A current - carrying wire of certain length is bent to form an arc that substends an angle `theta` at the centre. Let B be the magnetic field at the centre due to this wire. The correct graph between B and `theta` is

Options

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Answer

To solve the problem, we need to determine the relationship between the magnetic field \( B \) at the center of an arc of a current-carrying wire and the angle \( \theta \) that the arc subtends at the center.

### Step-by-Step Solution:

1. **Understanding the Arc Length**:
The length \( L \) of the arc is given by the formula:
\[ L = r \cdot \theta \]
where \( r \) is the radius of the arc and \( \theta \) is in radians.

2. **Expressing the Radius in Terms of Length and Angle**:
Rearranging the above equation gives:
\[ r = \frac{L}{\theta} \]

3. **Magnetic Field Due to the Arc**:
The magnetic field \( B \) at the center of the arc due to a current \( I \) flowing through it is given by:
\[ B = \frac{\mu_0 I}{2r} \cdot \frac{\theta}{2\pi} \]
where \( \mu_0 \) is the permeability of free space.

4. **Substituting for Radius**:
Substitute \( r = \frac{L}{\theta} \) into the equation for \( B \):
\[ B = \frac{\mu_0 I}{2 \cdot \frac{L}{\theta}} \cdot \frac{\theta}{2\pi} \]
Simplifying this gives:
\[ B = \frac{\mu_0 I \cdot \theta}{4\pi L} \]

5. **Identifying the Relationship**:
From the equation \( B = k \cdot \theta \) where \( k = \frac{\mu_0 I}{4\pi L} \), we can see that \( B \) is directly proportional to \( \theta \).

6. **Considering the Range of \( \theta \)**:
As \( \theta \) increases, \( B \) also increases linearly. However, if we consider the case where \( \theta \) approaches zero, \( B \) will also approach zero.

7. **Graphing the Relationship**:
The graph of \( B \) versus \( \theta \) will be a straight line passing through the origin (0,0) and will have a positive slope.

### Conclusion:
The correct graph between \( B \) and \( \theta \) is a straight line that passes through the origin.