Two coaxial long solenoids of equal length have current, `i_(1), i_(2)`, number of turns per unit length `n_(1), n_(2)` and radius `r_(1), r_(2)` respectively. If `n_(1)i_(1)=n_(2)i_(2)` and the two solenoids carry current in opposite sense, the magnetic energy stored per unit length is `[r_(2)gt r_(1)]`

Options

A. `(mu_(0))/(2)n_(1)^(2)i_(1)^(2)pi(r_(2)^(2)-r_(1)^(2))`
B. `mu_(0)n_(1)^(2)i_(1)^(2)pi(r_(2)^(2)-r_(1)^(2))`
C. `(mu_(0))/(2)n_(1)^(2)i_(1)^(2)pir_(1)^(2)`
D. `(mu_(0))/(2)n_(2)^(2)i_(2)^(2)pi r_(2)^(2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the Magnetic Fields
For each solenoid, the magnetic field inside can be expressed as:
- For solenoid 1:
\[ B_1 = \mu_0 n_1 i_1 \]
- For solenoid 2:
\[ B_2 = \mu_0 n_2 i_2 \]

### Step 2: Consider the Direction of Currents
Since the currents in the two solenoids are in opposite directions, the net magnetic field \( B \) in the overlapping region will be:
\[ B = B_1 - B_2 = \mu_0 n_1 i_1 - \mu_0 n_2 i_2 \]

### Step 3: Use the Given Condition
We are given that:
\[ n_1 i_1 = n_2 i_2 \]
This implies:
\[ B = \mu_0 n_1 i_1 - \mu_0 n_2 i_2 = \mu_0 (n_1 i_1 - n_2 i_2) = 0 \]
Thus, the net magnetic field in the region between the two solenoids is zero.

### Step 4: Identify the Regions of Magnetic Field
1. **Region 1 (inside solenoid 1)**: The magnetic field is \( B_1 \).
2. **Region 2 (between solenoids)**: The net magnetic field is zero.
3. **Region 3 (inside solenoid 2)**: The magnetic field is \( B_2 \).

### Step 5: Calculate the Magnetic Energy Stored
The magnetic energy density \( u \) is given by:
\[ u = \frac{B^2}{2\mu_0} \]
For the regions where the magnetic field exists (Region 1 and Region 3), we need to calculate the energy stored per unit length.

#### For Region 1:
\[ u_1 = \frac{B_1^2}{2\mu_0} = \frac{(\mu_0 n_1 i_1)^2}{2\mu_0} = \frac{\mu_0 n_1^2 i_1^2}{2} \]

#### For Region 3:
\[ u_2 = \frac{B_2^2}{2\mu_0} = \frac{(\mu_0 n_2 i_2)^2}{2\mu_0} = \frac{\mu_0 n_2^2 i_2^2}{2} \]

### Step 6: Total Energy Stored Per Unit Length
The total energy stored per unit length \( U \) is the sum of the energy stored in both regions:
\[ U = u_1 + u_2 = \frac{\mu_0 n_1^2 i_1^2}{2} + \frac{\mu_0 n_2^2 i_2^2}{2} \]

### Step 7: Substitute the Given Condition
Since \( n_1 i_1 = n_2 i_2 \), we can substitute \( n_2 i_2 \) with \( n_1 i_1 \):
\[ U = \frac{\mu_0 n_1^2 i_1^2}{2} + \frac{\mu_0 n_1^2 i_1^2}{2} = \mu_0 n_1^2 i_1^2 \]

### Step 8: Express in Terms of Area
The area \( A \) of the annular region between the two solenoids is:
\[ A = \pi (r_2^2 - r_1^2) \]
Thus, the energy stored per unit length can be expressed as:
\[ U = \frac{\mu_0 n_1^2 i_1^2}{2} \cdot A = \frac{\mu_0 n_1^2 i_1^2 \pi (r_2^2 - r_1^2)}{2} \]

### Final Result
The magnetic energy stored per unit length is:
\[ \frac{\mu_0 n_1^2 i_1^2 \pi (r_2^2 - r_1^2)}{2} \]