The horizontal component of earth's magnetic field at a place is $3.5 \times 10^{-5} T$. A very long straight conductor carrying current of √2 A in the direction from South east to North West is placed.
The force per unit length experienced by the conductor is × $10^{-6}$ N/m.
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Solution
$\begin{aligned} B_H & =3.5 \times 10^{-5} \mathrm{~T} \\ F & =i \ell B \sin \theta, \mathrm{i}=\sqrt{2} A \\ \frac{F}{\ell} & =i B \sin \theta=\sqrt{2} \times 3.5 \times 10^{-5} \times \frac{1}{\sqrt{2}} \\ & =35 \times 10^{-6} \mathrm{~N} / \mathrm{m}\end{aligned}$
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