Two identical current carrying coaxial loops, carry current I in an opposite sense. A simple amperian loop passes through both of them once, calling the loop as C, which of the option is correct ?

Options

A. `underset(c)(oint) vecB.vecdl=pm2mu_(0)J`
B. the value of `oint vecB.vecdl` is independent of sense of C.
C. there may be a point on C where B and dl are parallel.
D. none of these

Answer

To solve the problem regarding two identical coaxial loops carrying current in opposite directions, we will follow these steps:

### Step 1: Understand the Setup
We have two identical coaxial loops. Let’s denote them as Loop 1 and Loop 2. Loop 1 carries a current \( I \) in an anticlockwise direction, while Loop 2 carries the same current \( I \) in a clockwise direction.

### Step 2: Define the Amperian Loop
We consider an Amperian loop \( C \) that passes through both loops. This loop is an imaginary closed path used to apply Ampere's circuital law.

### Step 3: Apply Ampere's Circuital Law
According to Ampere's circuital law:
\[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enc}} \]
where \( I_{\text{enc}} \) is the net current enclosed by the Amperian loop.

### Step 4: Determine the Enclosed Current
Since Loop 1 carries current \( I \) in one direction and Loop 2 carries current \( I \) in the opposite direction, the net current enclosed by the Amperian loop \( C \) is:
\[ I_{\text{enc}} = I - I = 0 \]

### Step 5: Calculate the Line Integral
Substituting \( I_{\text{enc}} = 0 \) into Ampere's law gives:
\[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 \cdot 0 = 0 \]
This implies that:
\[ \oint \mathbf{B} \cdot d\mathbf{l} = 0 \]

### Step 6: Analyze the Options
Now, we need to analyze the given options based on our findings:
- **Option A** states that \( \oint \mathbf{B} \cdot d\mathbf{l} = \pm 2\mu_0 \), which is incorrect since we found it to be 0.
- **Option B** states that the value of \( \oint \mathbf{B} \cdot d\mathbf{l} \) is independent of the sense of current, which is correct because regardless of the direction of the currents, the net current is still zero.
- **Option C** suggests there exists a point on \( C \) where \( \mathbf{B} \) and \( d\mathbf{l} \) are parallel, which is incorrect as the magnetic fields from both loops cancel each other out.
- **Option D** states that none of the above options are correct, which is also incorrect since Option B is correct.

### Conclusion
The correct answer is **Option B**.

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