A thin wire of length `l` is carrying a constant current. The wire is bent to form a circular coil. If radius of the coil, thus formed, is equal to R and number of turns in it is equal to `n`, then which of the following graphs represent (s) variation of magnetic field induction (B) at centre of the coil

Options

A.
B.
C.
D.

Answer

To solve the problem, we need to determine the variation of the magnetic field induction (B) at the center of a circular coil formed by bending a thin wire of length \( L \) carrying a constant current \( I \). The coil has \( n \) turns and a radius \( R \).

### Step-by-Step Solution:

1. **Understanding the Length of the Wire**:
The total length of the wire is given as \( L \). When the wire is bent to form a circular coil with \( n \) turns, the relationship between the length of the wire and the radius of the coil can be expressed as:
\[ L = n \cdot (2\pi R) \]
This equation states that the total length of the wire \( L \) is equal to the number of turns \( n \) multiplied by the circumference of each turn \( (2\pi R) \).

2. **Expressing Radius in Terms of \( L \) and \( n \)**:
From the equation above, we can express the radius \( R \) in terms of \( L \) and \( n \):
\[ R = \frac{L}{2\pi n} \]

3. **Magnetic Field at the Center of the Coil**:
The magnetic field \( B \) at the center of a circular coil with \( n \) turns carrying a current \( I \) is given by the formula:
\[ B = \frac{\mu_0 n I}{2R} \]
where \( \mu_0 \) is the permeability of free space.

4. **Substituting for \( R \)**:
Now, substituting the expression for \( R \) from step 2 into the magnetic field formula:
\[ B = \frac{\mu_0 n I}{2 \left(\frac{L}{2\pi n}\right)} = \frac{\mu_0 n I \cdot 2\pi n}{2L} \]
Simplifying this gives:
\[ B = \frac{\mu_0 \pi I n^2}{L} \]

5. **Identifying the Relationship**:
From the equation \( B = k n^2 \) (where \( k = \frac{\mu_0 \pi I}{L} \)), we see that the magnetic field \( B \) is proportional to the square of the number of turns \( n \). This indicates that as the number of turns increases, the magnetic field increases quadratically.

6. **Graph Representation**:
The relationship \( B \propto n^2 \) suggests that if we plot \( B \) against \( n \), the graph will be a parabola opening upwards. Therefore, the correct graph representing the variation of magnetic field induction at the center of the coil is a parabola.

### Conclusion:
The correct option is the graph that represents a parabola facing upwards, indicating that the magnetic field induction \( B \) increases with the square of the number of turns \( n \).