A straight horizontal wire carrying a current of 2.0 ext A has a...

For the magnetic force to balance the weight of the wire, the magnetic force must act upwards and its magnitude must be equal to the gravitational force.

The magnetic force on a current-carrying wire is given by $F_{m} = I L B \sin\theta$.

For the minimum magnetic field, $\theta = 90^{{}^{{}^{\circ}}}$, so $\sin\theta = 1$.

Equating the magnetic force to the weight of the wire:

$I L B = m g$

Given:

$I = 2.0 \text{ A}$

$m = 10 \text{ mg} = 10 \times 10^{-6} \text{ kg} = 10^{-5} \text{ kg}$

$L = 1.0 \text{ m}$

$g = 9.8 \text{ m/s}^{2}$

Substituting the values:

$2.0 \times 1.0 \times B = 10^{-5} \times 9.8$

$B = \dfrac{9.8 \times 10^{-5}}{2.0}$

$B = 4.9 \times 10^{-5} \text{ T}$