A long straight wire carrying current of 30 A rests on a table. Another wire AB of length 1 m, mass 3 g carries the same current but in the opposite direction, the wire AB is free to slide up and down. The height upto which AB will rise is

Options

A. 0.6 cm
B. 0.7 cm
C. 0.4 cm
D. 0.5 cm

Explanation

To solve the problem of how high the wire AB will rise, we will follow these steps:

### Step 1: Understand the Forces Acting on Wire AB
The wire AB is subjected to two main forces:
1. The gravitational force acting downward, which is given by \( F_g = mg \).
2. The magnetic force acting upward due to the interaction with the long straight wire PQ, which is given by the formula for the magnetic force between two parallel currents.

### Step 2: Calculate the Gravitational Force
Given:
- Mass of wire AB, \( m = 3 \, \text{g} = 3 \times 10^{-3} \, \text{kg} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

The gravitational force \( F_g \) can be calculated as:
\[ F_g = mg = (3 \times 10^{-3} \, \text{kg})(9.8 \, \text{m/s}^2) = 2.94 \times 10^{-2} \, \text{N} \]

### Step 3: Calculate the Magnetic Force
The magnetic force \( F_m \) between two parallel wires carrying current can be calculated using the formula:
\[ F_m = \frac{\mu_0 I_1 I_2 L}{2\pi h} \]
Where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space)
- \( I_1 = I_2 = 30 \, \text{A} \) (currents in both wires)
- \( L = 1 \, \text{m} \) (length of wire AB)
- \( h \) is the height we want to find.

### Step 4: Set the Forces Equal in Equilibrium
At equilibrium, the magnetic force equals the gravitational force:
\[ F_m = F_g \]
Substituting the expressions for the forces:
\[ \frac{\mu_0 I_1 I_2 L}{2\pi h} = mg \]

### Step 5: Rearranging to Solve for Height \( h \)
Rearranging the equation to solve for \( h \):
\[ h = \frac{\mu_0 I_1 I_2 L}{2\pi mg} \]

### Step 6: Substitute the Known Values
Substituting the known values into the equation:
\[ h = \frac{(4\pi \times 10^{-7} \, \text{T m/A})(30 \, \text{A})(30 \, \text{A})(1 \, \text{m})}{2\pi (3 \times 10^{-3} \, \text{kg})(9.8 \, \text{m/s}^2)} \]

### Step 7: Simplify the Equation
\[ h = \frac{(4 \times 10^{-7})(30)(30)}{2(3 \times 10^{-3})(9.8)} \]
Calculating the numerator:
\[ 4 \times 30 \times 30 = 3600 \times 10^{-7} \]
Calculating the denominator:
\[ 2 \times 3 \times 10^{-3} \times 9.8 = 58.8 \times 10^{-3} \]
Thus:
\[ h = \frac{3600 \times 10^{-7}}{58.8 \times 10^{-3}} = \frac{3600}{58.8} \times 10^{-4} \approx 61.22 \times 10^{-4} \, \text{m} = 0.6122 \, \text{cm} \]

### Final Result
The height up to which wire AB will rise is approximately:
\[ h \approx 0.6122 \, \text{cm} \approx 0.6 \, \text{cm} \]