At some location on earth, the horizontal component of earth’s… | JEE Main 2019 (10 Jan, Shift 2)

Q10JEE Main 2019 · 10 JAN · SHIFT 2Difficult

Earth's Magnetism

At some location on earth, the horizontal component of earth’s magnetic field is 18 × 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° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is -

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Answer: (d)

Needle rests at $45^\circ$ (the dip), so $B_V = B_H\tan 45^\circ = 18 \times 10^{-6}\ \text{T}$ .
When horizontal, $B_V$ gives a couple $\tau = mB_V L$ about the midpoint.
Force $F$ at one end (arm $L/2$ ): $F\frac{L}{2} = mB_V L \Rightarrow F = 2mB_V$
$F = 2 \times 1.8 \times 18 \times 10^{-6} = 6.48 \times 10^{-5}\ \text{N}$
So the answer is (d) $6.5 \times 10^{-5}$ N.

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