A straight wire having mass of 1.2 kg and length of 1m carries a current of 5 A. If the wire is suspended in mid-air by a uniform horizontal magnetic field, then the magnitude of field is

Options

A. 0.65T
B. 1.53T
C. 2.4T
D. 3.2T

Explanation

To solve the problem, we need to find the magnitude of the magnetic field (B) that can suspend a straight wire carrying a current. Here are the steps to derive the solution:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Wire:**
The wire is suspended in a magnetic field, and the forces acting on it are:
- The gravitational force (weight) acting downward: \( F_g = mg \)
- The magnetic force acting on the wire due to the current in the magnetic field: \( F_m = BIL \)

2. **Set Up the Equation:**
For the wire to be in equilibrium (suspended in mid-air), the magnetic force must balance the weight of the wire. Thus, we can write:
\[ F_m = F_g \]
Substituting the expressions for the forces, we have:
\[ BIL = mg \]

3. **Rearrange to Solve for B:**
To find the magnetic field (B), we rearrange the equation:
\[ B = \frac{mg}{IL} \]

4. **Substitute the Given Values:**
We are given the following values:
- Mass of the wire, \( m = 1.2 \, \text{kg} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
- Current, \( I = 5 \, \text{A} \)
- Length of the wire, \( L = 1 \, \text{m} \)

Now substituting these values into the equation:
\[ B = \frac{(1.2 \, \text{kg})(10 \, \text{m/s}^2)}{(5 \, \text{A})(1 \, \text{m})} \]

5. **Calculate the Magnetic Field:**
Performing the calculation:
\[ B = \frac{12 \, \text{kg m/s}^2}{5 \, \text{A}} = \frac{12}{5} = 2.4 \, \text{T} \]

6. **Final Answer:**
The magnitude of the magnetic field required to suspend the wire is:
\[ B = 2.4 \, \text{T} \]