A magnetic compass needl oscillates 30 times per minute at a place where the dip is `45^(@)` and 40 times per minute where the dip is `30^(@)` if `B_(1)` and `B_(2)` are respectively the total magnetic field due to the earth at the two places then the ratio `B_(1)//B_(2)` is best given by

Options

A. 1.8
B. 0.7
C. 3.6
D. 2.2

Answer

To solve the problem, we need to find the ratio of the total magnetic fields \( B_1 \) and \( B_2 \) at two different locations where the compass needle oscillates at different frequencies and dip angles.

### Step 1: Understand the relationship between frequency and magnetic field
The frequency of oscillation \( F \) of a magnetic compass needle is given by the formula:
\[ F = \frac{1}{2\pi} \sqrt{\frac{mB}{I}} \]
where:
- \( m \) is the magnetic moment,
- \( B \) is the magnetic field,
- \( I \) is the moment of inertia.

### Step 2: Set up the equations for the two locations
For the first location (where the dip angle is \( 45^\circ \) and frequency is \( 30 \) oscillations per minute):
\[ F_1 = 30 \text{ oscillations/min} = \frac{30}{60} \text{ oscillations/sec} = 0.5 \text{ Hz} \]
Using the formula:
\[ 0.5 = \frac{1}{2\pi} \sqrt{\frac{m B_1 \cos(45^\circ)}{I}} \]
Squaring both sides gives:
\[ (0.5 \cdot 2\pi)^2 = \frac{m B_1 \cos(45^\circ)}{I} \]

For the second location (where the dip angle is \( 30^\circ \) and frequency is \( 40 \) oscillations per minute):
\[ F_2 = 40 \text{ oscillations/min} = \frac{40}{60} \text{ oscillations/sec} = \frac{2}{3} \text{ Hz} \]
Using the formula:
\[ \frac{2}{3} = \frac{1}{2\pi} \sqrt{\frac{m B_2 \cos(30^\circ)}{I}} \]
Squaring both sides gives:
\[ \left(\frac{2}{3} \cdot 2\pi\right)^2 = \frac{m B_2 \cos(30^\circ)}{I} \]

### Step 3: Simplify the equations
From the first location:
\[ \frac{(0.5 \cdot 2\pi)^2 I}{m \cos(45^\circ)} = B_1 \]
From the second location:
\[ \frac{\left(\frac{2}{3} \cdot 2\pi\right)^2 I}{m \cos(30^\circ)} = B_2 \]

### Step 4: Find the ratio \( \frac{B_1}{B_2} \)
Now, we can find the ratio \( \frac{B_1}{B_2} \):
\[ \frac{B_1}{B_2} = \frac{(0.5 \cdot 2\pi)^2 \cos(30^\circ)}{\left(\frac{2}{3} \cdot 2\pi\right)^2 \cos(45^\circ)} \]
Canceling out common terms:
\[ \frac{B_1}{B_2} = \frac{(0.5)^2 \cos(30^\circ)}{\left(\frac{2}{3}\right)^2 \cos(45^\circ)} \]
Substituting the values of \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \) and \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ \frac{B_1}{B_2} = \frac{(0.5)^2 \cdot \frac{\sqrt{3}}{2}}{\left(\frac{2}{3}\right)^2 \cdot \frac{1}{\sqrt{2}}} \]

### Step 5: Calculate the ratio
Calculating the values:
\[ = \frac{0.25 \cdot \frac{\sqrt{3}}{2}}{\frac{4}{9} \cdot \frac{1}{\sqrt{2}}} \]
\[ = \frac{0.25 \cdot \sqrt{3} \cdot 9 \sqrt{2}}{4} \]
\[ = \frac{0.25 \cdot 9 \sqrt{6}}{4} \]
\[ = \frac{9 \sqrt{6}}{16} \]

### Final Step: Approximate the value
Calculating this gives us the ratio \( \frac{B_1}{B_2} \approx 0.7 \).

### Conclusion
Thus, the ratio \( \frac{B_1}{B_2} \) is approximately \( 0.7 \).

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