Thus, $$\frac{T}{{T'}} = \frac{{2\pi \sqrt {\frac{I}{{MH}}} }}{{2\pi \sqrt {\frac{I}{{MH'}}} }}$$
Given, $$T = 4\,\sec ,T' = 2\,\sec ,$$
So, $$\frac{4}{2} = \sqrt {\frac{{H'}}{H}} $$
$$ \Rightarrow H' = 4H$$

$$\eqalign{ & {\text{Here,}}\,\,r = 30\,cm = 0.3\,cm \cr & {\text{we}}\,{\text{know}}\,\frac{{{\mu _0}M}}{{4\pi {r^3}}} = {B_H} = 3.6 \times {10^{ - 5}} \cr & \Rightarrow M = \frac{{3.6 \times {{10}^{ - 5}}}}{{{{10}^{ - 7}}}}{\left( {0.3} \right)^3} = 9.7\,A{m^2} \cr} $$

$$\eqalign{ & \tau = MB\sin \theta ,\tau = iAB\sin {90^ \circ } \cr & \therefore A = \frac{\tau }{{iB}} = \frac{1}{2}\left( {BC} \right)\left( {AD} \right) \cr & {\text{But}}\,\frac{1}{2}\left( {BC} \right)\left( {AD} \right) \cr & = \frac{1}{2}\left( l \right)\sqrt {{l^2} - {{\left( {\frac{l}{2}} \right)}^2}} = \frac{{\sqrt 3 }}{4}{l^2} \cr & \Rightarrow \frac{{\sqrt 3 }}{4}{\left( l \right)^2} = \frac{\tau }{{Bi}} \cr & \therefore l = 2{\left( {\frac{\tau }{{\sqrt 3 B.i}}} \right)^{\frac{1}{2}}} \cr} $$

Magnetic field due to a bar magnet in the broad-side on position is given by
$$B = \frac{{{\mu _0}}}{{4\pi }}\frac{M}{{{{\left[ {{r^2} + \frac{{{\ell ^2}}}{4}} \right]}^{\frac{3}{2}}}}};M = m\ell .$$
After substituting the values and simplifying we get
$$B = 6 \times {10^{ - 5}}A - m$$

$$\eqalign{ & \therefore M = \frac{{4\pi }}{{{\mu _0}}}\frac{{{B_H}{{\left( {{r^2} - {\ell ^2}} \right)}^2}}}{{2r}} \cr & = {10^7} \times \frac{{\left( {0.34 \times {{10}^{ - 4}}} \right).{{\left[ {{{\left( {0.40} \right)}^2} - {{\left( {0.15} \right)}^2}} \right]}^2}}}{{2 \times 0.40}} \cr & = 8.0\,A{m^2} \cr} $$
The pole strength of the magnet is,
$$m = \frac{M}{{2\ell }} = \frac{{8.0}}{{0.30}} = 26.7\,Am$$

NOTE : Electro magnet should be amenable to magnetisation & demagnetization.
∴ retentivity should be low & coercivity should be low.

According to Curie's law, magnetic susceptibility $$\left( {{\chi _m}} \right)$$  of a paramagnetic substance is inversely proportional to absolute temperature $$\left( T \right)$$  i.e.
$${\chi _m} \propto \frac{1}{T}$$

$$H = B\cos \theta ,V = B\sin \theta $$
Here $$B$$ = earth’s magnetic field
$$\eqalign{ & \theta = {\text{angle}}\,{\text{of}}\,{\text{dip}} = {90^ \circ }\,{\text{at}}\,{\text{north}}\,{\text{field}} \cr & \Rightarrow H = B\cos {90^ \circ } = 0 \cr & V = B\sin {90^ \circ } = B \cr & \Rightarrow V > > H \cr} $$

The time period of a bar magnet in a magnetic field is given by
$$T = 2\pi \sqrt {\left( {\frac{I}{{MB}}} \right)} $$
where, $$I$$ is moment of inertia of bar magnet, $$M$$ is magnetic moment and $$B$$ is magnetic induction. When mass is made 4 times, moment of inertia $$I$$ becomes 4 times (as $$I = m{r^2},I \propto m$$   ). From the above equation of time period $$T \propto \sqrt I .$$   So, $$T$$ becomes twice as mass is quadrupled.

$${\chi _d} < {\chi _p} < {\chi _f}$$
For diamagnetic substance $${\chi _d}$$ is small negative $$\left( {{{10}^{ - 5}}} \right)$$
For paramagnetic substances $${\chi _p}$$ is small and positive $$\left( {{{10}^{ - 3}}\,{\text{to}}\,{{10}^{ - 5}}} \right)$$
For ferromagnetic substances $${\chi _f}$$ is very large $$\left( {{{10}^{ 3}}\,{\text{to}}\,{{10}^{ 5}}} \right)$$