A wire carrying current of 5 A is wound on a toroid in 4,000 turns, Calculate the magnetic field inside the core of toroid and outside the toroid if its inner radius is 15 cm and the outer radius is 18 cm.

Answer

To solve the problem, we need to calculate the magnetic field inside and outside a toroid. The toroid has a current of 5 A flowing through it and is wound with 4000 turns. The inner radius is 15 cm and the outer radius is 18 cm.

### Step-by-Step Solution:

1. **Identify the parameters:**
- Current (I) = 5 A
- Number of turns (N) = 4000
- Inner radius (r1) = 15 cm = 0.15 m
- Outer radius (r2) = 18 cm = 0.18 m

2. **Calculate the average radius (r') for the magnetic field inside the toroid:**
\[ r' = \frac{r1 + r2}{2} = \frac{0.15 + 0.18}{2} = \frac{0.33}{2} = 0.165 \text{ m} \]

3. **Use Ampere's Circuital Law to find the magnetic field inside the toroid:**
According to Ampere's Circuital Law:
\[ B \cdot L = \mu_0 \cdot I_{\text{enclosed}} \]
where \(L\) is the length of the path (which is \(2 \pi r'\)), and \(I_{\text{enclosed}} = N \cdot I\).

Thus, we can write:
\[ B \cdot (2 \pi r') = \mu_0 \cdot (N \cdot I) \]

4. **Rearranging the equation for B:**
\[ B = \frac{\mu_0 \cdot N \cdot I}{2 \pi r'} \]

5. **Substituting the values:**
- The permeability of free space, \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\)
- Substitute \(N = 4000\), \(I = 5\), and \(r' = 0.165\):
\[ B = \frac{(4\pi \times 10^{-7}) \cdot 4000 \cdot 5}{2 \pi \cdot 0.165} \]

6. **Calculating B:**
\[ B = \frac{(4 \times 10^{-7}) \cdot 4000 \cdot 5}{2 \cdot 0.165} \]
\[ = \frac{8 \times 10^{-4}}{0.33} \approx 2.424 \times 10^{-3} \, \text{T} \approx 24.24 \, \text{mT} \]

7. **Determine the magnetic field outside the toroid:**
For points outside the toroid, the enclosed current is zero because the current flowing in one direction is equal to the current flowing in the opposite direction. Therefore, according to Ampere's Law:
\[ B_{\text{outside}} = 0 \]

### Final Answers:
- Magnetic field inside the toroid: \(B \approx 24.24 \, \text{mT}\)
- Magnetic field outside the toroid: \(B = 0 \, \text{T}\)