WBCHSE Class 12 Physics Magnetic Properties Of Materials Short Question And Answers

Magnetic Effect Of Current And Magnetism

Magnetic Properties Of Materials Short Question And Answers

Question 1. Answer the following questions regarding Earth’s magnetism.

1. Name the three independent quantities conventionally used to specify Earth’s magnetic field.
2. The angle of dip at a location in southern India is about 18 0. Would you expect a greater or smaller dip angle in Britain?
3. If a map of magnetic field lines is prepared at Melbourne in Australia, would the lines seen; go into the ground or come out of the ground?
4. In which direction would a compass be free to move in the vertical plane point, if located right on the geomagnetic north or south pole?

Answer:

1.

1. Angle of dip
2. Angle of declination
3. Horizontal component of earth’s magnetic field.

2. Britain is situated far north compared to India. So the angle of the dip will be greater than 18°.

3. Australia is situated in the southern hemisphere. So the lines offeree would come out of the ground in Melbourne.

4. The geomagnetic north pole is a south pole. So the north pole of a compass needle would point vertically downwards. Similarly, the south pole of the needle would point vertically downwards at the geomagnetic south pole.

Question 2. The earth’s core is known to contain iron. Yet geologists do not regard this as a source of earth’s magnetism. Why?
Answer:

Iron present in the earth’s core is in the molten state. The temperature of this molten iron is much higher than the Curie point. So this iron cannot have any ferromagnetism.

Question 3. The age of the earth is 4 to 5 billion years. Geologists believe that during this period earth’s magnetism has changed, even reversing its direction several times. How can geologists know about Earth’s field in such a distant past?
Answer:

The magnetic field of the earth gets weakly recorded on some rocks during their solidification. Geologists trace the geomagnetic history, of the earth by analyzing these rocks.

Question 4. A short bar maghet placed on a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the center of the magnet. The earth’s magnetic field at the place is 0.36G. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null points (i.e., 14 cm ) from the center of the magnet (at null points, field due to a magnet is equal and opposite to. the horizontal component of earth’s magnetic field.)
Answer:

Axial magnetic field Bax is equal to the geomagnetic field (= 0.36) at the null points. Magnetic field at the equatorial line due to die magnet,

⇒ \(B_{\mathrm{eq}}=\frac{B_{\mathrm{ax}}}{2}=\frac{0.36}{2}=0.18 \mathrm{C}\)

∴ The total magnetic field at a point 14 cm away on the equatorial line,

B = B_eq + B_ax

= 0.36 + 0.18

= 0.54 G

Question 5. Where will the new null points be located if the bar magnet in the previous Example is rotated through 180°?
Answer:

The new null point will be located on the perpendicular bisector of the axis of the magnet.

The magnetic field at a distance rax on the axis,

⇒ \(B_H=\frac{\mu_0}{4 \pi} \cdot \frac{2 p_m}{r_{\mathrm{ax}}^3}\)

The magnetic field at a distance req on the perpendicular bisector of the axis,

⇒ \(B_H=\frac{\mu_0}{4 \pi} \cdot \frac{p_m}{r_{\mathrm{eq}}^3}\)

∴ \(\frac{\mu_0}{4 \pi} \cdot \frac{2 p_m}{r_{\mathrm{ax}}^3}=\frac{\mu_0}{4 \pi} \cdot \frac{p_m}{r_{\mathrm{eq}}^3}\) [∵ at null point both the fields are equal]

∴ \(r_{\mathrm{eq}}=\frac{r_{\mathrm{ax}}}{\sqrt[3]{2}}=\frac{14}{\sqrt[3]{2}}\)

= 11.1 cm.

Question 6. A short bar magnet of magnetic moment 5.25 x 10^-2 J.T^-1 is placed with its axis perpendicular to the earth’s magnetic field. At what distance from the center of the magnet, the resultant field is inclined at 45° with the earth’s, field on

1. Its normal bisector
2. Its axis.

The magnitude of the earth’s magnetic field at the place is 0.42 G. Ignore the length of the magnet in comparison to the” distances involved.
Answer:

1. B_eq = BH = 0.42 x 10^-4 T [∵ resultant field is inclined at an angle of 45º
to earth’s magnetic field]

⇒ \(B_{\mathrm{eq}}=\frac{\mu_0}{4 \pi} \cdot \frac{p_m}{r^3}\)

∵ \(r=\left(\frac{\mu_0}{4 \pi} \cdot \frac{p_m}{B_{\mathrm{eq}}}\right)^{\frac{1}{3}}=\left(\frac{10^{-7} \times 5.25 \times 10^{-2}}{0.42 \times 10^{-4}}\right)^{\frac{1}{3}} \mathrm{~m}\)

= 5cm

2. \(B_{\mathrm{ax}}=\frac{\mu_0}{4 \pi} \cdot \frac{2 p_m}{r^{\prime 3}}\left[\text { Here, } B_{\mathrm{ax}}=B_H\right]\)

∴ \(r^{\prime}=\left(\frac{\mu_0}{4 \pi} \cdot \frac{2 p_m}{B_H}\right)^{\frac{1}{3}}=\left(\frac{10^{-7} \times 2 \times 5.25 \times 10^{-2}}{0.42 \times 10^{-4}}\right)^{\frac{1}{3}} \mathrm{~m}\)

= 6.3cm

Question 7. Why is diamagnetism almost independent of temperature?
Answer:

Almost none of the molecules of a diamagnetic material act’ as magnetic dipoles. So thermal motion of the molecules does not affect their magnetic properties. Thus diamagnetism is almost independent of temperature.

Question 8. Why does a paramagnetic sample display greater magnetization (for the same magnetizing field) when cooled?
Answer:

When a paramagnetic sample is placed in a magnetic field the magnetic dipoles of the sample align themselves along the magnetic field. At higher temperatures, the internal energy of the dipoles increases as a result of which this alignment is destroyed. Thus, at lower temperatures, the sample shows greater magnetization’ due to better dipole alignment.

Question 9. If a food uses bismuth for its core, will the field in the core be greater or less than when the core is empty?
Answer:

As bismuth is a diamagnetic material, the field in the core will be slightly less when bismuth is used.

Question 10. Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields?
Answer:

The permeability of ferromagnetic material depends upon the magnetic field. It is more for the lower magnetic field.

Question 11. The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece. If the material is to go through repeated cycles of magnetization, which piece will dissipate greater heat energy?
Answer:

Heat dissipated per second is directly proportional to the area of the hysteresis loop of the material. Hence carbon steel will dissipate more energy than soft iron.

Question 12. ‘A system displaying a hysteresis loop such as a ferromagnet is a device for storing memory.’ Explain the meaning of the statement.
Answer:

Magnetization of a ferromagnetic material indicates the number of magnetic cycles undergone by the sample in an applied magnetic field. Thus it can be said that the sample stores the history of its magnetization as its memory’.

Question 13. What type of ferromagnetic material is used as the memory store of modem computers?
Answer:

Normally ferrites, which are specially treated barium iron oxides, are used.

Question 14. Suggest a method to shield certain regions of space from magnetic fields.
Answer:

The space has to be enclosed by rings of soft iron or some other ferromagnetic material. The magnetic lines of force would pass through the material and keep the space inside free from the magnetic field.

Question 15. A long straight horizontal cable carries a current of 2.5 A in the direction from 10° south of west to 10° north of east The magnetic meridian of the place happens to be 10° west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G and the angle of dip is zero. Locate the line of the neutral point. (Ignore the thickness of the cable.)
Answer:

As the angle of dip, θ = 0°

∴ BH = Bcosθ

= 0.33 x 1

= 0.33 G

If a is the distance of the neutral point, then,

⇒ \(B_H=\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{a}\) [∵ magnetic field at neutral point = horizontal component of earth’s magnetic field)

∴ \(a=\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{B_H}\)

= \(\frac{10^{-7} \times 2 \times 2.5}{0.33 \times 10^{-4}} \mathrm{~m}\)

= 1.5cm

Applying the thumb rule it is seen that the direction of the magnetic field is upwards from the tire cable. Thus, the neutral point is situated on a line above the cable at a distance of 1.5 cm and parallel to the cable.

Question 16. The magnetic moment vectors μ_s and μ_l associated with the intrinsic spin angular momentum \(\vec{s}\) and orbital angular momentum \(\vec{l}\), respectively, of on electron are predicted by quantum theory to be given by

⇒ \(\mu_s=-\left(\frac{e}{m}\right) s, \mu_l=-\left(\frac{3}{2 m}\right) l\)

Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.
Answer:

From the definition of \(\overrightarrow{\mu_l} \text { and } \vec{l}\)

⇒ \(\mu_l=I A=\left(\frac{e}{t}\right) \cdot \pi r^2\)

and, \(l=m v r=m \cdot \frac{2 \pi r^2}{t}\)

where an electron of mass m and charge -e completes one rotation in an orbit of radius r in time t.

∴ \(\frac{\mu_l}{l}=\frac{e}{2 m}\)

The electron has a negative charge, therefore, μ₁ and l are parallel and opposite to each other and both are perpendicular to the orbit

∴ \(\mu_l=-\frac{e}{2 m} \cdot l\)

This relation is obtained from classical physics.

But the relation \(\mu_s=-\left(\frac{e}{m}\right) s\) cannot be established without quantum theory.

Question 17. A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions Is 60° and one of the fields has a magnitude of 1.2 x 10^-2T. If the dipole comes to stable equilibrium at an angle of 15° with this field, what is the magnitude of the other field?
Answer:

Let the other field be B₂.

The dipole makes an angle (60° – 15°) = 45° with B₂.

∵ p_mB₁sinθ₁ = p_mB₂sinθ₂ [p_m = dipole moment, tarque = \(\vec{p}_m \times \vec{B}\)]

∴ \(B_2=B_1 \frac{\sin \theta_1}{\sin \theta_2}=1.2 \times 10^{-2} \times \frac{\sin 15^{\circ}}{\sin 45^{\circ}}\)

= 4.39 x 10^-3 T

Question 18. A wire is kept horizontally at a place in the northern hemisphere of the earth. In what direction will force act on the wire due to the vertical component of the earth’s magnetic field, if electric current flows through the wire from south to north?
Answer:

When a wire is placed horizontally in the northern hemisphere and the current in it flows from south to north then according to Fleming’s left-hand rule, the direction of the force acting on the wire will be along the west. The vertical component of the earth’s magnetic field is along \(\otimes\) mark.

Question 19. A copper wire of length l meter is bent to form a circular loop. If i ampere current flows through the loop, find out the magnitude of the magnetic moment of the loop.
Answer:

The length of a copper wire = l, let the radius be r when the wire is bent to form a circular coil.

∴ \(2 \pi r=l \text { or, } r=\frac{l}{2 \pi}\)

∴ Area of the circular coil,

⇒ \(A=\pi r^2=\pi \times \frac{l^2}{4 \pi^2}=\frac{l^2}{4 \pi}\)

∴ Magnetic moment of the circular wire = \(i A=\frac{i l^2}{4 \pi}\)

Question 20. Suppose that the source of Earth’s magnetism is a magnetic dipole placed at the center of the Earth. Find the moment of this magnetic dipole if the strength of the earth’s magnetic field at the equator is 4 x 10^-5T. Given, radius of the earth = 6.4 x 10⁶ m and \(\frac{\mu_0}{4 \pi}=10^{-7} \mathrm{~T} \cdot \mathrm{m} \cdot \mathrm{A}^{-1}\)
Answer:

The strength of the Earth’s magnetic field at the equator

= 4 x 10^-5T

Radius of the earth, R = 6.4 x 10⁶ m

∴ \(B=\frac{\mu_0}{4 \pi} \cdot \frac{p_m}{R^3}\) [pm = the moment of the magnetic dipole of the earth]

or, \(4 \times 10^{-5}=10^{-7} \cdot \frac{p_m}{\left(6.4 \times 10^6\right)^3}\)

or, \(p_m=\frac{4 \times 10^{-5} \times\left(6.4 \times 10^6\right)^3}{10^{-7}}\)

= 1.05 x 10²³ A.m²

Question 21. On what physical quantity does the magnetic moment of an electron revolving in an orbit depend?
Answer:

The magnetic moment of an electron depends on the specific charge of the electron \(\frac{e}{m}\) and the angular momentum of rotation of the electron \(\vec{L}\).

Question 22. A current-carrying loop behaves as a magnetic dipole.
Answer:

The torque acting on an electric dipole of the moment \(\vec{p}\) in an electric field \(\vec{E}, \vec{\tau}=\vec{p} \times \vec{E}\)

The torque acting on a loop of area A carrying current I in a magnetic field \(\vec{B}, \vec{\tau}=I \vec{A} \times \vec{B}\)

Hence, it can be said that a current-carrying loop behaves as a magnetic dipole.

Question 23. A circular coil of N turns and diameter d carries a current I. It is unwound and rewound to make another coil of diameter 2d, current I remaining the same. Calculate the ratio of the magnetic moments of the new coil and the original coil
Answer:

For the original coil, magnetic moment,

⇒ \(p_1=N I A=N I \cdot \frac{\pi d^2}{4}\)

For the new coil, the diameter becomes 2d, i.e., the circumference is doubled.

So the number of turns becomes \(\frac{N}{2}\). Magnetic moment,

⇒ \(p_2=\frac{N I}{2} \cdot \frac{\pi(2 d)^2}{4}=N I \cdot \frac{\pi d^2}{4} \cdot 2=2 p_1\)

∴ \(\frac{p_2}{p_1}=\frac{2}{1}\)

Question 24. A circular coil of closely wound N turns and radius r carries a current of 1.

1. The magnetic field at its center,
2. The magnetic moment of this coil

Answer:

1. \(B=\frac{\mu_0 N I}{2 r}\)

2. \(p_m=N I A=\pi N I r^2\)

Question 25. At a place, the horizontal component of the earth’s magnetic field is B, and the angle of dip is 60°. What is the value of the horizontal component of the earth’s magnetic field at the equator?
Answer:

Given, the horizontal component of the earth’s magnetic field, H = B.

The angle of dip, θ = 60°.

H = VcosB [V = vertical component of earth’s magnetic field]

= Vcos60°

= \(\frac{V}{2}\)

∴ V = 2H

At equator, θ = 0°

∴ \(H_{\text {eq }}=V \cos 0^{\circ} \text { or, } H_{\text {eq }}=2 H\)

Question 26. A bar magnet of magnetic moment 6 J.T^-1 is aligned at 60° with a uniform external magnetic field of 0.44 T. Calculate

1. The work done in turning the magnet to align its magnetic moment
   1. Normal to the magnetic field
   2. Opposite to the magnetic field
2. The torque on the magnet in the final orientation in case (2).

Answer:

Here, m = 6J.T^-1,θ₁ = 60° , B = 0.44 T

1. Work done in turning the magnet,

W = -mB(cosθ₂ – cosθ₁)

1. θ₁ = 60° , θ₂ = 90°

∴ W = -6 x 0.44 (cos90°- cos60°)

⇒ \(-6 \times 0.44\left(0-\frac{1}{2}\right)=1.32 \mathrm{~J}\)

2. θ₁ = 60°, θ₂ = 180°

W = -6 x 0.44 (cos180° – cos60°)

⇒ \(-6 \times 0.44\left(-1-\frac{1}{2}\right)\)

= 3.96J

2. The torque on the magnet in the final orientation in case (2),

⇒ \(\tau=m B \sin \theta=m B \sin 180^{\circ}=0\)