A conducting wire of length 0.1 m and mass $3 × 10^{-3} kg$, carrying a current of 5 A, is suspended freely in a horizontal direction in a uniform magnetic field. The magnitude of magnetic field perpendicular to the length of conductor is

0.06 T

The correct answer is Option (1) → 0.06 T

Magnetic force on a current-carrying wire: $F = B I L$

For horizontal suspension, magnetic force balances weight: $B I L = mg \;\Rightarrow\; B = \frac{mg}{I L}$

Given: $m = 3 \times 10^{-3} \, \text{kg}$, $g = 9.8 \, \text{m/s²}$, $I = 5 \, \text{A}$, $L = 0.1 \, \text{m}$

$B = \frac{3 \times 10^{-3} \cdot 9.8}{5 \cdot 0.1} = \frac{0.0294}{0.5} \approx 0.0588 \, \text{T}$

Magnetic field B ≈ 0.059 T