A straight horizontal conductor of length `L` meter and mass `m kg` carries a current `'I'` ampere. The minimum magnetic induction which must exist in the region to balance its weight

Options

A. `mg//iL`
B. `iL//mg`
C. `mgL//i`
D. `mL//ig`

Answer

To solve the problem of finding the minimum magnetic induction (magnetic field strength) required to balance the weight of a straight horizontal conductor carrying a current, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Conductor:**
- The conductor has a weight acting downwards due to gravity, which is given by the formula:
\[ W = mg \]
where \( m \) is the mass of the conductor and \( g \) is the acceleration due to gravity.

2. **Determine the Magnetic Force Acting on the Conductor:**
- When a current \( I \) flows through the conductor placed in a magnetic field \( B \), it experiences a magnetic force \( F_B \) given by:
\[ F_B = IBL \]
where \( L \) is the length of the conductor.

3. **Set Up the Equation for Equilibrium:**
- For the conductor to be in equilibrium (not falling), the upward magnetic force must balance the downward weight:
\[ F_B = W \]
Therefore, we can write:
\[ IBL = mg \]

4. **Solve for the Magnetic Induction \( B \):**
- Rearranging the equation to solve for \( B \), we get:
\[ B = \frac{mg}{IL} \]

### Final Result:
The minimum magnetic induction required to balance the weight of the conductor is:
\[ B = \frac{mg}{IL} \]