The number of turns and radius of cross-section of the coil of a tangent galvanometer are doubled. The reduction factor K will be

Options

A. k
B. 2k
C. 4k
D. `k /4 `

Answer

To solve the problem, we need to analyze how the reduction factor \( K \) of a tangent galvanometer changes when both the number of turns \( n \) and the radius \( r \) of the coil are doubled.

### Step-by-Step Solution:

1. **Understanding the Reduction Factor \( K \)**:
The reduction factor \( K \) for a tangent galvanometer is given by the formula:
\[ K = \frac{2 \pi r B_h}{\mu_0 n} \]
where:
- \( r \) is the radius of the coil,
- \( B_h \) is the horizontal component of the magnetic field,
- \( \mu_0 \) is the permeability of free space,
- \( n \) is the number of turns.

2. **Initial Values**:
Let the initial radius be \( r \) and the initial number of turns be \( n \). Thus, the initial reduction factor can be expressed as:
\[ K = \frac{2 \pi r B_h}{\mu_0 n} \]

3. **Doubling the Radius and Number of Turns**:
When the radius is doubled, it becomes \( 2r \), and when the number of turns is doubled, it becomes \( 2n \).

4. **New Reduction Factor \( K' \)**:
The new reduction factor \( K' \) can be calculated as:
\[ K' = \frac{2 \pi (2r) B_h}{\mu_0 (2n)} = \frac{2 \pi \cdot 2r B_h}{2 \mu_0 n} \]
Simplifying this expression gives:
\[ K' = \frac{2 \pi r B_h}{\mu_0 n} = K \]

5. **Conclusion**:
From the above calculations, we find that the new reduction factor \( K' \) is equal to the original reduction factor \( K \). Therefore, the reduction factor remains unchanged when both the number of turns and the radius of the coil are doubled.

### Final Answer:
The reduction factor \( K \) will remain the same, so:
\[ K' = K \]