A long solenoid has 200 turns per cm and carries a current `i`. The magnetic field at its centre is `6.28xx10^(-2) weber//cm^(2)`. Another long soloenoid has 100 turns per cm and it carries a current `(i)/(3)`. The value of the magnetic field at its centre is

Options

A. (a)`1.05xx10^(-4)Wb//m^(2)`
B. (b)`1.05xx10^(-2)Wb//m^(2)`
C. (c)`1.05xx10^(-5)Wb//m^(2)`
D. (d)`1.05xx10^(-5)Wb//m^(2)`

Explanation

To solve the problem step by step, 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 meters),
- \( I \) is the current flowing through the solenoid.

### Step 1: Analyze the first solenoid
The first solenoid has:
- Number of turns \( n_1 = 200 \) turns/cm = \( 200 \times 100 = 20000 \) turns/m (since 1 cm = 0.01 m).
- Current \( I_1 = i \).
- Magnetic field \( B_1 = 6.28 \times 10^{-2} \) Weber/cm² = \( 6.28 \times 10^{-2} \times 10^{-4} = 6.28 \times 10^{-6} \) Weber/m² (since 1 cm² = \( 10^{-4} \) m²).

Using the formula for the magnetic field:

\[ B_1 = \mu_0 n_1 I_1 \]

Substituting the known values:

\[ 6.28 \times 10^{-6} = \mu_0 (20000) (i) \]

### Step 2: Analyze the second solenoid
The second solenoid has:
- Number of turns \( n_2 = 100 \) turns/cm = \( 100 \times 100 = 10000 \) turns/m.
- Current \( I_2 = \frac{i}{3} \).

Using the formula for the magnetic field:

\[ B_2 = \mu_0 n_2 I_2 \]

Substituting the known values:

\[ B_2 = \mu_0 (10000) \left(\frac{i}{3}\right) \]

### Step 3: Relate the two equations
From the first solenoid, we can express \( \mu_0 i \):

\[ \mu_0 i = \frac{6.28 \times 10^{-6}}{20000} \]

Now substituting this into the equation for \( B_2 \):

\[ B_2 = \mu_0 (10000) \left(\frac{i}{3}\right) = \frac{6.28 \times 10^{-6}}{20000} \times 10000 \times \frac{1}{3} \]

### Step 4: Simplify the expression
Now simplifying \( B_2 \):

\[ B_2 = \frac{6.28 \times 10^{-6}}{20000} \times 10000 \times \frac{1}{3} = \frac{6.28 \times 10^{-6}}{2} \times \frac{1}{3} \]

Calculating the numerical value:

\[ B_2 = \frac{6.28 \times 10^{-6}}{6} = 1.04667 \times 10^{-6} \text{ Weber/m}^2 \]

### Step 5: Final result
Thus, the magnetic field at the center of the second solenoid is:

\[ B_2 \approx 1.05 \times 10^{-6} \text{ Weber/m}^2 \]