A current or 0.5 A produces a deflection of 60° in a tangent galvanometer. The cunent that produces a deflection of 30° in the same galvanometer is

Options

A. `(0.5)2A`
B. `(0.5)/3A`
C. `(0.5)/(sqrt2)A`
D. `(0.5)/(sqrt3)A`

Answer

To solve the problem, we will use the relationship between the deflection angle and the current in a tangent galvanometer. The tangent of the angle of deflection is directly proportional to the current flowing through the galvanometer.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Current \( I_1 = 0.5 \, \text{A} \) produces a deflection \( \theta_1 = 60^\circ \).
- We need to find the current \( I_2 \) that produces a deflection \( \theta_2 = 30^\circ \).

2. **Use the Relationship:**
The relationship between the current and the angle of deflection in a tangent galvanometer is given by:
\[ \tan(\theta) \propto I \]
This can be expressed as:
\[ \frac{\tan(\theta_1)}{\tan(\theta_2)} = \frac{I_1}{I_2} \]

3. **Substitute the Known Values:**
Plugging in the values we have:
\[ \frac{\tan(60^\circ)}{\tan(30^\circ)} = \frac{0.5}{I_2} \]

4. **Calculate the Tangents:**
We know:
- \( \tan(60^\circ) = \sqrt{3} \)
- \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \)

Thus, substituting these values:
\[ \frac{\sqrt{3}}{\frac{1}{\sqrt{3}}} = \frac{0.5}{I_2} \]

5. **Simplify the Equation:**
The left-hand side simplifies to:
\[ \sqrt{3} \times \sqrt{3} = 3 \]
Therefore, we have:
\[ 3 = \frac{0.5}{I_2} \]

6. **Solve for \( I_2 \):**
Rearranging the equation gives:
\[ I_2 = \frac{0.5}{3} \]
Thus,
\[ I_2 = \frac{1}{6} \, \text{A} \approx 0.167 \, \text{A} \]

### Final Answer:
The current that produces a deflection of \( 30^\circ \) in the same galvanometer is \( \frac{1}{6} \, \text{A} \) or approximately \( 0.167 \, \text{A} \).

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