At some location the horizontal component of earth's magnetic field...

The horizontal and vertical components of earth's magnetic field (${B}_{H} and {B}_{V}$) are related as

$\dfrac{{B}_{V}}{{B}_{H}} = tan\theta$

Here, $\theta = 45 °$ and ${B}_{H} = 18 \times {10}^{- 6}T$

$⇒ {B}_{V} = {B}_{H}\tan 45 °$

$⇒ {B}_{V} = {B}_{H} = 18 \times {10}^{- 6}T \left(∵ \tan 45 ° = 1\right)</p></blockquote><blockquote><p>Now, when$ the external forceF$is applied, to keep the needle stays in horizontal position is shown below,</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/d24e4fa9-330e-4dc2-b590-8a875bed2160-image.png" style="width: 305px; height: 191px;" /></p><p>Taking torque at point$P$, we get</p><p>$m{B}_{V} \times 2l = Fl$</p><p>$∴ F = 2 \times m{B}_{V}$</p><p>Substituting the given values, we get</p><p>$F = 2 \times 1 . 8 \times 18 \times {10}^{- 6}$</p><p>$F = 6 . 48 \times {10}^{- 5} = 6 . 5 \times {10}^{- 5}N$.</p></blockquote>$