Magnetism and Matter

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 o

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

Magnitude of Earth’s magnetic field at that place, H = 0.42 G = 0.42 $\times {10}^{-4}$ 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 $\pi \times {10}^{-7}$ T m $A^{-1}$ and M

5.14 The magnetic field on the axis of the magnet at a distance $d_1$ = 14 cm, can be written as

$B_1$ = $\frac{{\mu }_0}{4\pi }\frac{2M}{d_1^3}$ = H ……………….(1)

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

If the bar magnet is turned through 180 $°$ , then the neutral point will lie on the equatorial line. Hence, the magnet

5.13 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

$B_1$ = $\frac{{\mu }_0}{4\pi }\frac{2M}{d^3}$ = H ……………….(1)

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

Total magnetic field, B = $B_1+B_2$ = H + $\frac{H}{2}$ = 0.36 + 0.18 = 0.54 G

Therefore, the magn

5.12 Magnetic moment of the bar magnet, M = 0.48 J/T

The distance, d = 10 cm = 0.1 m

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

B = $\frac{{\mu }_0}{4\pi }\frac{2M}{d^3}$ , where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$ $\pi \times {10}^{-7}$

B = $\frac{4\pi \times {10}^{-7}\times 2\times 0.48}{4\pi \times {0.1}^3}$ = 0.96 $\times {10}^{-6}$ T = 0.96 G

The magnetic field is along S –

5.11 Given:

The angle of declination, $\theta$ = 12 $°$

The angle of dip, $\beta$ = 60 $°$

Horizontal component of Earth’s magnetic field, $B_H$ = 0.16 G

If Earth’s magnetic field be B, we can relate B and $B_H$ as $B_H=B\cos ?\beta$

B = $\frac{B_H}{\cos ?\beta }$ = $\frac{0.16}{\cos ?60°}$ = 0.32 G

Therefore, the Earth’s magnetic field lies in the vertical plane, 12west of the geographic mer

5.10 Given:

Horizontal component of Earth’s magnetic field, $B_H$ = 0.35 G

Angle made by the needle with the horizontal plane, $\beta$ = 22 $°$

If the Earth’s magnetic field at that location be B,

then $B_H$ = B $\cos ?\beta$

B = $\frac{B_H}{\cos ?\beta }$ = $\frac{0.35}{\cos ?22°}$ = 0.377 G

Therefore, the strength of Earth’s magnetic field at that location is 0.377 G

5.9 Number of turns, n = 16

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

Cross-section of the coil, A = $\pi r^2$ = $\pi \times {0.1}^2$ $m^2$ = 0.0314 $m^2$

Current in the coil, I = 0.75 A

Magnetic field strength, B = 5.0 $\times {10}^{-2}$ T

Frequency of oscillation of the coil, $\nu$ = 2/s

Magnetic moment, M = NIA = 16 $\times 0.75\times 0.0314$ = 0.377 A $m^2$

Frequency is given by the relat

5.8 (a) Number of turns, n = 2000

Area of cross-section, A = 1.6 $\times {10}^{-4}$ $m^2$

Current, I = 4.0 A

The magnetic moment along the axis of the solenoid is calculated as

M = nAI = 2000 $\times$ 1.6 $\times {10}^{-4}\times 4$ = 1.28 A $m^2$

(b) Magnetic field, B = 7.5 $\times {10}^{-2}$ T

Angle between magnetic field and the axis of the solenoid, $\theta$ = 30 $°$

Torque $\tau =MB\sin ?\theta$

=

5.7 Magnetic moment, M = 1.5 J/T

Magnetic field strength, B = 0.22 T

Initial angle between the axis and the magnetic field, ${\theta }_1$ = 0 $°$ and the final angle, ${\theta }_2$ = 90 $°$

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

W = -MB( cos ${\theta }_2-$ cos ${\theta }_1$ ) = - 1.5 $\times 0.22(\cos ?90°$ - cos 0 $°$