A 2.5 m long straight wire having mass of 500 g is suspended in mid air by a uniform horizontal magnetic field B. If a current of 4 A is passing through the wire then the magnitude of the field is `("Take g "10 m s^(-2))`

Options

A. 0.5T
B. 0.6T
C. 0.25T
D. 0.8T

Explanation

To solve the problem, we need to find the magnitude of the magnetic field \( B \) that can suspend a wire of length \( L \) and mass \( m \) when a current \( I \) is flowing through it. The magnetic force acting on the wire must equal the weight of the wire for it to be suspended in mid-air.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Wire**:
The wire experiences two main forces: the magnetic force \( F_m \) acting upwards and the gravitational force \( F_g \) acting downwards. For the wire to be suspended in mid-air, these forces must be equal:
\[ F_m = F_g \]

2. **Express the Gravitational Force**:
The gravitational force \( F_g \) can be calculated using the formula:
\[ F_g = mg \]
where:
- \( m = 500 \text{ g} = 0.5 \text{ kg} \) (convert grams to kilograms)
- \( g = 10 \text{ m/s}^2 \)

Thus,
\[ F_g = 0.5 \text{ kg} \times 10 \text{ m/s}^2 = 5 \text{ N} \]

3. **Express the Magnetic Force**:
The magnetic force \( F_m \) on a current-carrying wire in a magnetic field is given by:
\[ F_m = BIL \]
where:
- \( B \) is the magnetic field strength,
- \( I = 4 \text{ A} \) (current),
- \( L = 2.5 \text{ m} \) (length of the wire).

4. **Set the Forces Equal**:
Since \( F_m = F_g \), we can set the equations equal to each other:
\[ BIL = mg \]

5. **Rearrange to Solve for \( B \)**:
Rearranging the equation gives us:
\[ B = \frac{mg}{IL} \]

6. **Substitute the Known Values**:
Now, substitute the values into the equation:
\[ B = \frac{(0.5 \text{ kg})(10 \text{ m/s}^2)}{(4 \text{ A})(2.5 \text{ m})} \]

7. **Calculate**:
\[ B = \frac{5 \text{ N}}{10 \text{ A m}} = 0.5 \text{ T} \]

### Final Answer:
The magnitude of the magnetic field \( B \) is \( 0.5 \text{ T} \) (Tesla).