Physics Ncert Solutions Class 12th

5.19 Number of horizontal wires in the telephone cable, n = 4

Current in each wire, I = 1.0 A

The Earth's magnetic field at a location, H = 0.39 G = 0.39 $*{10}^{-4}$ T

Angle of dip at the location, $\beta$ = 35 $°$

Angle of declination, $\theta \approx$ 0 $°$

For a point 4.0 cm below the cable, r = 4.0 cm = 0.04 m

The horizontal component of earth's magnetic field can be written as

$H_H$ = H $\cos ?\beta$ - B, where

B = Magnetic field at 4 cm due to current I in four wires = 4 $*\frac{{\mu }_{0}I}{2\pi r}$

${\mu }_0$ = Permeability of free space = 4 $\pi *{10}^{-7}$ T m $A^{-1}$

Then B = 4 $*\frac{4\pi *{10}^{-7}*1}{2\pi *0.04}$ = 2 $*{10}^{-5}$ T = 0.2 $*{10}^{-4}$ T = 0.2 G

$H_H$ = H $\cos ?\beta$ – B = 0.39 $*{10}^{-4}$ $\cos ?\beta$ - 0.2 $*{10}^{-4}$ = 0.12 $*{10}^{-4}$ T= 0.12 G

The vertical component of Earth's magnetic field is given as

$H_V$ = H $\sin ?\beta$ = 0.39 $*{10}^{-4}$ = 0.22 $*{10}^{-4}$ T = 0.22 G

The angle made by the field with its horizontal component is given as:

$\theta ={\tan }^{-1}?\frac{H_V}{H_H}$ = ${\tan }^{-1}?\frac{0.22}{0.12}$ = 61.39 $°$

The resultant magnetic field at the point is given as

H = $\sqrt{{H_H}^2+{H_V}^2}$ = $\sqrt{{0.12}^2+{0.22}^2}$ = 0.25 G

For a point 4.0 cm above the cable, r = 4.0 cm = 0.04 m

The horizontal component of earth's magnetic field can be written as

$H_H$ = H $\cos ?\beta$ + B, where

B = Magnetic field at 4 cm due to current I in four wires = 4 $*\frac{{\mu }_{0}I}{2\pi r}$

${\mu }_0$ = Permeability of free space = 4 $\pi *{10}^{-7}$ T m $A^{-1}$

Then B = 4 $*\frac{4\pi *{10}^{-7}*1}{2\pi *0.04}$ = 2 $*{10}^{-5}$ T = 0.2 $*{10}^{-4}$ T = 0.2 G

$H_H$ = H $\cos ?\beta$ + B = 0.39 $*{10}^{-4}$ + 0.2 $*{10}^{-4}$ = 0.52 $*{10}^{-4}$ T= 0.52 G

The vertical component of Earth's magnetic field is given as

$H_V$ = H $\sin ?\beta$ = 0.39 $*{10}^{-4}\sin ?35°$ = 0.22 $*{10}^{-4}$ T = 0.22 G

The angle made by the field with its horizontal component is given as:

$\theta ={\tan }^{-1}?\frac{H_V}{H_H}$ = ${\tan }^{-1}?\frac{0.22}{0.52}$ = 22.93 $°$

The resultant magnetic field at the point is given as

H = $\sqrt{{H_H}^2+{H_V}^2}$ = $\sqrt{{0.52}^2+{0.22}^2}$ = 0.56 G $\frac{{\mu }_{0}2\pi nI}{4\pi r}$

5.17 (a) The hysteresis curve of a ferromagnetic material is given below.

(b)The dissipated heat energy is directly proportional to the area of a hysteresis loop. A carbon steel piece has a greater hysteresis curve area. Hence, it dissipates greater heat energy.

(c) The value of magnetisation is memory or record of hysteresis loop cycles of magnetisation. These bits of information correspond to the cycle of magnetisation. Hysteresis loops can be used for storing information.

(d) Ceramic is used for coating magnetic tapes in cassette players and for building memory stores in modern computers.

(e) A certain region of space can be shielded from magnetic fields if it is surrounded by soft iron rings. In such arrangements, the magnetic lines are drawn out of the region.

5.16 (a) Due to random thermal motion of molecules, the alignment of dipoles gets disrupted at high temperatures. On cooling, this disruption is reduced. Hence, a paramagnetic sample displays greater magnetisation when cooled.

(b) The induced dipole moment in a diamagnetic substance is always opposite to the magnetising field. Hence, the internal motion of atoms (which is related to the temperature) does not affect the diamagnetism of a material.

(c) Bismuth is a diamagnetic substance. Hence, a toroid with a bismuth core has a magnetic field slightly greater than toroid whose core is empty.

(d) The permeability of ferromagnetic materials is not independent of the applied magnetic field. It is greater for a lower field and vice versa.

(e) The permeability of a ferromagnetic material is not less than one. It is always greater than one. Hence, magnetic field lines are always nearly normal to the surface of such materials at every point.

(f) The maximum possible magnetisation of a paramagnetic sample can be of the same order of magnitude as the magnetisation of a ferromagnet. This requires high magnetising fields for saturation.

5.15 Magnetic moment of the bar magnet, M = 5.25 $*{10}^{-2}$ J/T

Magnitude of Earth's magnetic field at that place, H = 0.42 G = 0.42 $*{10}^{-4}$ T

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

B = $\frac{{\mu }_0}{4\pi }\frac{M}{R^3}$ , where

${\mu }_0$ = Permeability of free space = 4 $\pi *{10}^{-7}$ T m $A^{-1}$ and M = magnetic moment

When the resultant field is inclined at 45 $°$ with earth's field, B = H

B = $\frac{{\mu }_0}{4\pi }\frac{M}{R^3}$ = H

$R^3$ = $\frac{{\mu }_0}{4\pi }\frac{M}{H}$ = $\frac{4\pi *{10}^{-7}*5.25\mathrm{}*{10}^{-2}}{4\pi *0.42\mathrm{}*{10}^{-4}}$ 1.25 $*{10}^{-4}$

R = 0.05 m = 5 cm

The magnetic field at a distance $R_2$ from the centre of the magnet on its axis is given as:

$B_2$ = $\frac{{\mu }_0}{4\pi }\frac{2M}{R_2^3}$ , where

${\mu }_0$ = Permeability of free space = 4 $\pi *{10}^{-7}$ T m $A^{-1}$ and M = magnetic moment

When the resultant field is inclined at 45 $°$ with earth's field, $B_2$ = H

$B_2$ = $\frac{{\mu }_0}{4\pi }\frac{2M}{R_2^3}$ = H

$R_2^3$ = $\frac{{\mu }_0}{4\pi }\frac{2M}{H}$ = $\frac{4\pi *{10}^{-7}*2*5.25\mathrm{}*{10}^{-2}}{4\pi *0.42\mathrm{}*{10}^{-4}}=$ 2.5 $*{10}^{-4}$

R = 0.0629 m = 6.3 cm