Explain with the help of diagram the terms (i) magnetic declination and (ii) angle of dip at a given place. 

(i) Magnetic declination at a place may be defined as the angle between its magnetic meridian and the earth geographic meridian at the place. 

(ii) Angle of dip at a place is defined as the angle between the direction of intensity of earth’s magnetic field (B_E) and the horizontal direction in magnetic meridian at that place.

(i) X is a diamagnetic substance.
    Y is a paramagnetic substance. 

(ii) For a diamagnetic substance, the field lines are repelled or expelled and the field inside the material is reduced when a diamagnetic bar is placed in an external field.

For a paramagnetic substance, the field lines get concentrated inside the material and the field is increased when a paramagnetic bar is placed in an external field.

Here,
Horizontal component, ${\mathrm{B}}_{\mathrm{H}} = 0.4 \times {10}^{-4}\mathrm{T}$
Angle of dip, $\mathrm{\delta } = {45}^{\mathrm{o}}$ 

Using the formula,   $\boxed{\frac{{\mathrm{B}}_{\mathrm{V}}}{{\mathrm{B}}_{\mathrm{H}}} = \tan \mathrm{\delta } }$ 

Putting values, we get

${\mathrm{B}}_{\mathrm{V}} = {\mathrm{B}}_{\mathrm{H}} \tan \text{'}\mathrm{\delta }$ 

$$\begin{gathered} {\mathrm{B}}_{\mathrm{V}} = 0.4 \times {10}^{-4} \tan 45° \\ {\mathrm{B}}_{\mathrm{V}} = 0.4 \times {10}^{-4}\mathrm{T} \end{gathered}$$ 

which is the required vertical component of Earth's magnetic field.

Using the below formula we can find the value of resulatant magnetic field.

$$\begin{gathered} \boxed{{\mathrm{B}}_{\mathrm{H}} = {\mathrm{B}}_{\mathrm{E}} \cos \mathrm{\delta } } \\ {\mathrm{B}}_{\mathrm{E}} = \frac{{\mathrm{B}}_{\mathrm{H}}}{\cos \mathrm{\delta }} \end{gathered}$$ 

$$\begin{gathered} {\mathrm{B}}_{\mathrm{E}} = \frac{0.4 \times {10}^{-4}}{\cos 45°} \\ = 0.4\sqrt{2} \times {10}^{-4}\mathrm{T} \end{gathered}$$ 

${\mathrm{B}}_{\mathrm{E}} = 0.5656 \times {10}^{-4}\mathrm{T}.$

Here,
Length of magnet, 2l = 10 cm = 0.1 m

Magnetic moment, m = 12 Am

Distance of point from the mid point of axial line, d = 20 cm =0.2 m 


Magnetic field on the axial line,
           $\boxed{{\mathrm{B}}_{\mathrm{ax}}= \frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }} \times \frac{2 \mathrm{md}}{({\mathrm{d}}^2-l^2{)}^2}}$ 

 i.e., $$\begin{gathered} {\mathrm{B}}_{a xial } = \frac{{\mathrm{\mu }}_{0}2.2\mathrm{m}l\mathrm{d}}{4\mathrm{\pi }({\mathrm{d}}^2-l^2{)}^2} = \frac{{\mathrm{\mu }}_{0}4\mathrm{m}l\mathrm{d}}{4\mathrm{\pi }({\mathrm{d}}^2-l^2{)}^2} \\ = \frac{4\mathrm{\pi }\times {10}^{-7}\times 4\times 12\times 0.05\times 0.2}{4\mathrm{\pi }{\left[(0.2{)}^2-(0.05{)}^2\right]}^2} \end{gathered}$$ 

$\therefore$         ${\mathrm{B}}_1 = 3.4 \times {10}^{-5}\mathrm{T}$ 

At the same distance, on equatorial line, we have 

$\boxed{{\mathrm{B}}_{\mathrm{eq}} = \frac{1}{2}{\mathrm{B}}_{\mathrm{ax}}}$  

  $$\begin{gathered} {\mathrm{B}}_2 = \frac{1}{2}\times 3.4\times {10}^{-5}\mathrm{T} \\ = 1.7 \times {10}^{-5}\mathrm{T} \end{gathered}$$

(i) For – 1 ≥ χ < 0, material is diamagnetic,
(ii) For 0 < χ < ε, material is paramagnetic,
(a)    Range of relative magnetic permeability of diamagnetic material is
0 ≤ μ_r < 1
Range of relative magnetic permeability of paramagnetic material is
0 < μ_r < 1 + ε
(b)    Behaviour of magnetic field lines when diamagnetic material is placed in an external field.

Behaviour of magnetic field lines when paramagnetic material is placed in an external field.