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NCERT · Class 12 · Physics

Chapter 5: Magnetism and Matter

Practice each question, then open the answer or worked solution.

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Questions

25 questions
Q5.11Written answer

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 is 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.

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Worked solution

Angle of declination,

\(\begin{array}{l}\theta = 12°\end{array} \)

Angle of dip,

\(\begin{array}{l}\delta = 60°\end{array} \)

The horizontal component of earth’s magnetic field,

\(\begin{array}{l}B_{H} = 0.16\; G\end{array} \)

Earth’s magnetic field at the given location = B

We can relate B and

\(\begin{array}{l}B_{H}\end{array} \)
\(\begin{array}{l}B_{H} = Bcos\delta\end{array} \)
\(\begin{array}{l}∴ B = \frac{B_{H}}{cos\delta}\end{array} \)

=

\(\begin{array}{l}\frac{0.16}{cos\; 60°}\end{array} \)

Earth’s magnetic field lies in the vertical plane,

\(\begin{array}{l}12°\end{array} \)
\(\begin{array}{l}60°\end{array} \)
Q5.12Written answer

A short bar magnet has a magnetic moment of 0.48 J T–1. Give the direction and magnitude 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.

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Worked solution

Magnetic moment of the bar magnet, M =

\(\begin{array}{l}0.48\;J\;T^{-1}\end{array} \)

(a) Distance, d = 10 cm = 0.1 m

The magnetic field at distance d, from the centre of the magnet on the axis, is given by the relation:

\(\begin{array}{l}B = \frac{\mu_{0}\;2M}{4\pi d^{3}}\end{array} \)

Where,

\(\begin{array}{l}\mu_{0}\end{array} \)
\(\begin{array}{l}4\pi \times 10^{-7}\; TmA^{-1}\end{array} \)
\(\begin{array}{l}∴ B = \frac{4\pi \times 10^{-7}\times 2\times 0.48}{4\pi \times (0.1)^{3}}\end{array} \)

=

\(\begin{array}{l}0.96\times 10^{-4}\; T = 0.96 \;G\end{array} \)

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 given as:

\(\begin{array}{l}B = \frac{\mu_{0}\times M}{4\pi \times d^{3}}\end{array} \)

=

\(\begin{array}{l}\frac{4\pi \times 10^{-7}\times 0.48}{4\pi (0.1)^{3}}\end{array} \)

= 0.48 G

The magnetic field is along with the N–S direction.

 

Q5.13Written answer

A short bar magnet placed in 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 centre of the magnet. The earth’s magnetic field at the place is 0.36 G, and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null–point, i.e. 14 cm, from the centre of the magnet? (At null points, the field due to a magnet is equal and opposite to the horizontal component of the earth’s magnetic field.)

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Worked solution

Earth’s magnetic field at the given place, H = 0.36 G

The magnetic field at a distance d, on the axis of the magnet, is given as:

\(\begin{array}{l}B_{1} = \frac{\mu_{0}2M}{4\pi d^{3}} = H\;\;\;\;\;\;\;\;\;\;…(i)\end{array} \)

Where,

\(\begin{array}{l}\mu_{0}\end{array} \)

M = Magnetic moment

The magnetic field at the same distance, d, on the equatorial line of the magnet, is given as:

\(\begin{array}{l}B_{2} = \frac{\mu_{0}M}{4\pi d^{3}} = \frac{H}{2}\;\;\;\;\;\;[Using\;equation\;(i)]\end{array} \)

Total magnetic field, B =

\(\begin{array}{l}B_{1} + B_{2}\end{array} \)

=

\(\begin{array}{l}H + \frac{H}{2}\end{array} \)

= 0.36 + 0.18 = 0.54 G

Hence, the magnetic field is 0.54 G in the direction of the earth’s magnetic field.

Q5.14Written answer

If the bar magnet in exercise 5.13 is turned around by 180°, where will the new null points be located?

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Worked solution

The magnetic field on the axis of the magnet at a distance

\(\begin{array}{l}d_{1}\end{array} \)
\(\begin{array}{l}B_{1} = \frac{\mu_{0}2M}{4\pi (d_{1})^{3}} = H\;\;\;\;\;\;\;…(1)\end{array} \)

Where,

M = Magnetic moment

\(\begin{array}{l}\mu_{0}\end{array} \)

H = Horizontal component of the magnetic field at

\(\begin{array}{l}d_{1}\end{array} \)

If the bar magnet is turned through 1800, then the neutral point will lie on the equatorial line.

Hence, the magnetic field at a distance

\(\begin{array}{l}d_{2}\end{array} \)
\(\begin{array}{l}B_{1} = \frac{\mu_{0}2M}{4\pi (d_{2})^{3}} = H\;\;\;\;\;\;\;…(2)\end{array} \)

Equating equations (1) and (2), we get:

\(\begin{array}{l}\frac{2}{(d_{1})^{3}} = \frac{1}{(d_{2})^{3}}\end{array} \)
\(\begin{array}{l}\left [ \frac{d_{2}}{d_{1}} \right ]^{3} = \frac{1}{2}\end{array} \)
\(\begin{array}{l}∴ d_{2} = d_{1}\times \left ( \frac{1}{2} \right )^{\frac{1}{3}}\end{array} \)

=

\(\begin{array}{l}14\times 0.794 = 11.1 cm\end{array} \)

The new null points will be located 11.1 cm on the normal bisector.

Q5.15Written answer

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? The 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.

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Worked solution

Magnetic moment of the bar magnet, M =

\(\begin{array}{l}5.25\times 10^{-2}\;J\;T^{-1}\end{array} \)

Magnitude of earth’s magnetic field at a place, H = 0.42 G =

\(\begin{array}{l}0.42\times 10^{-4}\; T\end{array} \)

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

\(\begin{array}{l}B = \frac{\mu_{0}M}{4\pi R^{3}}\end{array} \)

Where,

\(\begin{array}{l}\mu_{0}\end{array} \)
\(\begin{array}{l}4\pi \times 10^{-7} TmA^{-1}\end{array} \)

When the resultant field is inclined at

\(\begin{array}{l}45°\end{array} \)
\(\begin{array}{l}∴ \frac{\mu_{0}M}{4\pi R^{3}} = H = 0.42\times 10^{-4}\end{array} \)
\(\begin{array}{l}R^{3} = \frac{\mu_{0}M}{0.42\times 10^{-4}\times 4\pi}\end{array} \)

=

\(\begin{array}{l}R^{3} = \frac{4\pi \times 10^{-7}\times 5.25\times 10^{-2}}{4\pi \times 0.42\times 10^{-4}}\end{array} \)

=

\(\begin{array}{l}12.5 \times 10^{-5}\end{array} \)
\(\begin{array}{l}∴R = 0.05\; m = 5\; cm\end{array} \)

 

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

\(\begin{array}{l}B’ = \frac{\mu_{0}2M}{4\pi R^{3}}\end{array} \)

The resultant field is inclined at

\(\begin{array}{l}45°\end{array} \)
\(\begin{array}{l}∴ B’ = H\end{array} \)
\(\begin{array}{l}\frac{\mu_{0}2M}{4\pi (R’)^{3}} = H\end{array} \)
\(\begin{array}{l}(R’)^{3} = \frac{\mu_{0}\;2M}{4\pi \times H}\end{array} \)

=

\(\begin{array}{l}\frac{4\pi \times 10^{-7}\times 2\times 5.25\times 10^{-2}}{4\pi\times 0.42\times 10^{-4}} = 25\times 10^{-5}\end{array} \)
\(\begin{array}{l}∴ R = 0.063\; m = 6.3\; cm\end{array} \)

 

Q5.16Written answer

Answer the following questions:
(a) Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled?
(b) Why is diamagnetism, in contrast, almost independent of temperature?
(c) If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty?
(d) Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields?
(e) Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at
every point.) Why?
(f) Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet?

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(a) The thermal motion reduces at a lower temperature, and the tendency to disrupt the alignment of the dipoles decreases.

(b) The dipole moment induced is always opposite to the magnetising field. Therefore, the internal motion of the atoms due to the temperature will not affect the magnetism of the material.
(c) Bismuth is a diamagnetic substance. Therefore,  a toroid with a bismuth core will have a field slightly less than when the core is empty.
(d) Permeability of the ferromagnetic material depends on the magnetic field. Permeability is greater for lower fields.
(e) Proof of this important fact (of much practical use) is based on boundary conditions of magnetic fields (B and H) at the interface of two media. (When one of the media has µ >> 1, the field lines meet this medium nearly normally.)
(f) Yes, apart from minor differences in the strength of the individual atomic dipoles of two different materials, a paramagnetic sample with saturated magnetisation will have the same order of magnetisation. But, of course, saturation requires impractically high magnetising fields.
Q5.17Written answer
Answer the following questions:
(a) Explain qualitatively on the basis of the domain picture the irreversibility in the magnetisation curve of a ferromagnet.
(b) 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 magnetisation, which piece will dissipate greater heat energy?
(c) ‘A system displaying a hysteresis loop, such as a ferromagnet, is a device for storing memory?’ Explain the meaning of this statement.
(d) What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player or building ‘memory stores’ in a modern computer?
(e) A certain region of space is to be shielded from magnetic fields. Suggest a method.
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Worked solution
(a)  The domain aligns in the direction of the magnetic field when the substance is placed in an external magnetic field. Some energy is spent in the process of alignment. When the external field is removed, the substance retains some magnetisation. The energy spent in the process of magnetisation is not fully recovered. It is lost in the form of heat. This is the basic cause for the irreversibility of the magnetisation curve of a ferromagnet substance.
(b) Carbon steel piece because heat lost per cycle is proportional to the area of the hysteresis loop.
(c) Magnetisation of a ferromagnet is not a single-valued function of the magnetising field. Its value for a particular field depends both on the field and the history of magnetisation (i.e., how many cycles of magnetisation it has gone through, etc.). In other words, the value of magnetisation is a record or memory of its cycles of magnetisation. If information bits can be made to correspond to these cycles, the system displaying such a hysteresis loop can act as a device for storing information.
(d) Ceramics (specially treated barium iron oxides), also called ferrites.
(e) Surrounding the region with soft iron rings. magnetic field lines will be drawn into the rings, and the enclosed space will be free of a magnetic field. But this shielding is only approximate, unlike the perfect electric shielding of a cavity in a conductor placed in an external electric field.
Q5.18Written answer
A long straight horizontal cable carries a current of 2.5 A in the direction 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 neutral points (ignore the thickness of the cable). (At neutral points, the magnetic field due to a current carrying the cable is equal and opposite to the horizontal component of earth’s magnetic field.)
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Worked solution
Current in the wire = 2.5 A
The earth’s magnetic field at a location, R= 0.33 G = 0.33 x 10-4 T
The angle of dip is zero, δ = 0
Horizontal component of earth’s magnetic field, BH= R cosδ = 0.33 x 10-4 Cos 0 = 0.33 x 10-4 T
Magnetic field due to a current carrying conductor, Bc = (μ0/2π) x (I/r)
Bc= (4π x 10-7/2π) x (2.5/r) = (5 x 10-7/r)
BH = Bc
0.33 x 10-4  = 5 x 10-7/r
r =  5 x 10-7/0.33 x 10-4 
= 0.015 m = 1.5 cm
Hence, neutral points lie on a straight line parallel to the cable at a perpendicular distance of 1.5cm.
Q5.19Written answer
A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction, east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is 35°. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?
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Worked solution
First, let us decide the direction which would best represent the situation.
We know that
BH = B cos δ
= 0.39 × cos 35o G
BH = 0.32G
Here,
BV = B sin δ
= 0.39 × sin 35o G
BV = 0.22G
It is given that the telephone cable carries a total current of 4.0 A in the direction from east to west. So the resultant magnetic field is 4.0 cm below.
Bwire = 
\(\begin{array}{l}=\frac{\mu _{0}}{4\pi }\frac{2I}{r}\end{array} \)
\(\begin{array}{l}=10^{-7}\times \frac{2\times 4}{4\times 10^{-2}}\end{array} \)
= 2 × 10-5 T
= 0.2G
Net magnetic field
\(\begin{array}{l}\\B_{net}=\sqrt{(B_{H}-B_{wire})^{2}+B_{V}^{2}} \\=\sqrt{(0.12)^{2}+(0.22)^{2}} \\=\sqrt{0.0144+0.0484}\end{array} \)
= 0.25G
The resultant magnetic field at points 4 cm below the cable 0.25 G.
Q5.20Written answer
A compass needle free to turn in a horizontal plane is placed at the centre of a circular coil of 30 turns and a radius of 12 cm. The coil is in a vertical plane, making an angle of 45° with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.
(a) Determine the horizontal component of the earth’s magnetic field at the location.
(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90° in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.
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Worked solution
Number of turns = 30
The radius of the coil = 12 cm
Current in the coil = 0.35 A
The angle of dip, δ = 450
(a) Horizontal component of earth’s magnetic field,
BH = B sinδ
B is the magnetic field strength due to the current in the coil
B = (μ0/4π) (2πnI/r)
= (4π x 10-7/4π) (2π x 30 x 0.35/0.12)
= 5.49 x 10-5 T
Therefore, BH = B sinδ
=  (5.49 x 10-5 ) sin 450 
= 3.88 x 10-5 T = 0.388 G
(b)  The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90° in the anticlockwise direction. The needle will point from east to west.