A conducting rod of 1 m length and 1 kg mass is suspended by two verticle wires through its ends. An external magnetic field of 2T is applied normal to the rod. Now the current to be passed through the rod so as to make the tension in the wire zero is
[Take `g = 10 ms^(-2)]`

Options

A. `15 A`
B. `5 A`
C. `1.5 A`
D. `2.5 A`

Answer

To solve the problem, we need to determine the current that must be passed through a conducting rod suspended by two vertical wires so that the tension in the wires becomes zero when an external magnetic field is applied.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Rod:**
- The weight of the rod (W) acting downwards is given by:
\[ W = mg \]
where \( m = 1 \, \text{kg} \) (mass of the rod) and \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).
- Therefore, the weight of the rod is:
\[ W = 1 \, \text{kg} \times 10 \, \text{m/s}^2 = 10 \, \text{N} \]

2. **Magnetic Force Acting on the Rod:**
- The magnetic force (F) acting on the rod when current (I) flows through it in a magnetic field (B) is given by:
\[ F = I \cdot L \cdot B \]
where \( L = 1 \, \text{m} \) (length of the rod) and \( B = 2 \, \text{T} \) (magnetic field strength).
- Thus, the magnetic force can be expressed as:
\[ F = I \cdot 1 \, \text{m} \cdot 2 \, \text{T} = 2I \, \text{N} \]

3. **Setting Up the Equilibrium Condition:**
- For the tension in the wires to be zero, the magnetic force must balance the weight of the rod:
\[ F = W \]
- Therefore, we have:
\[ 2I = 10 \]

4. **Solving for Current (I):**
- Rearranging the equation gives:
\[ I = \frac{10}{2} = 5 \, \text{A} \]

### Final Answer:
The current that must be passed through the rod to make the tension in the wires zero is:
\[ \boxed{5 \, \text{A}} \]