A long solenoid with 40 turns per cm carries a current of 1 A. The magnetic energy stored per unit volume is..... J `m^(-3)`.

Options

A. `3.2pi`
B. `32pi`
C. `1.6pi`
D. `6.4pi`

Answer

To find the magnetic energy stored per unit volume in a long solenoid, we can follow these steps:

### Step 1: Determine the number of turns per meter
The problem states that the solenoid has 40 turns per centimeter. To convert this to turns per meter, we use the conversion factor that 1 meter = 100 centimeters.

\[ n = 40 \, \text{turns/cm} \times 100 \, \text{cm/m} = 4000 \, \text{turns/m} \]

### Step 2: Use the formula for magnetic energy density
The magnetic energy density \( u \) (energy per unit volume) in a solenoid is given by the formula:

\[ u = \frac{1}{2} \frac{B^2}{\mu_0} \]

where \( B \) is the magnetic field inside the solenoid and \( \mu_0 \) is the permeability of free space.

### Step 3: Find the magnetic field \( B \) in the solenoid
The magnetic field inside a long solenoid is given by:

\[ B = \mu_0 n I \]

where:
- \( n \) is the number of turns per unit length (in turns/m),
- \( I \) is the current (in Amperes).

Substituting the values we have:

\[ B = \mu_0 (4000 \, \text{turns/m}) (1 \, \text{A}) \]

### Step 4: Substitute \( B \) into the energy density formula
Now, we can substitute \( B \) back into the energy density formula:

\[ u = \frac{1}{2} \frac{(\mu_0 n I)^2}{\mu_0} \]

This simplifies to:

\[ u = \frac{1}{2} \mu_0 n^2 I^2 \]

### Step 5: Substitute known values
The value of \( \mu_0 \) is approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \). Now substituting \( n = 4000 \, \text{turns/m} \) and \( I = 1 \, \text{A} \):

\[ u = \frac{1}{2} (4\pi \times 10^{-7}) (4000)^2 (1)^2 \]

Calculating \( (4000)^2 \):

\[ (4000)^2 = 16 \times 10^6 \]

Now substituting this back:

\[ u = \frac{1}{2} (4\pi \times 10^{-7}) (16 \times 10^6) \]

### Step 6: Simplify the expression
\[ u = 2\pi \times 10^{-7} \times 16 \times 10^6 \]
\[ u = 32\pi \times 10^{-1} \, \text{J/m}^3 \]

### Step 7: Final result
Thus, the magnetic energy stored per unit volume is:

\[ u = 3.2\pi \, \text{J/m}^3 \]

### Final Answer
The magnetic energy stored per unit volume is \( 3.2\pi \, \text{J/m}^3 \).

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