Given, the mass of the wire $=200\mathrm{ }\mathrm{g}=0.2\mathrm{ }\mathrm{kg}$

Length of the wire $1.5\mathrm{ }\mathrm{m}$

Current $i=2 A$

Magnetic field $B=?$

The force acting on the current-carrying wire in a uniform magnetic field

$F=\mathrm{B}\mathrm{i}\mathrm{l}\sin \theta$

$F=\mathrm{B}\mathrm{i}\mathrm{l}$ $\left(\because \theta ={90}^o\right)$

Weight of the wire $w=mg=0.2\times 9.8\mathrm{ }\mathrm{N}$

In the position of the suspension

$\mathrm{B}\mathrm{i}\mathrm{l}=mg$

$B=\frac{mg}{il}=\frac{0.2\times 9.8}{2\times 1.5}=0.65\mathrm{ }\mathrm{T}$