. Hence, the magnetic induction at, the axis point of M Cos θ is given by

Also, the given point P lies on the equatorial-line of component M Sin θ. Hence, the magnetic induction at the equatorial point of M Sin θ is given by

Let B₁ and B₂ be represented by sides PQ and PT of completed parallelogram PQRT. The diagonal PR represents the resultant magnetic induction in magnitude and direction.

This is the magnitude of the resultant induction B at point P.

Let ∝ be the angle made by the resultant B with the direction of OP

This is the angle made by B with OP.  Hence, the total inclination of the resultant induction B with the axis of the dipole is  ( θ + ∝ )

Special cases:

Case 1: If P is a point on the axis of the dipole, then θ = 0° or θ = 180° and Cos θ = ± 1

Case – 2: If P is a point on the equator of the dipole, then θ = 90° and Cos θ = 0

Magnetic Potential at Any Point Due to a Short Bar Magnet:

The magnetic potential at a point in a magnetic field is defined as the work done in moving unit north pole from infinity to that point. It is denoted by ‘V’ and its S.I. unit is J/Am or Wb/m.

In free space, the magnetic potential at a point due to the magnetic pole of strength ‘m’ units and at a distance, r is given by

Consider a short magnetic dipole NS.  Let M be the magnetic moment of the dipole

M = m x 2l ………………(1)

The direction of M is along the axis from S-pole to N-pole.

Consider a point ‘P’ near the dipole at distance ‘r’ from its centre O. i.e. OP = r. Let ‘θ’ be the angle between the line joining the point from the centre O and the axis of the dipole (angle between OP and SN).

Now the magnetic potential due to the North Pole of a magnetic dipole is given by

The magnetic potential due to the North Pole of a magnetic dipole is given by

Since the magnetic potential is a scalar quantity, the resultant potential at a point P is given by

Case 1: If P is a point on the axis of the dipole, then θ = 0° or θ = 180° and Cos θ = ± 1

Case – 2: If P is a point on the equator of the dipole, then θ = 90° and Cos θ = 0

Hence V_equator = 0

Science > Physics > Magnetism > Magnetic Induction and Magnetic Potential at any Point

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In this article, we shall study Biot-Savart’s Law and its application to find the magnetic induction at a point due to current carrying infinitely long conductor and the magnetic induction at the centre of a current-carrying circular coil.

Magnetic Induction:

The magnetic induction at any point in the magnetic field is defined as the magnetic flux passing through the unit area at that point. It is denoted by letter “B”. It is a vector quantity. Its S.I. unit is Wb/m² or tesla (T).

Mathematically, B = ∅ /A

Where B = Magnetic induction, ∅= Magnetic flux

A = Area through which magnetic flux is passing

Magnetic Induction Due to Current-Carrying Conductor:

Biot Savart’s law (Laplace’s Law):

The magnetic induction at a point near current-carrying conductor is directly proportional to (a) the current in the element (I), (b) the length of the element (dl), (c) the sine of the angle between the element and the straight line joining the element to the point, and inversely proportional to the square of the distance between the element and the point.

Explanation:

Consider a conductor of any shape in the form of wire. Let ‘i’ be the current through the conducting wire. Let ‘dl’ be a small element of the conductor, then the quantity i. dl is known as the current element of the conductor. Let P be the point at which magnetic induction due to current carrying conductor is to be found. Let be the position vector of point  P with respect to the current element i.dl. Let θ be the angle between dl and r, then by Biot Savart’s law the magnetic induction due to the small element dl is given by

dB ∝ i           …………….. (1)

dB ∝ dl               ……………(2)

dB ∝ sinθ        ……………(3)

dB ∝ 1/r²            ……………(4)

From equations (1), (2), (3) and (4)

This is a mathematical expression for the Biot-Savart’s law.

This is a mathematical expression for the Biot-Savart’s law in the vector form.

The total magnetic induction due to current in the whole conductor is given by

Magnetic Induction at a Point Near an Infinitely Long Straight Current-Carrying Conductor:

Let us consider infinitely long straight conductor carrying current i. Let P be the point at which magnetic induction due to the current-carrying conductor is to be found. Let us consider a small element dl of the conductor at O. Let vector AP or vector r be the position vector of point  P with respect to the current element. Let θ be the angle between the position vector of the point P and the current element. Let us draw PC perpendicular to the length of the conductor is drawn such that PC = R  and OC = l.  Now, angle COP = π – θ

Let dB be the magnetic induction at point ‘P’ due to the current-carrying conductor it is given by Biot Savart’s law.

This is an expression for magnetic induction at the point near the infinitely long straight conductor.

Magnetic Induction at Centre of Current-Carrying Circular Coil:

Consider a current-carrying coil having radius R and having a single turn. Let i be the current through this circular coil.  Let O be the centre of the circular coil at which magnetic induction is to be found.  Let dl be the small element of the current-carrying coil at P, let θ be the angle between the current element and line joining the current element with point O.  In this case, θ = 90⁰.

Magnetic induction at point O due to current element i. dl is given by Biot-Savart’s law

The total magnetic induction at “O” can be found by integrating both sides.

This is an expression for magnetic induction at the centre of the current-carrying circular coil.

Previous Topic: Magnetic Effect of Electric Current

Next Topic: Ampere’s Law

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In this article, we shall study magnetic induction at the point on the magnetic axis and magnetic equator of bar magnet.

Magnetic Induction at a Point on Axis of Bar Magnet:

The line passing through the poles of a bar magnet is called the axis of the magnet. Consider a bar magnet having pole strengths +m & -m and magnetic length to ‘2l’. The magnetic dipole moment vector is given by

M = m × 2l …….. (1)

Its direction is from the south pole to the north pole.

Consider point P on the axis of the magnet at a distance of ‘r’ from the centre of magnet O.

Consider the north pole. Magnetic induction at P due to the north pole is given by

The direction of magnetic induction is away from the north pole and along the axis of the magnet.

Consider the south pole. Magnetic induction at a point on the axis due to the south pole is given by

The direction of magnetic induction is towards the south pole along the axis of the magnet. Let B be the resultant magnetic induction at P

Then,    B = B₁ +  B₂ ………….(4)

This is an expression for magnetic induction at a point on the axis of a bar magnet.

For short bar magnet, l is very less than r. (l << r), hence l can be neglected. (i.e.  l = 0)

This is an expression for magnetic induction at a point on the axis of the short bar magnet.

Magnetic Induction at a Point on Equator of Bar Magnet:

The perpendicular bisector of the segment joining the north pole and south pole of a bar magnet is called equator of the magnet.

Consider a bar magnets having pole strength +m & -m & m.l. 2l the magnetic dipole movement vector is given by

M = m × 2l ………….. (1)

The direction of the magnetic dipole moment is from the south pole to north pole.

Let P be the point on the equator of a bar magnet at a distance of r from the centre of magnet O.

Consider the north pole. Magnetic induction at P due to the north pole is given by

Consider the south pole. Magnetic induction at P due to the south pole is given by

Resolving the magnetic induction B₁ & B₂ along the axis of the magnet and the along the equator of the magnet. The components B₁sinθ and B₂sinθ are equal & opposite hence cancel each other. The component B₁ cos θ and B₂ cos θ are in the same direction where they reinforce (support) each other. Let B be the resultant magnetic induction at P then

This is an expression for Magnetic induction at a point on the equator of the bar magnet.

For short bar magnet (l << r). l is small so can be neglected. (l = 0)

This is an expression for magnetic induction at a point on the equator of a short bar magnet. Its direction is from the north pole to south pole.

Numerical Problems:

Example – 01:

Find the magnetic induction at a point distant 10 cm on the axis of a short bar magnet of moment 0.2 Am².

Given: Distance from centre = r = 10 cm = 0.1 m, Magnetic moment = 0.2 Am². Point on axis.

To Find: B =?

Solution:

Ans: The magnetic induction at the point is 4 x 10^-5 T

Example – 02:

Find the magnetic induction at a point distant 20 cm on the equator of a short bar magnet of moment 5 Am².

Given: Distance from centre = r = 20 cm = 0.2 m, Magnetic moment = 5 Am²., The point on the equator.

To Find: Magnetic induction= B =?

Solution:

Ans: The magnetic induction at the point is 6.25 x 10^-5 T

Example – 03:

Find the magnetic induction at a point 0.5 m from either pole on the equator of a bar magnet of moment 5 Am².

Given: Distance from either pole = 0.5 m, Magnetic moment = 5 Am².

To Find: B =?

Solution:

As the point is equidistant from either pole it is on the equator of the bar magnet

∴ r² + l² = (0.5)² = 0.25

Ans: The magnetic induction at the point is 4 x 10^-6 T

Example – 04:

Find magnetic induction due to a short bar magnet at a point distant 10 cm a) on its axis and b) on its equator, given that the magnetic dipole moment of the magnet is 0.25 Am².

Given: Distance from centre = r = 10 cm = 0.1 m, Magnetic moment = 0.25 Am².

To Find: B_axis =? B_equator =?

Solution:

Ans: The magnetic induction at the point on the axis is 5 x 10^-5 T and

that on the equator is 2.5 x 10^-5 T

Example – 05:

A bar magnet has pole strength 10 Am and a magnetic length of 5 cm. Find B at equidistant point of 10 cm from either pole.

Given: Distance from either pole = 10 cm = 0.1 m, pole strength = m = 10 Am, magnetic length = 2l = 5 cm = 0.05 m

To Find: B =?

Solution:

Magnetic Moment = M = m . 2l = 10 x 0.05 = 0.5 Am²

As the point is equidistant from either pole it is on the equator of the bar magnet

∴ r² + l² = (0.1)² = 0.01

Ans: The magnetic induction at the point is 5 x 10^-6 T

Example – 06:

Find magnetic induction due to a short bar magnet at a point distant 20 cm a) on its axis and b) on its equator, given that the magnetic dipole moment of the magnet is 0.5 Am².

Given: Distance from centre = r = 20 cm = 0.2 m, Magnetic moment = 0.5 Am².

To Find: B_axis =? B_equator =?

Solution:

Ans: The magnetic induction at the point on axis is 1.25 x 10^-5 T and

that on equator is 6.25 x 10^-6 T

Example – 07:

Find the magnetic induction at a point distant 8 cm on the equator from the centre of a short bar magnet of moment 0.2 JT^-1.

Given: Distance from centre = r = 8 cm = 0.08 m, Magnetic moment = 0.2 JT^-1, Point on equator.

To Find: B =?

Solution:

Ans: The magnetic induction at the point is 3.91 x 10^-5 T

Example – 08:

Find the magnetic induction at a point distant 10 cm from the centre of the magnet on the axis of a short bar magnet of moment 0.24 JT^-1.

Given: Distance from centre = r = 10 cm = 0.1 m, Magnetic moment = 0.24 JT^-1. Point on axis.

To Find: B =?

Solution:

Ans: The magnetic induction at the point is 4.8 x 10^-5 T

Example – 09:

Find the magnetic induction due to a bar magnet of magnetic induction 0.5 Am² at a point on its axis at a distance of 15 cm from the nearest pole. The magnetic length of magnet is 10 cm.

Given: Magnetic moment = M = 0.5 Am², distance from nearest pole = 15 cm = 0.15 m, magnetic length = 2 l = 10 cm, l = 5 cm = 0.05 m, distance of point from centre = r = 0.15 + 0.05 = 0.20 m,  μ_o/4π = 10^-7 Wb/Am.

To Find: B_axis =?

Solution:

Ans: The magnetic induction at the point is 1.422 x 10^-5 T

Example – 10:

The magnetic induction at a point on the equator of a magnetic dipole at a distance of 10 cm from its centre is 5 x 10^-5 Wb/m². Calculate the magnetic moment of the magnet.

Given: Distance from centre = r = 10 cm = 0.1 m, Magnetic moment = B = 5 x 10^-5 Wb/m^2, Point on equator.

To Find: Magnetic moment = M =?

Solution:

Ans: The magnetic moment is 0.5 Am²

Example – 11:

The magnetic induction at a point on the axis at a distance 20 cm from the centre of the magnet is 1.5 x 10^-5 Wb/m². Find the magnetic induction at a point on the equator at the same distance from the centre.

Given: distance of point from centre of magnet = r = 20 cm for both the points. Magnetic induction = B_axis = 1.5 x 10^-5 Wb/m²,

To Find: B_equator =?

Solution:

Ans: Magnetic induction at a point on equator is 7.5 x 10^-6 Wb/m².

Example – 12:

The strength of magnetic field at a point on axis of a magnetic dipole at a distance of 10 cm from its centre is 4 x 10^-5 Wb/m². Calculate the magnetic moment of the magnet.

Given: Distance from centre = r = 10 cm = 0.1 m, Magnetic moment = B = 4 x 10^-5 Wb/m^2, Point on axis.

To Find: M =?

Solution:

Ans: The magnetic moment is 0.2 Am²

Example – 13:

The strength of magnetic field at point P on the axis of short bar magnet is equal to the magnetic induction at point Q on the equatorial line. Find the ratio of the distances from the centre of magnet.

Given: B_axis = B_equator,

To Find: ratio of distances r_P:r_Q = ?

Solution:

Ans: The required ratio of distances is 2^1/3: 1

Example – 14:

Earth’s magnetic field may be imagined to be due to a magnetic dipole located at the centre of the earth. If the magnetic field at a point on the magnetic equator is 3 x 10^-5 Wb/m². What is the magnetic moment of such a magnet? What is the value of earth’s magnetic field at the north pole of the earth? Radius of the earth = 6400 Km.

Given: B = 3 x 10^-5 Wb/m², Point on Equator, distance from centre = r = 6400 km = 6.4 x 10⁶ m

To Find: Magnetic moment = M =? Magnetic field at the north pole =?

Solution:

Ans: The magnetic moment of magnet is 7.86 x 10²² Am²

and magnetic induction at north pole is 6 x 10^-5 T.

Example – 15:

A magnet has magnetic length 0.10 m and pole strength 12 Am. Find the magnitude of the magnetic field B at a point on its axis at a distance of 0.02 m from centre.

Given: Magnetic length = 2l = 0.10 m, l = 0.05 m, pole strength = 12 Am, distance from the centre = r = 0.02 m, μ_o/4π = 10^-7 Wb/Am.

To Find: B_axis =?

Solution:

Ans: Magnetic induction on the axis is 3.4 x 10^-5 T.

Example – 16:

Two short magnets A and B of magnetic moments M₁ = 2.7 Am² and M₂ = 3.2 Am² respectively are kept as shown. Find the resultant magnetic field due to the magnets at P. r₁ = 30 cm and r₂ = 40 cm.

Given: M₁ = 2.7 Am², M₂ = 3.2 Am², r₁ = 30 cm = 0.3 m, and r₂ = 40 cm = 0.4 m.

Science > Physics > Magnetism > Magnetic Induction at a Point on the Axis and the Equator

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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 study the methods of magnetization, methods of demagnetization and Induced Magnetization. The process of converting iron or its alloys into a magnet is called magnetization.

Methods of Magnetization:

Single Touch Method:

Keep the bar which is to be magnetized on a wooden table. Take a strong permanent magnet and bring one of its poles (say north pole) to one end of the rod and gently rub the magnet on the rod from one end to the other. When the other end is reached, lift away the magnet from the rod and bring back to the starting end. Repeat the process several times.

The rod will get magnetized such that its starting end will act as the north pole and the other end as the south pole. If the south pole of the magnet is used for magnetization, then the rod will get magnetized such that its starting end will act as the south pole and the other end as the north pole.

Divided Touch Method:

Keep the bar which is to be magnetized on a top of two permanent magnets. Take another two strong permanent magnets and bring their opposite poles and touch them in the middle of the rod and gently rub the two magnets on the rod moving away from each other towards the end. When ends are reached lift away the magnets and bring back them again at the starting position. Repeat the process several times.

The rod will get magnetized such that the end where the south pole leaves the rod becomes north pole and the end where the north pole leaves the rod becomes south pole. The use of supporting magnets increases the strength of magnetization.

Double Touch Method:

Keep the bar which is to be magnetized on a top of two permanent magnets. Two permanent magnets separated by a piece of wood or cork are held together such that their opposite poles are together. This combination is placed on the rod at the centre and then moved to and fro without lifting it. The rod gets magnetized such that the opposite pole that to the nearest stroking magnet is created on its ends. The use of supporting magnets increases the strength of magnetization.

Electrical Method:

Wound an insulated copper wire around the rod forming a coil. Pass a strong direct electric current through the coil for some time. The rod gets magnetized. Such magnets are called electromagnets, The end of the rod at which current enters in anticlockwise direction becomes north pole and the other end of the rod becomes south pole.

Methods of Demagnetization:

The process used to destroy the magnetic properties of the material is called demagnetization. Demagnetization can be done

• by rough handling of the magnet.
• by continuously dropping it from a height or by hammering.
• by heating the magnet to high temperature and allowing it to cool, by keeping it in the east-west direction. This breaks the orderly alignment of molecules of the magnet.
• passing high frequency alternating electric current. This breaks the orderly alignment of molecules of the magnet.
• A single magnet left on its own loses magnetism due to induction between molecules.
• When a magnet is kept near another magnet of similar strength with their like poles facing each other, both get demagnetized due to induction in a few days.

Induced Magnetism:

When a permanent magnet is brought near a magnetic material like iron without touching it, the iron rod behaves like a magnet and attracts iron filings, clips towards itself. When the magnet is removed, the iron rod loses its magnetic property and the attracted iron filings and clips fall down. Thus magnetic material (iron rod) behaves like a magnet so long it is kept near the magnet.

The magnetism acquired by a magnetic material when it is kept near a magnet is called induced magnetism. The process in which a piece of magnetic material acquires the magnetic properties temporarily due to the presence of another magnet near it is called magnetic induction.

If we consider north pole of inducer magnet, then the end of the rod near to the north pole becomes south pole and the far end of the rod becomes north pole. Thus magnetic pole induces an opposite polarity on the near end and similar polarity on the farther end of the iron rod.

Characteristics of Induced Magnetism:

• Induced magnetism has no independent existence.
• Induced magnetism is temporary.
• Inducer magnetic pole induces an opposite polarity on the near end and similar polarity on the farther end of the iron rod.
• Induced magnetism precedes attraction.

Factors Affecting Magnetic Induction:

• The strength of the inducing magnet. More powerful magnets can produce a high induced magnetism in magnetic substance.
• Magnetic induction is inversely proportional to the distance between the inducing magnet and magnetic substance.
• It depends on the nature of the magnetic substance.

Notes:

• Induced magnetism in soft iron is temporary. Hence it is used for making temporary magnets.
• Induced magnetism in steel is permanent. Hence it is used for making permanent magnets.
• The degree of magnetization is steel is less than that in soft iron.
• Steel retains magnetic properties for a long time because its retentivity is high.
• soft iron retains magnetic properties for a short time because its retentivity is weak.

Induced magnetism precedes attraction:

When a permanent magnet is brought near a magnetic material like iron without touching it, the iron rod behaves like a magnet. Inducer magnetic pole induces an opposite polarity on the near end and similar polarity on the farther end of the iron rod. If we consider north pole of inducer magnet, then the end of the rod near to the north pole becomes south pole and the far end of the rod becomes north pole. Thus the pole of inducer magnet and pole produce on the rod are unlike, hence they attract each other. Thus induced magnetism precedes attraction.

Induced magnetism has no independent existence:

When a permanent magnet is brought near a magnetic material like iron rod without touching it, the iron rod behaves like a magnet. When the magnet is removed magnetism in the iron rod disappears. Thus the existence of induced magnetism in the rod is associated with the presence of an inducer magnet. Thus Induced magnetism has no independent existence.

Induced magnetism is temporary:

When a permanent magnet is brought near a magnetic material like iron rod without touching it, the iron rod behaves like a magnet. When the magnet is removed magnetism in the iron rod disappears. Thus the existence of induced magnetism in the rod is associated with the presence of an inducer magnet. Thus induced magnetism is temporary.

Experiment to Show That Induced Magnetism is Temporary:

Bring anyone pole of a bar magnet near small iron nails. A chain of nails gets formed. Because magnetism is induced in one of the nails near to the pole and it acts as a magnet and attracts remaining nails towards it. When the magnet is moved away (removed). The induced magnetism is lost and all the nails fall down. This shows that the induced magnetic is temporary.

Previous Topic: Properties of Magnet

Next Topic: Magnetic Field and Magnetic Lines of Force

Science > Physics > Magnetism > Magnetization and Demagnetization

The post Magnetization, Demagnetization, and Induced Magnetization appeared first on The Fact Factor.