What is the magnitude of force on a wire of length `0*04m` placed inside a solenoid near its centre, making an angle of `30^@` with its axis? The wire carries a current of `12A` and the magnetic field due to the solenoid is of magnitude `0*25T`.

Explanation

To find the magnitude of the force on a wire placed inside a solenoid, we can use the formula for the magnetic force on a current-carrying conductor in a magnetic field:

\[ F = I \cdot L \cdot B \cdot \sin(\theta) \]

Where:
- \( F \) is the force on the wire (in Newtons),
- \( I \) is the current through the wire (in Amperes),
- \( L \) is the length of the wire (in meters),
- \( B \) is the magnetic field strength (in Teslas),
- \( \theta \) is the angle between the wire and the magnetic field (in degrees).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Length of the wire, \( L = 0.04 \, \text{m} \)
- Current through the wire, \( I = 12 \, \text{A} \)
- Magnetic field strength, \( B = 0.25 \, \text{T} \)
- Angle with the axis, \( \theta = 30^\circ \)

2. **Convert the Angle to Radians (if necessary):**
- In this case, we can use the sine of the angle directly in degrees.

3. **Calculate \( \sin(\theta) \):**
- \( \sin(30^\circ) = 0.5 \)

4. **Substitute the Values into the Formula:**
\[ F = I \cdot L \cdot B \cdot \sin(\theta) \]
\[ F = 12 \, \text{A} \cdot 0.04 \, \text{m} \cdot 0.25 \, \text{T} \cdot 0.5 \]

5. **Perform the Calculation:**
\[ F = 12 \cdot 0.04 \cdot 0.25 \cdot 0.5 \]
\[ F = 12 \cdot 0.01 = 0.12 \, \text{N} \]

### Final Answer:
The magnitude of the force on the wire is \( 0.12 \, \text{N} \).