A horizontal wire carries 200A current below which another wire of linear density `20 xx 10^(-3)` Kg/m carrying a current is kept at 2cm distance. If the wire kept below hangs in air, then the current in the wire is

Options

A. 9.8 A
B. 98 A
C. 980 A
D. 9800 A

Answer

To solve the problem, we need to find the current \( I_2 \) in the second wire that is hanging in air. The forces acting on the second wire must be balanced, meaning the upward magnetic force due to the first wire must equal the downward gravitational force acting on the second wire.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Second Wire:**
- The downward force is the weight of the second wire, given by:
\[ F_{\text{down}} = mg \]
- The upward force is the magnetic force due to the first wire, given by:
\[ F_{\text{up}} = \frac{\mu_0 I_1 I_2 l}{2 \pi r} \]
where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space)
- \( I_1 = 200 \, \text{A} \) (current in the first wire)
- \( I_2 \) is the current in the second wire
- \( l \) is the length of the wire (assumed to be the same for both wires)
- \( r = 2 \, \text{cm} = 0.02 \, \text{m} \) (distance between the wires)

2. **Set Up the Equation for Force Balance:**
Since the forces are balanced, we can set them equal to each other:
\[ mg = \frac{\mu_0 I_1 I_2 l}{2 \pi r} \]

3. **Express Mass in Terms of Linear Density:**
The mass \( m \) of the second wire can be expressed in terms of its linear density \( \lambda \):
\[ m = \lambda l \]
where \( \lambda = 20 \times 10^{-3} \, \text{kg/m} \).

4. **Substituting Mass into the Force Balance Equation:**
Substitute \( m \) into the force balance equation:
\[ \lambda l g = \frac{\mu_0 I_1 I_2 l}{2 \pi r} \]

5. **Cancel \( l \) from Both Sides:**
Since \( l \) appears on both sides, we can cancel it out:
\[ \lambda g = \frac{\mu_0 I_1 I_2}{2 \pi r} \]

6. **Rearranging the Equation to Solve for \( I_2 \):**
Rearranging gives:
\[ I_2 = \frac{2 \pi r \lambda g}{\mu_0 I_1} \]

7. **Substituting Known Values:**
- \( r = 0.02 \, \text{m} \)
- \( \lambda = 20 \times 10^{-3} \, \text{kg/m} \)
- \( g = 9.8 \, \text{m/s}^2 \)
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)
- \( I_1 = 200 \, \text{A} \)

Plugging in these values:
\[ I_2 = \frac{2 \pi (0.02) (20 \times 10^{-3}) (9.8)}{4\pi \times 10^{-7} \times 200} \]

8. **Simplifying the Expression:**
- The \( \pi \) cancels out:
\[ I_2 = \frac{2 (0.02) (20 \times 10^{-3}) (9.8)}{4 \times 10^{-7} \times 200} \]
- Simplifying further:
\[ I_2 = \frac{0.0008 \times 9.8}{8 \times 10^{-5}} = \frac{7.84 \times 10^{-3}}{8 \times 10^{-5}} = 98 \, \text{A} \]

### Final Answer:
The current in the second wire \( I_2 \) is \( 98 \, \text{A} \).

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