Q.A magnetic needle is suspended freely so that it can rotate in the magnetic meridian. In order to keep it in the horizontal position, a weight of 0.1 g is kept on one end of the needle. If the pole strength of this needle is 20 A m, find the value of the vertical component of the earth's magnetic field. ($g=9.8$ m s$^{-2}$)

[!TLDR] Treating this as a torque balance of the suspended needle (the magnetic couple on its two poles, balanced by the weight's torque at the far end) gives $B_V=\dfrac{wg}{2q_m}\approx2.45\times10^{-5}$ T. [!ANSWER] $B_V\approx2.45\times10^{-5}$ T by this torque-balance deriv …

A needle "suspended freely so that it can rotate in the magnetic meridian" is pivoted at its centre and free to tip about a horizontal axis through that centre -- the same kind of setup used for a dip needle. With pole strength $q_m$ at each end, separated by full length $2l$, the vertical component $B_V$ of Earth's field exerts a force $q_mB_V$ on each pole, in opposite senses (since the two poles are $+q_m$ and $-q_m$); these two forces, acting at the two ends (each a distance l from the pivot), form a couple of total torque $\tau_{mag}=2q_mlB_V$ (equivalently, $\tau_{mag}=mB_V$ using the full moment $m=q_m\times2l$ -- exactly the $\tau=mB\sin\theta$ result of section 11.2, with $\theta=90^\circ$ between the horizontal needle and the vertical field). This magnetic torque is balanced by the counter-torque of the added weight w, placed at one end (a distance l from the pivot): $\tau_{weight}=wgl$. Setting the two torques equal, $2q_mlB_V=wgl$, the lever arm l cancels, giving $B_V=\dfrac{wg}{2q_m}$. Substituting $w=0.1\ \text{g}=1\times10^{-4}$ kg, $g=9.8$ m s$^{-2}$, $q_m=20$ A m: $B_V=\dfrac{(1\times10^{-4})(9.8)}{2\times20}=\dfrac{9.8\times10^{-4}}{40}=2.45\times10^{-5}$ T. This rigorous torque-balance result differs from the textbook's printed a …

Set up a torque balance for the freely-pivoted needle: the magnetic couple on its two poles due to $B_V$ (total torque $mB_V=2q_mlB_V$) equals the …

Treating this as a simple single-pole FORCE balance ($wg=q_mB_V$, ignoring the needle's other pole and the lever-arm/torque nature of the setup) rather than a full torque balance of the suspended dipole -- a uniform field can only ever exert a net TORQUE on a dipole, never a net …