A current of `10` ampere is flowing in a wire of length `1.5m`. A force of `15 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^@`

Explanation

To solve the problem, we need to find the angle \( \theta \) between the magnetic field and the direction of the current using the formula for the force acting on a current-carrying conductor in a magnetic field.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Current, \( I = 10 \, \text{A} \)
- Length of the wire, \( L = 1.5 \, \text{m} \)
- Magnetic field strength, \( B = 2 \, \text{T} \)
- Force acting on the wire, \( F = 15 \, \text{N} \)

2. **Use the Formula for Magnetic Force:**
The magnetic force \( F \) on a current-carrying wire is given by the formula:
\[ F = BIL \sin \theta \]
where:
- \( F \) is the force,
- \( B \) is the magnetic field strength,
- \( I \) is the current,
- \( L \) is the length of the wire,
- \( \theta \) is the angle between the magnetic field and the current.

3. **Rearrange the Formula to Solve for \( \sin \theta \):**
We can rearrange the formula to isolate \( \sin \theta \):
\[ \sin \theta = \frac{F}{BIL} \]

4. **Substitute the Known Values:**
Now, substitute the values into the equation:
\[ \sin \theta = \frac{15 \, \text{N}}{(2 \, \text{T})(10 \, \text{A})(1.5 \, \text{m})} \]

5. **Calculate the Denominator:**
Calculate the product in the denominator:
\[ BIL = 2 \times 10 \times 1.5 = 30 \, \text{T.A.m} \]

6. **Calculate \( \sin \theta \):**
Now substitute this back into the equation:
\[ \sin \theta = \frac{15}{30} = \frac{1}{2} \]

7. **Determine \( \theta \):**
To find \( \theta \), we take the inverse sine:
\[ \theta = \sin^{-1}\left(\frac{1}{2}\right) \]
From trigonometric values, we know:
\[ \theta = 30^\circ \]

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

---