If the current (I) flowing through a circular coil, its radius (R) and number of turns (N) in it are each doubled, magnetic flux density at its centre becomes :

Options

A. two times
B. four times
C. eight times
D. sixteen times

Explanation

To solve the problem, we need to understand how the magnetic flux density (B) at the center of a circular coil is affected when the current (I), radius (R), and number of turns (N) are all doubled.

### Step-by-Step Solution:

1. **Understand the Formula for Magnetic Flux Density**:
The magnetic flux density (B) at the center of a circular coil is given by the formula:
\[ B = \frac{\mu_0 N I}{2R} \]
where:
- \( B \) = magnetic flux density
- \( \mu_0 \) = permeability of free space (a constant)
- \( N \) = number of turns in the coil
- \( I \) = current flowing through the coil
- \( R \) = radius of the coil

2. **Identify the Changes**:
According to the problem, the current (I), radius (R), and number of turns (N) are each doubled:
- New current \( I' = 2I \)
- New radius \( R' = 2R \)
- New number of turns \( N' = 2N \)

3. **Substitute the New Values into the Formula**:
Substitute the new values into the magnetic flux density formula:
\[ B' = \frac{\mu_0 N' I'}{2R'} \]
Substituting the new values:
\[ B' = \frac{\mu_0 (2N) (2I)}{2(2R)} \]

4. **Simplify the Expression**:
Simplify the expression:
\[ B' = \frac{\mu_0 \cdot 2N \cdot 2I}{4R} \]
\[ B' = \frac{\mu_0 N I}{2R} \cdot 2 \]
Notice that \( \frac{\mu_0 N I}{2R} \) is the original magnetic flux density \( B \):
\[ B' = 2B \]

5. **Conclusion**:
Thus, when the current, radius, and number of turns are each doubled, the magnetic flux density at the center of the coil becomes:
\[ B' = 2B \]
Therefore, the answer is that the magnetic flux density at its center becomes **two times** its initial value.

### Final Answer:
The magnetic flux density at its center becomes **2B** (two times the initial value).