A current of 5 ampere is flowing in a wire of length 1.5 metres. A force of `7.5 N` acts on it when it is placed in a uniform magnetic field of 2 Tesla. The angle between the magnetic field and the direction of the current is

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To find the angle between the magnetic field and the direction of the current, we can use the formula for the magnetic force acting on a current-carrying conductor placed in a magnetic field. The formula is given by:

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

Where:
- \( F \) = force acting on the wire (in Newtons)
- \( I \) = current flowing through the wire (in Amperes)
- \( L \) = length of the wire (in meters)
- \( B \) = magnetic field strength (in Tesla)
- \( \theta \) = angle between the current direction and the magnetic field direction

### Step-by-Step Solution:

1. **Identify the given values:**
- Current, \( I = 5 \, \text{A} \)
- Length of the wire, \( L = 1.5 \, \text{m} \)
- Force, \( F = 7.5 \, \text{N} \)
- Magnetic field strength, \( B = 2 \, \text{T} \)

2. **Substitute the values into the formula:**
\[ 7.5 = 5 \cdot 1.5 \cdot 2 \cdot \sin(\theta) \]

3. **Calculate the product on the right side:**
\[ 5 \cdot 1.5 = 7.5 \]
\[ 7.5 \cdot 2 = 15 \]
So, we have:
\[ 7.5 = 15 \cdot \sin(\theta) \]

4. **Rearranging the equation to solve for \( \sin(\theta) \):**
\[ \sin(\theta) = \frac{7.5}{15} \]
\[ \sin(\theta) = 0.5 \]

5. **Finding the angle \( \theta \):**
The angle whose sine is \( 0.5 \) is:
\[ \theta = 30^\circ \]

### Final Answer:
The angle between the magnetic field and the direction of the current is \( 30^\circ \).