A wire is wound on a long rod of material of relative permeability `mu_(r) = 4000` to make a solenoid. If the current through the wire is 5 A and number of turns per unit length is 1000 per metre, then the magnetic field inside the solenoid is

Options

A. `4 pi mT`
B. `8 pi mT`
C. `4 pi T`
D. `8 pi T`

Explanation

To find the magnetic field inside the solenoid, we can use the formula:

\[ B = \mu N I \]

where:
- \( B \) is the magnetic field inside the solenoid,
- \( \mu \) is the permeability of the material,
- \( N \) is the number of turns per unit length,
- \( I \) is the current flowing through the wire.

### Step 1: Calculate the permeability \( \mu \)

The permeability \( \mu \) is given by:

\[ \mu = \mu_0 \mu_r \]

where:
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \),
- \( \mu_r \) is the relative permeability of the material, which is given as \( 4000 \).

Substituting the values:

\[ \mu = (4\pi \times 10^{-7}) \times 4000 \]

### Step 2: Calculate \( \mu \)

Now we can calculate \( \mu \):

\[ \mu = 4\pi \times 10^{-7} \times 4000 \]
\[ \mu = 16\pi \times 10^{-4} \, \text{T m/A} \]

### Step 3: Substitute values into the magnetic field formula

Now, we can substitute the values of \( \mu \), \( N \), and \( I \) into the magnetic field formula:

Given:
- \( N = 1000 \, \text{turns/m} \)
- \( I = 5 \, \text{A} \)

So,

\[ B = \mu N I \]
\[ B = (16\pi \times 10^{-4}) \times 1000 \times 5 \]

### Step 4: Calculate \( B \)

Now, let's calculate \( B \):

\[ B = 16\pi \times 10^{-4} \times 1000 \times 5 \]
\[ B = 16\pi \times 5 \times 10^{-1} \]
\[ B = 80\pi \times 10^{-1} \]
\[ B = 8\pi \, \text{T} \]

### Final Answer

Thus, the magnetic field inside the solenoid is:

\[ B = 8\pi \, \text{T} \]