A loop of flexible conducting wire of length l lies in magnetic field B which is normal to the plane of loop. A current l is passed through the loop. The tension developed in the wire to open up is

Options

A. `pi/2 Bil`
B. `(Bil)/2`
C. `(Bil)/(2pi)`
D. `Bil`

Explanation

To find the tension developed in a flexible conducting wire loop lying in a magnetic field when a current is passed through it, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a loop of flexible conducting wire of length \( l \) in a magnetic field \( B \) that is normal (perpendicular) to the plane of the loop.
- A current \( I \) flows through the loop.

2. **Determine the Radius of the Loop**:
- The circumference of the loop is equal to the length of the wire, so we have:
\[ l = 2\pi R \]
- From this, we can express the radius \( R \) as:
\[ R = \frac{l}{2\pi} \]

3. **Consider the Forces Acting on the Loop**:
- The magnetic force \( dF_m \) acting on a small segment \( dl \) of the wire due to the current in a magnetic field is given by:
\[ dF_m = I \cdot dl \cdot B \]
- The direction of this force is vertical (upward) due to the right-hand rule.

4. **Analyze the Tension in the Wire**:
- Let \( T \) be the tension in the wire. The tension has both vertical and horizontal components. For a small angle \( d\theta \), the vertical components of tension on either side of the loop will be:
\[ T \sin\left(\frac{d\theta}{2}\right) + T \sin\left(\frac{d\theta}{2}\right) = 2T \sin\left(\frac{d\theta}{2}\right) \]

5. **Equate the Forces for Equilibrium**:
- For the loop to be in equilibrium, the total magnetic force must balance the vertical components of the tension:
\[ dF_m = 2T \sin\left(\frac{d\theta}{2}\right) \]
- Substituting \( dF_m \):
\[ I \cdot dl \cdot B = 2T \sin\left(\frac{d\theta}{2}\right) \]

6. **Express \( dl \)**:
- The length \( dl \) can be expressed in terms of the radius \( R \) and the angle \( d\theta \):
\[ dl = R \cdot d\theta \]

7. **Substitute and Simplify**:
- Substitute \( dl \) into the force equation:
\[ I \cdot (R \cdot d\theta) \cdot B = 2T \sin\left(\frac{d\theta}{2}\right) \]
- For small angles, we can approximate \( \sin\left(\frac{d\theta}{2}\right) \approx \frac{d\theta}{2} \):
\[ I \cdot (R \cdot d\theta) \cdot B = 2T \cdot \frac{d\theta}{2} \]
- This simplifies to:
\[ I \cdot R \cdot B = T \]

8. **Substitute \( R \)**:
- Now substitute \( R = \frac{l}{2\pi} \):
\[ T = I \cdot \left(\frac{l}{2\pi}\right) \cdot B \]
- Thus, the tension \( T \) in the wire is:
\[ T = \frac{I \cdot l \cdot B}{2\pi} \]

### Final Answer:
The tension developed in the wire to open up the loop is:
\[ T = \frac{I \cdot l \cdot B}{2\pi} \]