A toroid has a core (non-ferromagnetic) of inner racius `25^(circ) cm` and outer radius 26 cm, around which 3,500 tums ofe wire are wound. If the cument in the wire is 11 A, then what is the magpetic field (in tesla) inside the core of the toroid?

Answer

To find the magnetic field inside the core of a toroid, we can use the formula:

\[ B = \frac{\mu_0 n I}{2 \pi r} \]

where:
- \(B\) is the magnetic field,
- \(\mu_0\) is the permeability of free space (\(4 \pi \times 10^{-7} \, \text{T m/A}\)),
- \(n\) is the number of turns per unit length,
- \(I\) is the current in amperes,
- \(r\) is the radius at which we want to find the magnetic field.

### Step 1: Calculate the average radius of the toroid
The average radius \(r\) can be calculated as:
\[ r = \frac{\text{inner radius} + \text{outer radius}}{2} = \frac{25 \, \text{cm} + 26 \, \text{cm}}{2} = \frac{51 \, \text{cm}}{2} = 25.5 \, \text{cm} = 0.255 \, \text{m} \]

### Step 2: Calculate the number of turns per unit length \(n\)
The total number of turns is given as 3500. The circumference of the average radius can be calculated as:
\[ \text{Circumference} = 2 \pi r = 2 \pi (0.255) \approx 1.600 \, \text{m} \]
Now, we can find \(n\):
\[ n = \frac{\text{Total number of turns}}{\text{Circumference}} = \frac{3500}{1.600} \approx 2187.5 \, \text{turns/m} \]

### Step 3: Substitute the values into the magnetic field formula
Now, we can substitute the values into the magnetic field formula:
\[ B = \frac{(4 \pi \times 10^{-7}) \times (2187.5) \times (11)}{2 \pi (0.255)} \]

### Step 4: Simplify the expression
We can simplify the expression:
\[ B = \frac{(4 \times 2187.5 \times 11)}{2 \times 0.255} \times 10^{-7} \]
Calculating the numerator:
\[ 4 \times 2187.5 \times 11 = 96350 \]
Now substituting back:
\[ B = \frac{96350}{0.51} \times 10^{-7} \approx 188,000 \times 10^{-7} = 1.88 \times 10^{-3} \, \text{T} \]

### Final Answer
The magnetic field inside the core of the toroid is approximately:
\[ B \approx 1.88 \, \text{mT} \, (\text{milliTesla}) \]

---