The total flux emanating from the closed surface S enclosing the dipole is zero. So,
\({ \Phi }_{ B }=\oint { \overset { \rightarrow }{ B } .d\overset { \rightarrow }{ A } } =0\)
Here the integral is taken over closed surface. Since no isolated magnetic pole (called magnetic monopole) exists, this integral is always zero,
\(\oint { \overset { \rightarrow }{ B } .d\overset { \rightarrow }{ A } } =0\)
This is similar to Gauss’s law in electrostatics.

 (a) a bar magnet cut into two pieces along its length:

When the bar magnet is cut along the axis into two pieces, new magnetic pole strength is \({ q }_{ m }^{ ' }=\frac { { q }_{ m } }{ 2 } \) but magnetic length does not change. So, the magnetic moment is
\({ p }_{ m }^{ ' }={ q' }_{ m }2l\)
\({ p }_{ m }^{ ' }=\frac { { q }_{ m } }{ 2 } 2l=\frac { 1 }{ 2 } ({ q }_{ m }2l)=\frac { 1 }{ 2 }p_ m\)
In vector notation, \(\vec{p}'_m=\frac{1}{2}\vec{p}_m\)
(b) a bar magnet cut into two pieces perpendicular to the axis:

When the bar magnet is cut perpendicular to the axis into two pieces, magnetic pole strength will not change but magnetic length will be halved. So the magnetic moment is
\({ p }_{ m }^{ ' }={ q }_{ m }\times \frac { 1 }{ 2 } (2l)=\frac { 1 }{ 2 } ({ q }_{ m }.2l)=\frac { 1 }{ 2 } { p }_{ m }\) 
In vector notation, \(\vec{p}'_m=\frac{1}{2}\vec{p}_m\) 

Density of the magnet is
Density = \(\frac { Mass }{ volume } \Rightarrow Volume=\frac { Mass }{ Density } \)
\(Volume=\frac { 200\times 1{ 0 }^{ -3 }kg }{ \left( 8\times 1{ 0 }^{ -3 }kg \right) \times 1{ 0 }^{ 6 }{ m }^{ -3 } } =25\times { 10 }^{ -6 }{ m }^{ 3 }\)
Magnitude of magnetic moment p_m = 2A m²
Intensity of magnetization,
\(I=\frac { magnetic\ moment }{ Volume } =\frac { 2 }{ 25\times { 10 }^{ -6 } } \)
M = 0.8 x 10⁵ Am^-1

Let  \(\vec{p}_m\)be the dipole and before switching ON the external magnetic field, there is no orientation. Therefore, the energy U = 0.
As soon as external magnetic field is switched ON, the magnetic dipole orient parallel (θ = 0^o) to the magnetic field with energy,
U_parallel = U_minimum = -p_mBcos 0
U_parallel = -p_mB
since cos 0^o = 1
Otherwise, the magnetic dipole orients anti-parallel (θ = 180^o) to the magnetic field with energy,
U _{anti-parallel} = U _maximum = -p_mBcos 180
\(\Rightarrow \) U_{anti-parallel} = P_mB
since cos 180^o = -1

Given magnetic moment 0.5 J T^-1 and distance r = 0.1 m
(a) When the point lies on the axial line of the bar magnet, the magnetic field for short magnet is given by
\({ { \vec B }_{ axial } } =\frac { { \mu }_{ ° } }{ 4\pi } \left( \frac { 2{ p }_{ m } }{ { r }^{ 3 } } \right) \hat { i } \)
\({ { \vec B }_{ axial } } =1{ 0 }^{ -7 }\times \left( \frac { 2\times 0.5 }{ { \left( 0.1 \right) }^{ 3 } } \right) =1\times { 10 }^{ -4 }\hat { i } \ T\)
Hence, the magnitude of the magnetic field along axial is B_axial = 1 x 10^-4 T and direction is towards South to North.
(b) When the point lies on the normal bisector (equatorial) line of the bar magnet, the magnetic field for short magnet is given by
\({ {\vec B }_{ equatorial } } =-\frac { { \mu }_{ ° } }{ 4\pi } \frac { { p }_{ m } }{ { r }^{ 3 } } \hat { i } \)
\({ {\vec B }_{ equatorial } } =-1{ 0 }^{ -7 }\left( \frac { 0.5 }{ { \left( 0.1 \right) }^{ 3 } } \right) \hat { i } =-0.5\times 1{ 0 }^{ -4 }\hat { i } \ T \)
Hence, the magnitude of the magnetic field along axial is B_equatorial = 0.5 x 10^-4 T and direction is towards North to South.
Note that magnitude of B_axial is twice that of magnitude of B_equatorial and the direction of B_axial and B_equatorial are opposite.