Two identical small bar magnets each of dipole moment \( 3\sqrt{5} \text{ J/T} \) 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 \( \alpha \times 10^{-3} \text{ T} \). The value of \( \alpha \) is _______. (\( \mu_0 = 4\pi \times 10^{-7} \text{ Tm/A} \))
Step 1: Understanding the Concept:
Point P lies midway between two perpendicular bar magnets. For the horizontal magnet, P lies exactly on its axial line. For the vertical magnet, P lies exactly on its equatorial line. Because the magnetic fields from these two configurations are perpendicular to each other, the net magnetic field is their vector sum via the Pythagorean theorem. Step 2: Key Formula or Approach:
Magnetic field on the axial line of a short dipole: $B_{axial} = \frac{\mu_0}{4\pi} \frac{2M}{r^3}$
Magnetic field on the equatorial line of a short dipole: $B_{equatorial} = \frac{\mu_0}{4\pi} \frac{M}{r^3}$
Net Field: $B_{net} = \sqrt{B_{axial}^2 + B_{equatorial}^2}$ Step 3: Detailed Explanation:
Let the horizontal magnet be Magnet 1 and the vertical magnet be Magnet 2.
Distance between centers is $10 \text{ cm}$. Since P is midway, the distance from each center to P is $r = 5 \text{ cm} = 0.05 \text{ m} = 5 \times 10^{-2} \text{ m}$.
Magnetic moment $M = 3\sqrt{5} \text{ J/T}$.
Calculate field due to Magnet 1 (Axial):
$B_1 = \frac{\mu_0}{4\pi} \frac{2M}{r^3}$
Calculate field due to Magnet 2 (Equatorial):
$B_2 = \frac{\mu_0}{4\pi} \frac{M}{r^3}$
Since Magnet 1 is horizontal, $B_1$ is horizontal. Magnet 2 is vertical, and its equatorial field at P is anti-parallel to its axis, meaning $B_2$ is vertical. Thus, the vectors are at $90^\circ$ to each other.
\[ B_{net} = \sqrt{B_1^2 + B_2^2} = \sqrt{\left(\frac{\mu_0}{4\pi}\frac{2M}{r^3}\right)^2 + \left(\frac{\mu_0}{4\pi}\frac{M}{r^3}\right)^2} \]
\[ B_{net} = \frac{\mu_0}{4\pi} \frac{M}{r^3} \sqrt{2^2 + 1^2} = \frac{\mu_0}{4\pi} \frac{M\sqrt{5}}{r^3} \]
Substitute the values:
$\frac{\mu_0}{4\pi} = 10^{-7} \text{ Tm/A}$.
$M = 3\sqrt{5}$.
$r = 5 \times 10^{-2} \text{ m} \implies r^3 = 125 \times 10^{-6} \text{ m}^3$.
\[ B_{net} = 10^{-7} \times \frac{(3\sqrt{5})(\sqrt{5})}{125 \times 10^{-6}} \]
\[ B_{net} = 10^{-7} \times \frac{3 \times 5}{125 \times 10^{-6}} = 10^{-7} \times \frac{15}{125} \times 10^6 \]
\[ B_{net} = \frac{15}{125} \times 10^{-1} = \frac{3}{25} \times 10^{-1} \]
Convert fraction to decimal: $\frac{3}{25} = 0.12$.
\[ B_{net} = 0.12 \times 10^{-1} \text{ T} = 0.012 \text{ T} = 12 \times 10^{-3} \text{ T} \]
We are given $B_{net} = \alpha \times 10^{-3} \text{ T}$.
Therefore, $\alpha = 12$. Step 4: Final Answer:
The value of $\alpha$ is 12.
