The tangents deflection produced in tan A and B positions by a short magnet at equal distances are in the ratio .

Answer

To solve the problem of finding the ratio of the tangents of deflection produced by a short magnet at equal distances, we will follow these steps:

### Step 1: Define the Tangent Deflections
We start by defining the tangent deflections at positions A and B. The tangent deflection at position A is given by:

\[ \tan A = \frac{\mu_0 \cdot 2M \cdot D}{4\pi (D^2 - L^2)^{2}} \]

And the tangent deflection at position B is given by:

\[ \tan B = \frac{\mu_0 \cdot M}{4\pi (D^2 + L^2)^{3/2}} \]

Where:
- \( \mu_0 \) is the permeability of free space,
- \( M \) is the magnetic moment,
- \( D \) is the distance between the two poles,
- \( L \) is the distance from one pole to the center of the magnet.

### Step 2: Set Up the Ratio
We need to find the ratio of \( \tan A \) to \( \tan B \):

\[ \frac{\tan A}{\tan B} = \frac{\frac{\mu_0 \cdot 2M \cdot D}{4\pi (D^2 - L^2)^{2}}}{\frac{\mu_0 \cdot M}{4\pi (D^2 + L^2)^{3/2}}} \]

### Step 3: Simplify the Ratio
Cancelling out common terms \( \mu_0 \) and \( 4\pi \):

\[ \frac{\tan A}{\tan B} = \frac{2D}{(D^2 - L^2)^{2}} \cdot (D^2 + L^2)^{3/2} \]

### Step 4: Factor Out Common Terms
Now, we can factor out \( D^2 \) from both the numerator and the denominator:

\[ \frac{\tan A}{\tan B} = \frac{2D \cdot (D^2 + L^2)^{3/2}}{(D^2 - L^2)^{2}} \]

### Step 5: Neglect Higher Order Terms
Assuming \( L \) is much smaller than \( D \) (i.e., \( \frac{L}{D} << 1 \)), we can neglect higher order terms:

\[ \frac{\tan A}{\tan B} \approx \frac{2D \cdot D^3}{D^4} = 2 \]

### Step 6: Conclusion
Thus, the ratio of the tangents of deflection at positions A and B is:

\[ \frac{\tan A}{\tan B} = 2 : 1 \]

### Final Answer:
The ratio of \( \tan A \) to \( \tan B \) is \( 2 : 1 \).

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