A fixed horizontal wire carries a current of 200 A. Another wire having a mass per unit length `10^(-2)` kg/m is placed below the first wire at a distance of 2 cm and parallel to it. How much current must be passed through the second wire if it floats in air without any support? What should be the direction of current in it?

Options

A. 25 A (direction of current is same to first wire)
B. 25 A (direction of current is opposite to first wire)
C. 49 A (direction of current is same to first wire)
D. 49 A (direction of current is opposite to first wire)

Answer

To solve the problem, we need to determine the current that must flow through the second wire so that it can float in air without any support. Here’s a step-by-step solution:

### Step 1: Understand the Forces Acting on the Second Wire
The second wire will float when the magnetic force acting on it due to the first wire equals its weight. The weight (W) of the second wire can be expressed as:
\[ W = m \cdot g \]
where:
- \( m \) is the mass of the second wire per unit length multiplied by its length \( L \),
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

### Step 2: Calculate the Weight of the Second Wire
Given that the mass per unit length of the second wire is \( 10^{-2} \, \text{kg/m} \):
\[ W = (10^{-2} \, \text{kg/m}) \cdot L \cdot g \]
\[ W = (10^{-2}) \cdot L \cdot 9.8 \]

### Step 3: Determine the Magnetic Force Acting on the Second Wire
The magnetic force \( F_m \) acting on the second wire due to the magnetic field created by the first wire can be expressed as:
\[ F_m = I_2 \cdot L \cdot B \]
where:
- \( I_2 \) is the current in the second wire,
- \( L \) is the length of the second wire,
- \( B \) is the magnetic field at the location of the second wire due to the first wire.

### Step 4: Calculate the Magnetic Field \( B \) Created by the First Wire
The magnetic field \( B \) at a distance \( r \) from a long straight wire carrying current \( I_1 \) is given by:
\[ B = \frac{\mu_0 I_1}{2 \pi r} \]
where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (the permeability of free space),
- \( I_1 = 200 \, \text{A} \) (the current in the first wire),
- \( r = 0.02 \, \text{m} \) (the distance between the wires).

Substituting the values:
\[ B = \frac{(4\pi \times 10^{-7}) \cdot 200}{2\pi \cdot 0.02} \]
\[ B = \frac{(4 \times 200 \times 10^{-7})}{0.04} \]
\[ B = \frac{800 \times 10^{-7}}{0.04} \]
\[ B = 2 \times 10^{-3} \, \text{T} \]

### Step 5: Set Up the Equilibrium Condition
For the second wire to float, the magnetic force must equal the weight:
\[ m \cdot g = I_2 \cdot L \cdot B \]
Substituting the expressions for weight and magnetic force:
\[ (10^{-2} \cdot L \cdot 9.8) = I_2 \cdot L \cdot (2 \times 10^{-3}) \]

### Step 6: Simplify and Solve for \( I_2 \)
Cancelling \( L \) from both sides (assuming \( L \neq 0 \)):
\[ 10^{-2} \cdot 9.8 = I_2 \cdot (2 \times 10^{-3}) \]
\[ I_2 = \frac{10^{-2} \cdot 9.8}{2 \times 10^{-3}} \]
\[ I_2 = \frac{9.8 \times 10^{-2}}{2 \times 10^{-3}} \]
\[ I_2 = 49 \, \text{A} \]

### Step 7: Determine the Direction of Current
Using the right-hand rule, if the current in the first wire is flowing in one direction, the current in the second wire must flow in the same direction for the magnetic force to act upwards (to balance the weight).

### Final Answer
The current that must be passed through the second wire to make it float is **49 A**, and the direction of the current should be **the same as that in the first wire**.

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