The magnetic field inside a toroidal solenoid of radius R is B. If the current through it is doubled and its radius is also doubled, keeping the number of turns same, the magnetic field produced by it will be

Options

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

Answer

To solve the problem, we need to analyze how the magnetic field inside a toroidal solenoid changes when the current and radius are modified.

### Step-by-Step Solution:

1. **Understand the Formula for Magnetic Field in a Toroidal Solenoid:**
The magnetic field \( B \) inside a toroidal solenoid is given by the formula:
\[ B = \frac{\mu_0 n I}{2 \pi r} \]
where:
- \( \mu_0 \) is the permeability of free space,
- \( n \) is the number of turns per unit length,
- \( I \) is the current flowing through the solenoid,
- \( r \) is the radius of the toroid.

2. **Identify Initial Conditions:**
Let the initial current be \( I \) and the initial radius be \( R \). The initial magnetic field \( B \) can be expressed as:
\[ B = \frac{\mu_0 n I}{2 \pi R} \]

3. **Change the Current and Radius:**
According to the problem, the current is doubled:
\[ I' = 2I \]
The radius is also doubled:
\[ r' = 2R \]
The number of turns \( N \) remains the same, so the number of turns per unit length \( n \) does not change.

4. **Calculate the New Magnetic Field:**
Substitute the new values into the magnetic field formula:
\[ B' = \frac{\mu_0 n I'}{2 \pi r'} = \frac{\mu_0 n (2I)}{2 \pi (2R)} \]
Simplifying this expression:
\[ B' = \frac{\mu_0 n (2I)}{4 \pi R} = \frac{1}{2} \cdot \frac{\mu_0 n I}{2 \pi R} \]
Thus, we find:
\[ B' = \frac{1}{2} B \]

5. **Conclusion:**
The new magnetic field \( B' \) is half of the original magnetic field \( B \):
\[ B' = \frac{B}{2} \]

### Final Answer:
The magnetic field produced by the toroidal solenoid after doubling the current and radius, while keeping the number of turns the same, will be \( \frac{B}{2} \).