A magnet hung at 45° with magnetic meridian makes an angle of 60° with the horizontal. The actual value of the angle of dip is - (a) tan ⁻¹(√(3/2)) (b) tan ⁻¹(√(6)) (c) tan ⁻¹(√(2/3)) (d) tan ⁻¹(√(1/2))

Answer: (a). Apparent dip: tan δ' = (tan δ)/(cos α) ⇒ tan δ = tan δ'cos α tan δ = tan 60°cos 45° = (√(3))/(√(2)) = √(3/2) So the answer is (a) tan ⁻¹(√(3/2)).

A magnet hung at 45° with magnetic meridian makes an angle of 60°… | JEE Main 2022 (27 Jul, Shift 1)

Q3JEE Main 2022 · 27 JUL · SHIFT 1Medium

Earth's Magnetism

A magnet hung at $45^\circ$ with magnetic meridian makes an angle of $60^\circ$ with the horizontal. The actual value of the angle of dip is -

Show answer and solution

Answer: (a)

Apparent dip: $\tan\delta' = \frac{\tan\delta}{\cos\alpha} \Rightarrow \tan\delta = \tan\delta'\cos\alpha$
$\tan\delta = \tan 60^\circ\cos 45^\circ = \frac{\sqrt{3}}{\sqrt{2}} = \sqrt{\frac{3}{2}}$
So the answer is (a) $\tan^{-1}\left(\sqrt{\frac{3}{2}}\right)$ .

Report a mistake

Earth's Magnetism question from JEE Main 2022 (27 Jul, Shift 1)

More questions on Earth's Magnetism