A straight wire of mass 300 g and length 1.5 m carries a current of 2 A. It is suspended in mid air by a uniform horizontal magnetic field. The magnitude of the magnetic field is

(Neglect the earth's magnetic field and $g = 10 m/s^2$)

1 T

The correct answer is Option (3) → 1 T

Given:

Mass of wire $m = 300 \, g = 0.3 \, kg$

Length of wire $L = 1.5 \, m$

Current $I = 2 \, A$

Gravitational acceleration $g = 10 \, m/s^2$

Weight of wire: $W = m g = 0.3 \times 10 = 3 \, N$

Magnetic force on wire: $F = B I L$

For suspension in mid air: $F = W$

$B I L = W$

$B = \frac{W}{I L} = \frac{3}{2 \times 1.5} = \frac{3}{3} = 1 \, T$

Answer: $1 \, Tesla$