Magnetic Field at Midway Point Between Two Bar Magnets

To find the magnetic field at the point $P$ midway between the two identical small bar magnets, we consider the contribution of each magnet separately.

1. Given Data:

• Dipole moment of each magnet, $M = 3\sqrt{5} \text{ J/T}$
• Center-to-center separation between magnets, $d = 10 \text{ cm} = 0.1 \text{ m}$
• Distance from point $P$ to the center of each magnet, $r = \frac{d}{2} = 5 \text{ cm} = 0.05 \text{ m} = 5 \times 10^{-2} \text{ m}$
• Permeability constant, $\frac{\mu_0}{4\pi} = 10^{-7} \text{ T}\cdot\text{m/A}$

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2. Magnetic Field due to the First Magnet ($\vec{B}_1$): Point $P$ lies on the axial line (end-on position) of the horizontal magnet (first magnet).

position: $B_1 = \frac{\mu_0}{4\pi} \frac{2M}{r^3}$

Substituting the given values: $B_1 = 10^{-7} \times \frac{2 \times 3\sqrt{5}}{(5 \times 10^{-2})^3}$ $B_1 = 10^{-7} \times \frac{6\sqrt{5}}{125 \times 10^{-6}} = \frac{6\sqrt{5}}{1.25} \times 10^{-3} \text{ T} = 4.8\sqrt{5} \times 10^{-3} \text{ T}$

The direction of $\vec{B}_1$ is along the axis of the magnet (horizontally to the right).

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3. Magnetic Field due to the Second Magnet ($\vec{B}_2$): Point $P$ lies on the equatorial line (broadside-on position) of the vertical magnet (second magnet).

The magnitude of the magnetic field at an equatorial point is given by: $B_2 = \frac{\mu_0}{4\pi} \frac{M}{r^3} = \frac{B_1}{2}$

Substituting the value of $B_1$: $B_2 = \frac{4.8\sqrt{5} \times 10^{-3}}{2} = 2.4\sqrt{5} \times 10^{-3} \text{ T}$

The direction of $\vec{B}_2$ is perpendicular to the line joining the centers (vertically upwards).

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4. Net Magnetic Field at Point $P$: Since the axes of the two magnets are perpendicular to each other, $\vec{B}_1$ and $\vec{B}_2$ are mutually perpendicular ($\theta = 90^\circ$).

The resultant magnetic field $B$ at point $P$ is: $B = \sqrt{B_1^2 + B_2^2}$ $B = \sqrt{(4.8\sqrt{5} \times 10^{-3})^2 + (2.4\sqrt{5} \times 10^{-3})^2}$ $B = 2.4\sqrt{5} \times 10^{-3} \times \sqrt{2^2 + 1^2}$ $B = 2.4\sqrt{5} \times \sqrt{5} \times 10^{-3} \text{ T}$ $B = 2.4 \times 5 \times 10^{-3} \text{ T} = 12 \times 10^{-3} \text{ T}$

Comparing this result with the given form $B = \alpha \times 10^{-3} \text{ T}$: $\alpha = 12$