When the number of turns in a toroidal coil is doubled, then the value of magnetic flux density will becomes-

Options

A. four times
B. eight times
C. half
D. double

Explanation

To solve the question regarding the effect of doubling the number of turns in a toroidal coil on the magnetic flux density, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Magnetic Flux Density (B):**
Magnetic flux density (B) is related to the magnetic field strength in a toroidal coil. The magnetic field inside a toroid can be expressed as:
\[ B = \mu_0 \cdot n \cdot I \]
where:
- \(B\) is the magnetic flux density,
- \(\mu_0\) is the permeability of free space,
- \(n\) is the number of turns per unit length,
- \(I\) is the current flowing through the coil.

2. **Define the Number of Turns (N):**
The number of turns \(N\) in the toroidal coil can be expressed as:
\[ n = \frac{N}{2\pi r} \]
where \(r\) is the radius of the toroid.

3. **Effect of Doubling the Number of Turns:**
If the number of turns \(N\) is doubled (i.e., \(N' = 2N\)), then the new number of turns per unit length becomes:
\[ n' = \frac{N'}{2\pi r} = \frac{2N}{2\pi r} = 2 \cdot \frac{N}{2\pi r} = 2n \]

4. **Substituting into the Magnetic Field Equation:**
Substitute \(n'\) back into the magnetic field equation:
\[ B' = \mu_0 \cdot n' \cdot I = \mu_0 \cdot (2n) \cdot I = 2 \cdot (\mu_0 \cdot n \cdot I) = 2B \]

5. **Conclusion:**
Therefore, when the number of turns in a toroidal coil is doubled, the magnetic flux density \(B\) also doubles. Thus, the final answer is:
\[ \text{The magnetic flux density becomes doubled.} \]

### Final Answer:
The value of magnetic flux density will become **doubled**.

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