The horizontal component of earth's magnetic field at a place is 3...

Horizontal component of magnetic field, ${B}_{H} = 3 . 5 \times {10}^{- 5}T$

Force acting on wire, $F = ilBsin\theta, \text{where}i = \sqrt{2}A$

Therefore, $\dfrac{F}{l} = iBsin\theta = \sqrt{2} \times 3 . 5 \times {10}^{- 5} \times \dfrac{1}{\sqrt{2}}$

$= 35 \times {10}^{- 6}N{m}^{- 1}$