Two identical small bar magnets each of dipole moment \(3\sqrt{5}\) J/T are placed at a center to center separation of \(10\) cm, with their axes perpendicular to each other as shown in figure. The value of magnetic field at the point P midway between the magnets is \(\alpha \times 10^{-3}\) T. The value of \(\alpha\) is ______.
(\(\mu_0=4\pi \times 10^{-7}\) Tm/A)

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: \(12\)