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:
The point P is located 5 cm (\( 0.05 \text{ m} \)) from the center of each magnet. Relative to one magnet, P is in an axial position. Relative to the other magnet, P is in an equatorial position. The net magnetic field is the vector sum of these two fields.
Step 2: Key Formula or Approach:
1. Axial field: \( B_A = \frac{\mu_0}{4\pi} \frac{2M}{r^3} \).
2. Equatorial field: \( B_E = \frac{\mu_0}{4\pi} \frac{M}{r^3} \).
3. Net field: \( B = \sqrt{B_A^2 + B_E^2} \).
Step 3: Detailed Explanation:
Given \( M = 3\sqrt{5} \text{ J/T} \) and \( r = 0.05 \text{ m} \).
Let \( K = \frac{\mu_0}{4\pi} \frac{M}{r^3} = 10^{-7} \frac{3\sqrt{5}}{(0.05)^3} = 10^{-7} \frac{3\sqrt{5}}{125 \times 10^{-6}} = \frac{3\sqrt{5}}{125} \times 10^{-1} \text{ T} \).
Then \( B_A = 2K \) and \( B_E = K \).
The fields are perpendicular, so the net field is:
\[ B = \sqrt{(2K)^2 + K^2} = \sqrt{5K^2} = K\sqrt{5} \]
Substitute \( K \):
\[ B = \left( \frac{3\sqrt{5}}{125} \times 10^{-1} \right) \sqrt{5} = \frac{3 \times 5}{125} \times 10^{-1} = \frac{15}{125} \times 0.1 \text{ T} \]
\[ B = 0.12 \times 0.1 = 0.012 \text{ T} = 12 \times 10^{-3} \text{ T} \]
Comparing with \( \alpha \times 10^{-3} \), we get \( \alpha = 12 \).
Step 4: Final Answer:
The value of \( \alpha \) is 12.
