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 is. 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 find the current that must be passed through the second wire so that it can float in the air without any support. We will follow these steps:

### Step 1: Understanding the Forces
The second wire will experience two forces:
1. The gravitational force acting downwards, \( F_g = mg \)
2. The magnetic force acting upwards due to the current in the first wire, \( F_m \)

For the second wire to float, these two forces must be equal:
\[ F_m = F_g \]

### Step 2: Calculate the Gravitational Force
The mass per unit length of the second wire is given as \( \lambda = 10^{-2} \) kg/m. The gravitational force can be calculated using:
\[ F_g = mg = \lambda L g \]
where \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)) and \( L \) is the length of the wire. Since we are looking for the force per unit length, we can consider \( L = 1 \, \text{m} \):
\[ F_g = (10^{-2} \, \text{kg/m}) \cdot (1 \, \text{m}) \cdot (9.8 \, \text{m/s}^2) = 9.8 \times 10^{-2} \, \text{N/m} \]

### Step 3: Calculate the Magnetic Force
The magnetic force on the second wire due to the first wire is given by:
\[ F_m = \frac{\mu_0}{4\pi} \cdot \frac{2 I_1 I_2}{R} \cdot L \]
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 (which we need to find)
- \( R = 2 \, \text{cm} = 0.02 \, \text{m} \)

### Step 4: Set the Forces Equal
Setting \( F_m = F_g \):
\[ \frac{\mu_0}{4\pi} \cdot \frac{2 I_1 I_2}{R} = \lambda g \]

### Step 5: Substitute Known Values
Substituting the known values into the equation:
\[ \frac{(4\pi \times 10^{-7})}{4\pi} \cdot \frac{2 \cdot 200 \cdot I_2}{0.02} = 10^{-2} \cdot 9.8 \]
This simplifies to:
\[ 10^{-7} \cdot \frac{400 I_2}{0.02} = 9.8 \times 10^{-2} \]

### Step 6: Solve for \( I_2 \)
Now, we can solve for \( I_2 \):
\[ 10^{-7} \cdot 20000 I_2 = 9.8 \times 10^{-2} \]
\[ 20000 I_2 = 9.8 \times 10^{5} \]
\[ I_2 = \frac{9.8 \times 10^{5}}{20000} \]
\[ I_2 = 49 \, \text{A} \]

### Step 7: Determine the Direction of Current
The direction of the current in the second wire must be the same as the current in the first wire for the two wires to attract each other.

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

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