Physics · 2019
JEE Main · 10 January 2019, Shift 2 · Q20
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…
At some location on earth the horizontal component of earth's magnetic field is $\displaystyle 18 \times 10^{-6} \mathrm{~T}$. At this location, magnetic needle of length $\displaystyle 0.12$ m and pole strength $\displaystyle 1.8$ Am is suspended from its mid-point using a thread, it makes $\displaystyle 45$° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is :
Official answer
(1)
$\displaystyle 6.5 \times 10^{-5} \mathrm{~N}$
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.