A long solenoid has 200 turns per cm and carries a current i. The magnetic field at its center is `6.28 xx 10^(-2) Wb m^(-2)`. Another along solenoid has 100 turns per cm and it carries a current i/3 . The value of the magnetic field at its center is……... `10^(-2) Wb m^(-2)`

Explanation

To solve the problem, we will use the formula for the magnetic field inside a long solenoid, which is given by:

\[ B = \mu_0 n I \]

where:
- \( B \) is the magnetic field,
- \( \mu_0 \) is the permeability of free space (a constant),
- \( n \) is the number of turns per unit length (in turns per meter),
- \( I \) is the current flowing through the solenoid.

### Step 1: Calculate the magnetic field for the first solenoid

Given:
- The first solenoid has \( n_1 = 200 \) turns/cm, which we convert to turns/m:
\[ n_1 = 200 \, \text{turns/cm} = 20000 \, \text{turns/m} \]
- The current flowing through it is \( I_1 = i \).
- The magnetic field at its center is given as \( B_1 = 6.28 \times 10^{-2} \, \text{Wb/m}^2 \).

Using the formula:
\[ B_1 = \mu_0 n_1 I_1 \]
we can express \( \mu_0 \) in terms of \( B_1 \):
\[ \mu_0 = \frac{B_1}{n_1 I_1} \]

### Step 2: Calculate the magnetic field for the second solenoid

For the second solenoid:
- The number of turns is \( n_2 = 100 \) turns/cm, which we convert to turns/m:
\[ n_2 = 100 \, \text{turns/cm} = 10000 \, \text{turns/m} \]
- The current flowing through it is \( I_2 = \frac{i}{3} \).

Using the same formula for the magnetic field:
\[ B_2 = \mu_0 n_2 I_2 \]

### Step 3: Substitute \( \mu_0 \) from Step 1 into Step 2

Substituting \( \mu_0 \) from Step 1 into the equation for \( B_2 \):
\[ B_2 = \left(\frac{B_1}{n_1 I_1}\right) n_2 I_2 \]

### Step 4: Plug in the values

Substituting the known values:
\[ B_2 = \left(\frac{6.28 \times 10^{-2}}{20000 \cdot i}\right) \cdot (10000) \cdot \left(\frac{i}{3}\right) \]

### Step 5: Simplify the expression

Now simplify:
\[ B_2 = \frac{6.28 \times 10^{-2} \cdot 10000}{20000 \cdot 3} \]
\[ B_2 = \frac{6.28 \times 10^{-2} \cdot 10000}{60000} \]
\[ B_2 = \frac{6.28 \times 10^{-2}}{6} = 1.04666667 \times 10^{-2} \, \text{Wb/m}^2 \]
Rounding this gives:
\[ B_2 \approx 1.05 \times 10^{-2} \, \text{Wb/m}^2 \]

### Final Answer
The value of the magnetic field at the center of the second solenoid is approximately:
\[ B_2 \approx 1.05 \times 10^{-2} \, \text{Wb/m}^2 \]