If a wire of length `l` meter placed in uniform magnetic field 1.5 Tesla at angle `30^(@)` with magnetic field. The current in a wire 10 amp. Then force on a wire will be

Options

A. 7.5 N
B. 1.5 N
C. 0.5 N
D. 2.5 N

Answer

To solve the problem, we will use the formula for the magnetic force on a current-carrying wire, which is given by:

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

Where:
- \( F \) is the force on the wire (in Newtons),
- \( I \) is the current (in Amperes),
- \( L \) is the length of the wire (in meters),
- \( B \) is the magnetic field strength (in Tesla),
- \( \theta \) 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, \( L = 1 \, \text{m} \) (assumed from the context),
- Magnetic field strength, \( B = 1.5 \, \text{T} \),
- Current in the wire, \( I = 10 \, \text{A} \),
- Angle with the magnetic field, \( \theta = 30^\circ \).

2. **Convert the Angle to Radians (if necessary):**
- In this case, we can use the sine function directly in degrees, so no conversion is needed.

3. **Calculate \( \sin(\theta) \):**
- \( \sin(30^\circ) = \frac{1}{2} \).

4. **Substitute the Values into the Formula:**
\[ F = I \cdot L \cdot B \cdot \sin(\theta) \]
\[ F = 10 \, \text{A} \cdot 1 \, \text{m} \cdot 1.5 \, \text{T} \cdot \frac{1}{2} \]

5. **Perform the Calculation:**
- First, calculate \( 10 \cdot 1 \cdot 1.5 \cdot \frac{1}{2} \):
\[ F = 10 \cdot 1.5 \cdot 0.5 = 10 \cdot 0.75 = 7.5 \, \text{N} \]

6. **Conclusion:**
- The force on the wire is \( F = 7.5 \, \text{N} \).

### Final Answer:
The force on the wire is **7.5 Newtons**.

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