JEE Main 2022 · 27 July · morning · Single correct · +4 / −1
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 and solution
Answer (a) tan ⁻¹(√(3/2))
Solution
Physics sir · 4 steps
Turning the needle away from the meridian weakens only the horizontal pull, so multiply tan 60° by cos 45°.
- What changes. The needle can only turn in the vertical plane it hangs in. That plane makes 45° with the magnetic meridian. The vertical component V acts fully in this plane, but only H cos 45° of the horizontal component lies along it.
- Apparent dip. The needle settles along the resultant of these two components. Its angle with the horizontal is the apparent dip δ', while the true dip δ has tan δ = V/H.
tan δ' = V/(H cos θ) = (tan δ)/(cos θ) ⇒ tan δ = tan δ'cos θ
- Put in numbers. Here δ' = 60° and θ = 45°. So tan δ = tan 60° × cos 45° = √3 × 1/√2 = √(3/2).
- So. The true angle of dip is δ = tan⁻¹(√(3/2)), which is option (a).
The idea to keep. Out of the meridian, the needle feels all of V but only H cos θ. So the apparent dip is always steeper than the true dip, and tan δ = tan δ'cos θ.
Answer from the NTA final answer key. Solution by PrepWiser.