JEE Main 2019 — Magnetic Properties Of Matter Question with Solution

Without applied forces, (in equilibrium position) the needle will stay in the resultant magnetic field of earth. Hence, the dip ' $\theta$ ' at this place is $45^{\circ }$ (given).


We know that, horizontal and vertical components of earth's magnetic field $(B_H$ and $B_V$ ) are related as

$$\frac{B_V}{B_H}=\tan \theta$$

Here, $\theta =45^{\circ }$ and $B_H=18\times 10^{-6}\mathrm{T}$

$$\Rightarrow B_V=B_H\tan 45^{\circ }$$

$$\Rightarrow B_V=B_H=18\times 10^{-6}\mathrm{T}\left(\because \tan 45^{\circ }=1\right)$$

Now, when the external force $F$ is applied, so as to keep the needle stays in horizontal position is shown below,


Taking torque at point $P$, we get

$$\begin{array}{l}m{B}_V\times 2l=Fl\\ \\ \therefore F=2\times m{B}_V\end{array}$$

Substituting the given values, we get

$$\begin{array}{rl}& =2\times 1.8\times 18\times 10^{-6}\\ & \\ & =6.48\times 10^{-5}=6.5\times 10^{-5}\mathrm{N}\end{array}$$