A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid air by a uniform horizontal magnetic field B perpendicular to length of wire. The magnitude of B is ($g = 10 ms^{-2}$)

0.67 T

The correct answer is Option (4) → 0.67 T

For wire to be suspended in air its weight is balanced by magnetic force.

$\Rightarrow F = ILB = mg $

$\Rightarrow B = \frac{mg}{IL} = \frac{0.2 \times 10}{2\times 1.5} = 0.67T$