A current of `10^(-5)A` produced a deflection of `10^(@)` in a moving coil galvanometer . A current of `10^(-6)amp` in the same galvanometer produces a deflection of `1^(@)`

Explanation

To solve the problem, we need to understand the relationship between the current flowing through the galvanometer and the deflection it produces. The deflection (θ) is directly proportional to the current (I) flowing through the galvanometer.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The deflection θ of the galvanometer is directly proportional to the current I flowing through it. This can be expressed as:
\[ \theta \propto I \]
or
\[ \theta = k \cdot I \]
where k is a constant.

2. **Setting Up the Known Values**:
From the problem, we have two sets of values:
- For \( I_1 = 10^{-5} \, A \), the deflection \( \theta_1 = 10^\circ \).
- For \( I_2 = 10^{-6} \, A \), we need to find \( \theta_2 \).

3. **Using the Proportionality**:
Since the deflection is directly proportional to the current, we can set up the ratio:
\[ \frac{\theta_1}{\theta_2} = \frac{I_1}{I_2} \]

4. **Substituting the Known Values**:
Plugging in the known values:
\[ \frac{10^\circ}{\theta_2} = \frac{10^{-5}}{10^{-6}} \]

5. **Simplifying the Ratio**:
Simplifying the right side:
\[ \frac{10^{-5}}{10^{-6}} = 10^{1} = 10 \]
Thus, we have:
\[ \frac{10^\circ}{\theta_2} = 10 \]

6. **Solving for \( \theta_2 \)**:
Rearranging the equation to find \( \theta_2 \):
\[ \theta_2 = \frac{10^\circ}{10} = 1^\circ \]

### Final Answer:
The deflection \( \theta_2 \) when a current of \( 10^{-6} \, A \) flows through the galvanometer is \( 1^\circ \).