At some location the horizontal component of earth's magnetic field is $$18 \times 10^{-6}$$ T. At this location, magnetic needle of length 0.12 m and pole strength 1.8 Am is suspended from its mid-point using a thread, it makes 45$$^{\circ}$$ angles with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is:

We have the horizontal component of earth’s magnetic field at the place as $$B_h = 18 \times 10^{-6}\ \text{T}$$.

The magnetic needle, whose north and south poles have equal pole strength $$m = 1.8\ \text{A m}$$ and are separated by a length $$L = 0.12\ \text{m}$$, is suspended from its mid-point. In the earth’s field it comes to rest making an angle of $$45^{\circ}$$ with the horizontal.

The inclination (dip) angle of the earth’s field is defined by the relation

$$\tan\delta = \frac{B_v}{B_h},$$

where $$B_v$$ is the vertical component and $$\delta$$ is the angle that the total field makes with the horizontal.

Here the needle itself makes $$45^{\circ}$$ with the horizontal, so $$\delta = 45^{\circ}$$. Hence

$$\tan 45^{\circ} = 1 = \frac{B_v}{B_h} \quad\Longrightarrow\quad B_v = B_h = 18 \times 10^{-6}\ \text{T}.$$

Each pole of the magnet experiences a vertical magnetic force due to $$B_v$$ given by the basic formula

$$F_{\text{mag}} = m\,B_v.$$

For the north pole this force is, say, downward; for the south pole it is upward, so the two equal and opposite forces form a couple. The line joining the poles is the needle of length $$L$$, and the pivot is at the centre. The torque produced by this couple about the mid-point is obtained from the definition of torque of a couple (force × distance between the forces):

$$\tau_{\text{mag}} = m\,B_v \times L.$$

We now want the needle to stay exactly horizontal. For that we have to apply an external vertical force $$F$$ at one end of the needle. This force, acting at a perpendicular distance $$\dfrac{L}{2}$$ from the mid-point, produces an opposite torque

$$\tau_{\text{ext}} = F \times \frac{L}{2}.$$

In equilibrium the net torque must vanish, so

$$\tau_{\text{ext}} = \tau_{\text{mag}}.$$

Substituting the expressions, we get

$$F \times \frac{L}{2} = m\,B_v \times L.$$

Solving this for $$F$$ gives

$$F = \frac{m\,B_v \times L}{\dfrac{L}{2}} = 2\,m\,B_v.$$

Now we substitute the numerical values:

$$F = 2 \times 1.8\ \text{A m} \times 18 \times 10^{-6}\ \text{T}.$$

First multiply the ordinary numbers:

$$1.8 \times 18 = 32.4,$$

and then include the factor of 2:

$$2 \times 32.4 = 64.8.$$

Hence

$$F = 64.8 \times 10^{-6}\ \text{N} = 6.48 \times 10^{-5}\ \text{N}.$$

Rounding to two significant figures gives

$$F \approx 6.5 \times 10^{-5}\ \text{N}.$$

Hence, the correct answer is Option C.