Telangana TSBIE TS Inter 2nd Year Physics Study Material 8th Lesson Magnetism and Matter Textbook Questions and Answers.

## TS Inter 2nd Year Physics Study Material 8th Lesson Magnetism and Matter

Very Short Answer Type Questions

Question 1.

A magnetic dipole placed in a magnetic field experiences a net force. What can you say about the nature of the magnetic field?

Answer:

For a magnetic dipole placed in a magnetic field, some net force is experienced It implies that force on the two poles of dipole is not equal.

This will happen only when magnetic dipole is in non-uniform magnetic field.

Question 2.

There is no question in text book. [TS Mar. 19, 17, May 14]

Question 3.

What happens to compass needles at the Earth’s poles? [TS Mar. 19, 17, May 14]

Answer:

When a compass is taken to earth poles say north pole then south pole of compass will adhere to north pole. It will align it self along magnetic meridian line.

Similarly when it is taken to south pole then north pole of compass is attracted by south pole and it will align itself along magnetic meridian line.

Question 4.

What do you understand by the ‘magnetisation’ of a sample?

Answer:

Magnetisation (T) :

It is the ratio of magnetic moment per unit volume.

I = (\(\frac{M}{V}\)) where M = the magnetic moments and V = volume of the given material.

Magnetic intensity is a vector, dimensions L^{-1} A.

Unit : Ampere/metre : Am^{-1}

Question 5.

What is the magnetic moment associated with a solenoid?

Answer:

Magnetic moment associated with a solenoid (M) = nIA. Where

n = Number of turns in solenoid;

I = Current through it;

A = Area vector

Question 6.

What are the units of magnetic moment, magnetic induction and magnetic field? [AP Mar. 16. May 17, 16; TS Mar. 16]

Answer:

1. Magnetic moment m is a vector. Unit A-m², dimensions L^{-2} A.

2. Magnetic induction (B) and magnetic field (B) are used with same meaning. Magnetic induction B is a vector.

Unit: Tesla (T), Dimension : MT^{-2}A^{-1}.

Question 7.

Magnetic lines form continuous closed loops. Why? [AP Mar. ’19, ’16, May ’18; TS May ’18, Mar. ’17]

Answer:

In magnetism magnetic monopole (single pole) is not existing. The simple possible way is to take a magnetic dipole. So the path a free magnetic needle or compass starts from north pole and terminates at south pole forms a loop.

Hence magnetic field lines are always closed loops.

Question 8.

Define magnetic declination. [TS Mar. 18, May 18, 17, 16; AP Mar. 18, 14, May 17, 16]

Answer:

Magnetic declination (D) :

The magnetic meridian at a place makes some angle (D) with true geographic north and south direction.

The angle between true geographic north to the north shown by magnetic compass is called “magnetic declination or simply declinations (D).”

Question 9.

Define magnetic inclination or angle of dip. [AP Mar. ’17, ’15; TS Mar. ’15]

Answer:

Magnetic inclination or angle of dip (I) :

It is the angle of total magnetic field BE at a given place with the surface of earth.

(OR)

The angle between horizontal to earth’s surface and net magnetic field of earth BE at that point.

Question 10.

Classify the following materials with regard to magnetism: Manganese, Cobalt, Nickel, Bismuth, Oxygen, Copper. [AP Mar. 19. 18. 17, 16, 15; TS Mar. 16. 15]

Answer:

Manganese : Paramagnetic substance

Cobalt : Ferromagnetic substance

Nickel : Ferromagnetic substance

Bismuth : Diamagnetic substance

Oxygen : Paramagnetic substance

Copper : Diamagnetic substance

Question 11.

In the magnetic meridian of a certain place, the horizontal component of the earth’s magnetic field is 0.26 G and the dip angle is 60°. What Is the magnetic field of the earth at this location?

Answer:

Given HE = 0.26 G; Dip angle = 60

But Dip angle = \(\frac{H_E}{B_E}\) = cos θ ⇒ B_{E} = H_{E} cos θ

∴ Magnetic field of earth = 0.26 × cos 60° = 2 × 0.26 = 0.52 G

Question 12.

Define Magnetisation of a sample. What is its SI unit?

Answer:

Magnetisation (I) : It is the ratio of net magnetic moment per unit volume.

I = \(\frac{m_{net}}{V}\) where m_{net} = the vectorial sum of magnetic moments of atoms in bulk material and V is volume of the given material.

Magnetic intensity is a vector, dimensions L^{-1} A.

Unit: Ampere/metre : A m^{-1}.

Question 13.

Define Magnetic susceptibility. Mention its unit. [AP Mar. ’15]

Answer:

Magnetic susceptibility (χ) :

It is a measure for the response of magnetic materials to an external field.

It is a dimensionless quantity.

Short Answer Questions

Question 1.

What are Ferromagnetic materials? Give examples. What happens to a ferromagnetic material at Curie temperature?

Answer:

Ferromagnetism:

- These substances are strongly attracted by magnets.
- The susceptibility (χ) is +ve and very large.
- Individual atoms of these substances will spontaneously align in a common direction over a small volume called domain.
- Size of domain is nearly 1 mm³ or a domain may contain nearly 10
^{11}atoms. - In these substances, magnetic field lines are very crowded.
- Every ferromagnetic substance will transform into paramagnetic substance at a temperature called Curie Temperature (T
_{c}).

Ex: Manganese, Iron, Cobalt, Nickel etc.

Effect of temperature on Ferromagnetic substances :

When ferromagnetic substances are heated upto Curie temperature, they will be converted into paramagnetic substances.

Question 2.

Derive an expression for the axial field of a solenoid of radius “r”, containing “n” turns per unit length and carrying current “I”.

Answer:

The behaviour of a magnetic dipole and a current carrying solenoid are similar.

Let a solenoid of radius ‘a’ and length 2l contains n turns and a current ‘I’ is passed through it.

Magnetic moment of solenoid (M) = nlA.

Consider a circular element of thickness dx of solenoid at a distance x from its centre. Choose any point ‘P’ on the axis of solenoid at a distance ‘r’ from centre of the axis.

Magnetic field at point P

This is similar to magnetic field at any point on the axial line of magnetic dipole.

Question 3.

The force between two magnet poles separated by a distance ‘d’ in air is ‘F. At what distance between them does the force become doubled?

Answer:

Force between two magnetic poles F = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{m_1m_2}}{\mathrm{d^2}}\)

When separation between the poles is reduced by √2 times their force between them is doubled.

Question 4.

Compare the properties of para, dia and ferromagnetic substances. [TS & AP June ’15]

Answer:

Paramagnetic substances | Diamagnetic substances | Ferromagnetic substances |

1. Feebly attracted by magnets. | 1. Repelled by magnets. | 1. Strongly attract by magnets. |

2. Susceptibility is +ve and nearly equals to one. χ = i |
2. Susceptibility is -ve and less than one. χ < 1 |
2. Susceptibility is + ve and large. χ > > 1 |

3. In a magnetic filed they move from weak field to strong field. | 3. They move from strong field to weak field. | 3. They move from weak field to strong field. |

4. They have individual atomic magnetic moments but total magnetic moment is zero. Ex: Aluminium, sodium etc. |
4. Individual atomic magnetic moment is zero. Ex: Bismuth, copper, lead. |
4. They have individual atomic magnetic moments. These atoms will form domains. Magnetic moment of all atoms in adomain is in same direction. Ex: Iron, cobalt, nickel. |

Question 5.

Explain the elements of the Earth’s magnetic field and draw a sketch showing the relationship between the vertical component, horizontal component and angle of dip.

Answer:

Earth’s magnetism :

The magnetic field of earth is believed to arise due to electrical currents produced by convective motion of metallic fluids in outer core of earth. This effect is also known as the “dynamo effect”.

- The magnetic north pole of earth is at a latitude of 79.74° N and at a longitude of 71.8° W. It is some where in North Canada.
- The magnetic south pole of earth is at 79.74° S and 108.22° E in the Antarctica.

Magnetic declination (D) :

The angle between true geographic north to the mag-netic north shown by magnetic compass is called “magnetic declination or simply declination (D).”

Angle of dip or inclination (I):

The angle of dip is the angle of total magnetic field BE at a given place with the surface of earth.

At a given place horizontal component of earth’s magnetic field H_{E} = B_{E} cos I.

Vertical component of earth’s magnetic field Z_{E} = B_{E} sin I.

Tangent of dip tan I = \(\frac{Z_E}{H_E}\)

Question 6.

Define retentivity and coercivity. Draw the hysteresis curve for soft iron and steel. What do you infer from these curves?

Answer:

Hysteresis loop :

Magnetic hysteresis loop a graph between magnetic field (B) and magnetic intensity (H) of a ferromagnetic substance. It is as shown in figure.

i) When applied magnetic field B is gradually increased then magnetic intensity in the material will also gradually increases and reaches a saturation point a’. It indicates that all atomic magnets of the sample are parallel to applied field.

ii) When applied magnetic field is gradually decreased to zero still then some magnetic intensity will remain in the material.

Retentivity or Remanence :

The magnetic intensity (H) of a material at applied magnetic field B – 0 is called “retentivity”. In hysteresis loop value of H on +ve Y-axis i.e., at B = 0 gives retentivity.

iii) When applied magnetic field (B) is reversed then magnetic intensity of the sample gradually decreases and finally it is magnetised in opposite direction upto saturation say point’d’.

Coercivity :

The -ve value of magnetic field (B^applied (i.e., in opposite direction of mag netisation) at which the magnetic intensity (H) inside the sample is zero is called “coer-civity”.

In hysteresis diagram the-value of B on -ve X-axis gives coercivity.

iv) When direction of magnetic field is reversed and gradually increased again we can reach the point of saturation a’.

Area of hysteresis loop is large for ferromagnetic substances with high permeability value.

Question 7.

If B is the magnetic field produced at the centre of adrcular coil of one turn of length L carrying current I then what is the manetic field at the centre of the same coil which is made into 10 turns?

Answer:

One turn coil means it is a circular loop.

For 1^{st} Case:

Magnetic field at the centre of a loop B_{1} = \(\frac{\mu_0}{2} \frac{\mathrm{I}}{\mathrm{r_1}}\)

Given that length of the wire = L.

For 2^{nd} Case :

Given wire is made into a coil of 10 turns ⇒ n = 10

∴ 2πr_{2}10 = L ⇒ r_{2} = \(\frac{L}{2 \pi}.\frac{1}{10}\)

For a coil of n turns magnetic field at its centre B_{2} = \(\frac{\mu_0}{2} \frac{\mathrm{nI}}{\mathrm{r_2}}\)

When same length of wire is made into a coil of n turns then B_{2} at centre = n² times previous value B_{1}.

⇒ B_{2} = n² B_{1} for given current T.

Question 8.

If the number of turns of a solenoid is doubled, keeping the other factors constant, how does the magnetic held at the axis of the solenoid change?

Answer:

A solenoid will produce almost uniform magnetic field (B) along its axis.

Magnetic field along the axis of a solenoid B = µ_{0}nI.

Where n is number of turns per unit length.

In our case number of turns of a solenoid is doubled keeping others as constant i.e., length of solenoid L is not changed and permeability p0, and current T not changed. So new number of turns n_{2} = 2n_{1}.

∴ New magnetic field at the same given point B_{2} = µ_{0}n_{2}I.

But n_{2} = 2n_{1}; ∴ B_{2} = µ_{0}2n_{1}I = 2B_{1}

When number of turns of a solenoid is doubled then magnetic field at the given point on the axis of solenoid will also double.

Long Answer Questions

Question 1.

Derive an expression for the magnetic field at a point on the axis of a current carrying circular loop.

Answer:

Consider a circular loop of radius R’ carrying a steady current i. Consider any point P’ on the axis of the coil (say X – axis).

From Biot – Savart’s law magnetic field at

The magnetic field at ‘P’ makes some angle ‘θ’ with X – axis. So resolve \(\mathrm{d} \bar{B}\) into components \(\mathrm{d} \bar{B}_x\) and \(\mathrm{d} \bar{B}\)⊥. Sum of \(\mathrm{d} \bar{B}\)⊥ is zero. Because \(\mathrm{d} \bar{B}\)⊥ component by an element d/ is cancelled by another diametrically opposite component. From fig \(\mathrm{d} \bar{B}_x\) = dB. cos θ. Where

Total magnetic field due to all elements on the circular loop

Question 2.

Prove that a bar magnet and a solenoid produce similar fields. (IMP)

Answer:

Magnetic field lines suggest that the behaviour of a current-carrying solenoid and a bar magnet are similar.

When a bar magnet is cut into two parts it will behave like two weak bar magnets. Similarly when a solenoid is cut into two parts and current is circulated through them they will also act as two solenoids of weak magnetic properties. Analogy between solenoid and bar magnet.

Let a solenoid of radius ‘a’ and length 2l contains n turns and a current ‘I’ is passed through it.

Magnetic moment of solenoid M = nlA.

Consider a circular element of thickness dx of solenoid at a distance x from its centre. Choose any point ‘P’ on the axis of solenoid at a distance r from centre of the axis.

∴ Total magnetic field B is obtained by integrating dB with in the limits -1 to 1.

This value is similar to magnetic field at any point on the axial line of magnetic dipole. Thus, a bar magnet and a solenoid produce ! similar magnetic fields.

Question 3.

A small magnetic needle is set into oscillations in a magnetic field B. Obtain an expression for the time period of oscillation.

Answer:

Let a small compass of magnetic moment m and moment of inertia ‘I’ is placed in a uniform magnetic field ’B’.

Let the compass is set into oscillation in a horizontal plane.

Torque on the needle is τ = MB sin θ.

Where 0 angle between M and B.

At equilibrium the magnitude of deflecting torque and restoring torque are equal.

∴ Restoring torque τ = I\(\frac{\mathrm{d}^2 \theta}{\mathrm{dt}^2}\) = – MB sin θ.

– ve sign indicates that restoring torque is in opposite direction of deflecting torque.

θ is small; sin θ = θ.

Magnetic needle in a uniform field

From principles of angular.

Simple harmonic motion \(\frac{MB}{I}\) = ω²

⇒ ω = \(\sqrt{\frac{MB}{I}}\)

∴ Time period of oscillation of a magnetic needle placed in a magnetic field T = 2π\(\sqrt{\frac{I}{MB}}\)

and magnetic field at that point B = 4π²\(\frac{I}{MT^2}\).

Question 4.

A bar magnet, held horizontally, is set into angular oscillations in the Earth’s magnetic field. It has time periods T_{1} and T_{2} at two places, where the angles of dip are θ_{1} and θ_{2} respectively. Deduce an expression for the ratio of the resultant magnetic fields at the two places.

Answer:

Let a bar magnet is held horizontally at a given place where earth’s magnetic field is B_{B}. When it is set into vibration in a horizontal plane it will oscillate with a time period T = 2π\(\sqrt{\frac{I}{MH_E}}\) ……….. (1)

Where H_{E} is horizontal component of earth’s magnetic field.

The relation between resultant magnetic field B_{E}, horizontal magnetic field H_{E} and angle of dip or inclination ‘I’ of earth’s magnetic field is H_{E} = B_{E} cos I. — (2)

Given at place ‘1’ angle of dip = θ_{1}

Given at place ‘2’ angle of dip = θ_{2}.

Question 5.

Define magnetic susceptibility of a material. Name two elements one having positive susceptibility and other having negative susceptibility. [AP Mar. ’15]

Answer:

Susceptibility χ :

The ratio of magnetisation of a sample (I) to the magnetic intensity (H) is called “susceptibility”.

Susceptibility χ = \(\frac{I}{H}\)

It is a dimensionless quantity.

It is a measure of how a magnetic material responds to external magnetic field.

For ferromagnetic materials susceptibility is +ve and χ >> 1 and χ is nearly in the order of 1000.

Ex: Iron, Cobalt, Nickel.

For paramagnetic substances χ is positive and nearly equals to one (χ ≅ 1).

Ex : Calcium, Aluminium, Platinum.

For diamagnetic substances χ is small and negative, magnetisation M and magnetic intensity H are in opposite direction.

Ex : Bismuth, Copper, Mercury, Gold.

Question 6.

Obtain Gauss’ Law for magnetism and explain it.

Answer:

Gauss Law in magnetism: Gauss law states that the net magnetic flux through any closed loop is zero.

\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{ds}}=0\)

Explanation :

Consider Gaussian surface I & II as shown in figure. In both cases the number of magnetic field lines entering the surface is equal to number of magnetic field lines leaving the surface.

The net magnetic flux is zero for both surfaces. This law is true for any surface of any shape. Consider an irregular Gaussian surface as shown in figure. Consider a smdl vector area element ∆S of closed surface ‘S’ placed in a Φ magnetic field B. Flux through ∆s is say ∆Φ.

When applied magnetic field B is gradually increased then magnetic intensity in the material will also gradually increases and reaches a saturation point ‘a’. It indicates that all atomic magnets of the sample are parallel to applied field.

Now flux (i.e., Number of magnetic field lines through unit area) through the element ∆S is given by ∆Φ_{B} = B . ∆S.

Let us divided the total Gaussian surface ‘S’ into number of small areas ∆S_{1}, ∆S_{2}, ∆S_{3}, ………… ∆S_{n}.

The total flux through the Gaussian surface

Where all’ term includes surface area of all surface elements ∆S_{1}, ∆S_{2}, ………… ∆S_{n}.

Magnetic monopole is not existing so there is no source or sinks of B’ in Gaussian surface. The simplest possible source is magnetic dipole i.e., bar magnet. Magnetic field lines of a bar magnet are closed curves of loops. So all the lines entering the Gaussian surface must leave from it.

Hence net magnetic flux through a closed surface is zero.

Question 7.

What do you understand by “hysteresis”? How does this property influence the choice of materials used in different appliances where electromagnets are used?

Answer:

Hysteresis loop :

Magnetic hysteresis loop is a graph between magnetic field (B) and magnetic intensity (H) of a ferromagnetic substance. It is as shown in figure.

i) When applied magnetic field B is gradually increased then magnetic intensity in the material will also gradually increases and reaches a saturation point ‘a’. It indicates that all atomic magnets of the sample are parallel to applied field.

ii) When applied magnetic field is gradually decreased to zero still then some magnetic intensity will remain in the material.

Retentivity or Remanence :

The magnetic intensity (H) of a material at applied magnetic field B = 0 is called “retentivity”. In hysteresis loop value of H on +ve Y-axis i.e., at B = 0 gives retentivity.

iii) When applied magnetic field (B) is reversed then magnetic intensity of the sample gradually decreases and finally it is magnetised in opposite direction upto saturation say point d’.

Coercivity :

The -ve value of magnetic field (B) applied (i.e., in opposite direction of magnetisation) at which the magnetic inten-sity (H) inside the sample is zero is called “coercivity”.

In hysteresis diagram the value of B on -ve X-axis gives coercivity.

iv) When direction of magnetic field is reversed and gradually increased again we can reach the point of saturation a’. Area of hysteresis loop is large for ferromagnetic substances with high permeability value.

Application of hysteresis in electromagnets. Hysteresis curve allows us to select suitable materials for permanent magnets and for electromagnets.

For materials to use as permanent magnets they must have high retentivity and high coercivity.

For electromagnets ferromagnetic substances of high permeability and low retentivity are used because when current is switched off it must loose magnetic properties quickly.

In case of transformers and telephone diaphragms they are subjected to prolonged AC cycles. For these applications the hysteresis curve of ferromagnetic materials must be narrow.

In this way hysteresis loop helps us to select magnetic materials for various applications.

Problems

Question 1.

What is the torque acting on a plane coil of “n” turns carrying a current “i” and having an area A, when placed in a constant magnetic field B?

Solution:

Number of turns = n ; Current = i;

Area of coil = A; Magnetic field = B.

Torque on a current carrying coil in a magnetic field τ = ni (\(\overline{\mathrm{A}}\times\overline{\mathrm{B}}\)) = niAB sin θ.

Question 2.

Acoilof 20 turns has an area of 800 mm² and carries a current of 0.5 A. If it is placed in a magnetic field of intensity 0.3 T with its plane parallel to the field, what is the torque that it experiences?

Solution:

Number of turns n = 200; Current i = 0.5 A;

Area A = 800 mm² = 800 × 10^{-6} m² (∵ 1 mm² = 10^{-6} m²) ;

Magnetic field B = 0.3 T.

Area of coil parallel to the field ⇒ Angle between area vector \(\overline{\mathrm{A}}\) and magnetic filed \(\overline{\mathrm{B}}\) =90°. Since area vector \(\overline{\mathrm{A}}\) is perpendicular to area of the coil.

∴ τ = niBA = 200 × 0.5 × 0.3 × 800 × 10^{-6}

= 3 × 800 × 10^{-6} = 2.4 × 10^{-3} N-m.

Question 3.

In the Bohr atom model the electrons move around the nucleus in circular orbits. Obtain an expression for the magnetic moment (µ) of the electron in a Hydrogen atom in terms of its angular momentum L.

Solution:

Charge of electron = e ;

Angular momentum = L;

Mass of electron = m_{e};

Magnetic moment of electron = µ = ?

When electron revolves in orbit current i = \(\frac{e}{T}\) ; Where time period T = 2π \(\frac{r}{υ}\)

∴ Current i = \(\frac{\mathrm{ev}}{2 \pi \mathrm{r}}\)

Magnetic moment of electron in orbit µ = iA

But m_{e} υr = L

∴ Magnetic moment of electron in orbit

µ = \(\frac{e}{2m_e}\)L

Question 4.

A solenoid of length 22.5 cm has a total of 900 turns and carries a current of 0.8 A. What is the magnetising field H near the centre and far away from the ends of the solenoid?

Solution:

Length of solenoid l = 22.5 cm = 22.5 × 10^{-2}m

Number of turns N = 900; Current I = 0.8 A.

a) Magnetising field near the centre H = ?

The behaviour of a solenoid is equal to that of a bar magnet.

Magnetic field due to a solenoid B = µ_{0}nI.

Magnetising field H = \(\frac{B}{\mu_0}\) = nI.

Where n = Nil.

∴ H = \(\frac{900}{22.5 \times 10^-2}\) × 0.8 = 3200 Am^{-1}.

A solenoid will give a uniform magnetic field along its axis.

Magnetising field far away from ends = 3200 Am^{-1}.

Question 5.

A bar magnet of length 0.1 m and with a magnetic moment of 5 Am² is placed in a uniform magnetic field of intensity 0.4 T, with its axis making an angle of 60° with the field. What is the torque on the magnet? [Mar. ’14]

Solution:

Length of bar magnet l = 0.1 m.;

Magnetic moment m = 5 Am².

Magnetic field B = 0.4 T ;

Angle with field θ = 60°.

Torque on the magnet τ = mB sin θ.

= 5 × 0.4 × sin 60°

= 5 × 0.4 × 0.8660 = 1.732 N-m.

Question 6.

If the Earth’s magnetic field at the equator is about 4 × 10^{-5} T, what is its approximate magnetic dipole moment? (radius ofi Earth = 6.4 × 10^{6} m)

Solution:

Radius of earth r = 6.4 × 10^{6} m ; Magnetic field near equator B = 4 × 10^{-5} T

Magnetic dipolemoment of earth = M.

Question 7.

The horizontal component of the earth’s magnetic field at a certain place is 2.6 × 10^{-5} T and the angle of dip is 60°. What is the magnetic field of the earth at this location?

Solution:

Horizontal component of earth’s magnetic field HE = 2.6 × 10^{-5} T

Angle of dip ‘I’ = 60°.

Earth’s magnetic field B_{E} = ?

Angle between B_{E} and H_{E} is called dip angle ‘I’.

H_{E} = B_{E} = cos θ ⇒ B_{E} = H_{E}/cos θ = 2.6 × 10^{-5}/ cos 60°

∴ B_{E} = 2.6 × 10^{-5}/ (0.5) = 5.2 × 10^{-5}T.

Question 8.

A solenoid, of insulated wire, is wound on a core with relative permeability 400. If the number of turns per metre is 1000 and the solenoid carries a current of 2A, calculate H, B and the magnetisation M.

Solution:

Relative permeability µ_{r} = 400 ;

Current I = 2A;

Number of turns / metre = n = 1000.

Magnetising force H = nl = 1000 × 2 = 2000 Am^{-1}.

Magnetic field along axis of solenoid B = ?

When solenoid is on a magnetic material B

= µ_{r} nl = 400 × 1000 × 2 = 8 × 10^{5} Am^{-1}

Intext Questions and Answer

Question 1.

A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10^{-2} J. What is the magnitude of magnetic moment of the magnet?

Solution:

Magnetic field strength, B = 0.25 T; Torque on the bar magnet, τ = 4.5 × 10^{-2} J

Angle between the bar magnet and the external magnetic field, θ = 30°

Torque is related to magnetic moment (M) as : τ = MB sin θ

Hence, the magnetic moment of the magnet is 0.36 J T^{-1}.

Question 2.

A short bar magnet of magnetic moment m = 0.32 J T^{-1} is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?

Solution:

Moment of the bar magnet, M = 0.32 J T^{-1 };

External magnetic field, B = 0.15 T

a) The bar magnet is aligned along the magnetic field. This system is considered as being instable equilibrium. Hence, the angle 0, between the bar magnet and the magnetic field is 0°.

Potential energy of the system

= -MB cos θ

= – 0.32 × 0.15 cos 0° = – 4.8 × 10^{-2} J

b) The bar magnet is oriented 180° to the magnetic field. Hence, it is in unstable equilibrium.

Potential energy = – MB cos θ ;

where θ = 180°

= -0.32 × 0.15 cos 180° = 4.8 × 10^{-2} J.

Question 3.

A closely wound solenoid of800 turns and area of cross-section 2.5 × 10^{-4} m² carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

Solution:

Number of turns in the solenoid, n = 800 ;

Area of cross-section, A = 2.5 × 10^{-4} m²

Current in the solenoid, I = 3.0 A

A current-carrying solenoid behaves as a bar magnet because a magnetic field develops along its axis, i.e., along its length. The magnetic moment associated with the given current-carrying solenoid is calculated as :

M = nIA = 800 × 3 × 2.5 × 10^{-4} = 0.6 J T^{-1}.

Question 4.

If the solenoid in exercise 8.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?

Solution:

Magnetic field strength, B = 0.25 T ;

Magnetic moment, M = 0.6 T^{-1}

The angle θ, between the axis of the solenoid and the direction of the applied field is 30°.

Therefore, the torque acting on the solenoid is given as : τ = MB sin θ

τ = 0.6 × 0.25 sin 30° = 7.5 × 10^{-2} J.

Question 5.

A bar magnet of magnetic moment 1.5 JT^{-1} lies aligned with the direction of a uniform magnetic field of 0.22 T.

a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment : (i) normal to die field direction, (ii) opposite to the field direction?

b) What is the torque on the magnet in cases (i) and (ii)?

Solution:

a) Magnetic moment, M = 1.5 J T^{-1} ;

Magnetic field strength, B = 0.22 T

i) Initial angle between the axis and the magnetic field, θ_{1} = 0°.

Final angle between the axis and the magnetic field, θ_{2} = 90°.

The work required to make the magnetic moment normal to the direction of magnetic field is given as :

W = -MB (cos θ_{2} – cos θ_{1})

∴ B = -1.5 × 0.22 (cos 90° – cos 0°)

= -0.33 (0 – 1) = 0.33 J.

ii) Initial angle between the axis and the magnetic field, θ_{1} = 0°.

Final angle between the axis and the magnetic field, θ_{2} = 180°.

The work required to make the magnetic moment opposite to the direction of magnetic field is given as :

W = – MB (cos θ_{2} – cos θ_{1})

∴ W = – 1.5 × 0.22 (cos 180° – cos 0°)

= -0.33 (-1 – 1) = 0.66 J.

b) For case CD : θ = θ_{2} = 90° then

Torque, τ = MB sin θ τ = 1.5 × 0.22 sin 90°

= 0.33 J

For case (ii): θ = θ_{2} = 180° then

Torque, τ = MB sin θ τ = MB sin 180° = 0 J

Question 6.

A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10^{-4} m², carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.

a) What is the magnetic moment associated with the solenoid?

b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10^{-2} T is set up at an angle of 30° with the axis of the solenoid?

Solution:

Number of turns on the solenoid, n = 2000,

Area of cross-section of the solenoid, A = 1.6 × 10^{-4} m²;

Current in the solenoid, I = 4 A

a) The magnetic moment along the axis of the solenoid is calculated as :

M = nAI = 2000 × 1.6 × 10^{-4} × 4 = 1.28 Am²

b) Magnetic field, B = 7.5 × 10^{-2} T

Angle between the magnetic field and the axis of the solenoid, 0 = 30°

Torque, τ = MB sin θ

∴ τ = 1.28 × 7.5 × 10^{-2} sin 30°

= 4.8 × 10^{-2} Nm.

Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is 4.8 × 10^{-2} Nm.

Question 7.

A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 8.0 × 10^{-2}T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s^{-1}. What is the moment of inertia of the coil about its axis of rotation?

Solution:

Number of turns in the circular coil, N = 16;

Radius of the coil, r = 10 cm = 0.1 m

Cross-section of the coil, A = πr² = π × (0.1)² m² ;

Current in the coil, I = 0.75 A

Magnetic field strength, B = 5.0 × 10^{-2} T ;

Frequency of oscillations of the coil, v = 2.0 s^{-1}

∴ Magnetic moment, M = NIA = Nlπr²

= 16 × 0.75 × π × (0.1)²

∴ M = 0.377 J T^{-1}

Hence, the moment of inertia of the coil about its axis of rotation is 1.19 × 10^{-4} kg m².

Question 8.

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.

Solution:

Horizontal component of earth’s magnetic field, B_{H} = 0.35 G

Angle made by the needle with the horizontal plane = Angle of dip = δ = 22°

Relation between B and B_{H} is B_{H} = B cos δ

Hence, the strength of earth’s magnetic field at the given location is 0.377 G.

Question 9.

At a certain location in Africa, a compass points 12° 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° above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location.

Solution:

Angle of declination, θ = 12°;

Angle of dip, δ = 60°

Horizontal component of earth’s magnetic field, B_{H} = 0.16 G

Earth’s magnetic field at the given location = B

Relation between B and BH is B_{H} = B cos θ

Earth’s magnetic field lies in the vertical plane, 12° West of the geographic meridian, making an angle of 60° (upward) with the horizontal direction. Its magnitude is 0.32 G.

Question 10.

A short bar magnet has a magnetic moment of 0.48 J T^{-1}. Give the direction and mag-nitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.

Solution:

Magnetic moment of the bar magnet,

M = 0.48 J T^{-1}.

a) Distance, d = 10 cm = 0.1 m

The magnetic field at distance d, from the centre of the magnet on the axis is

The magnetic field is along the S – N direction.

b) The magnetic field at a distance of 10 cm (i.e., d = 0.1 m) on the equatorial line of the magnet is

The magnetic field is along the N – S direction.

Question 11.

A short bar magnet of magnetic moment 5.25 × 10^{-2} J T^{-1} is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on

a) its normal bisector and

b) its axis. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.

Solution:

Magnetic moment of the bar magnet,

M = 5.25 × 10^{-2} J T^{-1}

Magnitude of earth’s magnetic field at a place, H = 0.42 G = 0.42 × 10^{-4} T

a) The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation :

B = \(\frac{\mu_0M}{4\pi R^3}\) ; When the resultant field is inclined at 45° with earth’s field, B = H

(OR) R = 0.05 m = 5 cm

b) The magnetic field at a distance R’ from the centre of the magnet on its axis is given as:

\(\frac{\mu_02M}{4\pi R^3}\)

B = The resultant field is inclined at 45° with earth’s field. ∴ B’ = H.