At some location on earth the horizontal... | Magnetism

JEE MAIN 2019

10-01-2019 S2

Question

At some location on earth the horizontal component of earth's magnetic field is $18 \times 10^{-6} \mathrm{~T}$. 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 $45^{\circ}$ angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is:

Select the correct option:

$1.3 \times 10^{-5} \mathrm{~N}$

$1.8 \times 10^{-5} \mathrm{~N}$

$6.5 \times 10^{-5} \mathrm{~N}$

$3.6 \times 10^{-5} \mathrm{~N}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution