A loop of flexible conducting wire lies in a magnetic field of 2.0 T with its plane perpendicular to the field. The length of the wire is 1 m. When a current of 1.1A is passed through the loop, it opens into a circle, then the tension developed in the wire is

Options

A. 0.15 N
B. 0.25 N
C. 0.35 N
D. 0.45 N

Answer

To solve the problem of finding the tension developed in a flexible conducting wire loop placed in a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a loop of flexible conducting wire with a total length of 1 meter.
- The loop is placed in a magnetic field of strength \( B = 2.0 \, \text{T} \) (Tesla).
- A current \( I = 1.1 \, \text{A} \) (Ampere) is passed through the loop, which causes it to open into a circle.

2. **Determine the Radius of the Circle**:
- The length of the wire is equal to the circumference of the circle formed when the wire opens up.
- The circumference \( C \) of a circle is given by:
\[ C = 2\pi r \]
- Setting the circumference equal to the length of the wire:
\[ 2\pi r = 1 \, \text{m} \]
- Solving for \( r \):
\[ r = \frac{1}{2\pi} \approx 0.159 \, \text{m} \]

3. **Calculating the Tension in the Wire**:
- The tension \( T \) in the wire can be calculated using the formula:
\[ T = B \cdot I \cdot r \]
- Substituting the known values:
\[ T = 2.0 \, \text{T} \cdot 1.1 \, \text{A} \cdot \frac{1}{2\pi} \]
- Calculating \( T \):
\[ T = 2.0 \cdot 1.1 \cdot \frac{1}{2\pi} \approx \frac{2.2}{6.283} \approx 0.35 \, \text{N} \]

4. **Final Result**:
- The tension developed in the wire is approximately \( 0.35 \, \text{N} \).

### Summary:
The tension developed in the wire when a current of 1.1 A is passed through it in a magnetic field of 2.0 T is approximately **0.35 N**.

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