Biot-Savart's law is angle dependent.

When the bar magnet is cut into two equal parts, as shown in the Fig. (a), P₁ behaves as N and P₂ behaves as S and magnetic moment of each part of magnet becomes \(\frac { M }{ 2 } \) . When pole P₂ is placed over pole P₁ as shown in Fig. (b), the Magnetic moment of the combination is zero i.e. \(\frac { M }{ 2 } -\frac { M }{ 2 } =0.\)

The system of magnets will be in stable equilibrium, if the net force on the system is zero and net torque on it is also zero. It will be possible, if the poles of the remaining two magnets is as shown in the figure.

Initially,
\(T=2\pi \sqrt { \frac { I }{ 3{ B }^{ ' } } } and \ finally, \ { T }^{ ' }=2\pi \sqrt { \frac { I }{ m(B+{ B }^{ ' } } } \)
where, B^' = horizontal magnetic field = \(24\mu \)T
and     B  = magnetic field due to downward conductor
 \(=\frac { { \mu }_{ 0 } }{ 4\pi } .\frac { 2i }{ a } \quad =18\mu \)T
\(\therefore \frac { { T }^{ ' } }{ T } =\sqrt { \frac { { B }^{ ' } }{ B+{ B }^{ ' } } } \Rightarrow \frac { { T }^{ ' } }{ 0.1 } =\sqrt { \frac { 24 }{ 18+24 } } \Rightarrow { T }^{ ' }=0.076\)s

Here, there is no significance of the given value of B and angle of dip as the vertical component of magnetic field at equator is zero.