12. 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.
Answer:
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 given by the relation:
\(B = \frac{\mu_{0}\;2M}{4\pi d^{3}}\)
Where,
μ0 = Permeability of free space = 4π × 10−7 TmA−1
\(∴ B = \frac{4\pi \times 10^{-7}\times 2\times 0.48}{4\pi \times (0.1)^{3}}\)
= 0.96 × 10−4
T = 0.96 G
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:
\(B = \frac{\mu_{0}\times M}{4\pi \times d^{3}}\)
= \(\frac{4\pi \times 10^{-7}\times 0.48}{4\pi (0.1)^{3}}\)
= 0.48 G
The magnetic field is along with the N – S direction.
13. 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, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)
Answer:
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:
\(B_{1} = \frac{\mu_{0}2M}{4\pi d^{3}} = H\) ......(i)
Where,
μ0 = Permeability of free space
M = Magnetic moment
The magnetic field at the same distance d, on the equatorial line of the magnet, is given as:
\(B_{2} = \frac{\mu_{0}M}{4\pi d^{3}} = \frac{H}2\) [Using equation (i)]
Total magnetic field, B = B1 + B2
= H + \(\frac{H}{2}\)
= 0.36 + 0.18 = 0.54 G
Hence, the magnetic field is 0.54 G in the direction of earth’s magnetic field.
14. If the bar magnet in exercise 5.13 is turned around by 180°, where will the new null points be located?
Answer:
The magnetic field on the axis of the magnet at a distance d1 = 14 cm, can be written as:
\(B_{1} = \frac{\mu_{0}2M}{4\pi (d_{1})^{3}} = H\) …(1)
Where,
M = Magnetic moment
μ0 = Permeability of free space
H = Horizontal component of the magnetic field at d1
If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial line.
Hence, the magnetic field at a distance d2, on the equatorial line of the magnet can be written as:
\(B_{1} = \frac{\mu_{0}2M}{4\pi (d_{2})^{3}} = H\) …(2)
Equating equations (1) and (2), we get:
\(\frac{2}{(d_{1})^{3}} = \frac{1}{(d_{2})^{3}}\)
\(\left [ \frac{d_{2}}{d_{1}} \right ]^{3} = \frac{1}{2}\)
\(∴ d_{2} = d_{1}\times \left ( \frac{1}{2} \right )^{\frac{1}{3}}\)
= 14 × 0.794 = 11.1 cm
The new null points will be located 11.1 cm on the normal bisector.
15. 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.
Answer:
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 ordinary bisector is given by:
\(B = \frac{\mu_{0}M}{4\pi R^{3}}\)
Where,
μ0 = Permeability of free space = 4π × 10−7 TmA−1
When the resultant field is inclined at 45° with earth’s field, B = H
\(∴ \frac{\mu_{0}M}{4\pi R^{3}} = H = 0.42\times 10^{-4}\)
\(R^{3} = \frac{\mu_{0}M}{0.42\times 10^{-4}\times 4\pi}\)
\(= R^{3} = \frac{4\pi \times 10^{-7}\times 5.25\times 10^{-2}}{4\pi \times 0.42\times 10^{-4}}\)
= 12.5 × 10−5
∴ R = 0.05 m = 5 cm
(b) The magnetic field at a distanced ‘R’ from the centre of the magnet on its axis is given as:
\(B’ = \frac{\mu_{0}2M}{4\pi R^{3}}\)
The resultant field is inclined at 45° with the earth’s field.
∴ B’ = H
\(\frac{\mu_{0}2M}{4\pi (R’)^{3}} = H\)
\((R’)^{3} = \frac{\mu_{0}\;2M}{4\pi \times H}\)
\(= \frac{4\pi \times 10^{-7}\times 2\times 5.25\times 10^{-2}}{4\pi\times 0.42\times 10^{-4}} = 25\times 10^{-5}\)
∴ R = 0.063 m = 6.3 cm
16. 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?
Answer:
(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 diamagnetic substance. Therefore, a toroid with bismith 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 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.
