Two identical small bar magnets each of dipole moment 3 5 J/T are...

Let the magnetic moment of each magnet be $M = 3\sqrt{5}$ J/T.

The distance of point P from the center of each magnet is $r = 5$ cm $= 0.05$ m.

For the first magnet (horizontal), point P lies on its axial line. The magnetic field at P due to the first magnet is:

$B_1 = \dfrac{\mu_0}{4\pi} \dfrac{2M}{r^3}$

The direction of $\vec{B}_1$ is along the axis of the first magnet.

For the second magnet (vertical), point P lies on its equatorial line. The magnetic field at P due to the second magnet is:

$B_2 = \dfrac{\mu_0}{4\pi} \dfrac{M}{r^3}$

The direction of $\vec{B}_2$ is parallel to the axis of the second magnet, which is perpendicular to the axis of the first magnet.

Since $\vec{B}_1$ and $\vec{B}_2$ are perpendicular to each other, the net magnetic field at P is:

$B = \sqrt{B_1^2 + B_2^2}$

Substituting the expressions for $B_1$ and $B_2$:

$B = \sqrt{\left(\dfrac{\mu_0}{4\pi} \dfrac{2M}{r^3}\right)^2 + \left(\dfrac{\mu_0}{4\pi} \dfrac{M}{r^3}\right)^2}$

$B = \dfrac{\mu_0}{4\pi} \dfrac{M}{r^3} \sqrt{2^2 + 1^2} = \dfrac{\mu_0}{4\pi} \dfrac{\sqrt{5}M}{r^3}$

Substituting the given values $\dfrac{\mu_0}{4\pi} = 10^{-7}$ T m/A, $M = 3\sqrt{5}$ J/T, and $r = 0.05$ m:

$B = 10^{-7} \times \dfrac{\sqrt{5} \times 3\sqrt{5}}{(0.05)^3}$

$B = 10^{-7} \times \dfrac{15}{125 \times 10^{-6}}$

$B = \dfrac{15 \times 10^{-1}}{125}$

$B = 0.1 \times 0.12$ T

$B = 0.012$ T

$B = 12 \times 10^{-3}$ T

Comparing this with the given value $\alpha \times 10^{-3}$ T, we get $\alpha = 12$.

Answer: $12$