JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Magnetic Properties of Matter

A magnet hung at

4 5^{\circ}

with magnetic meridian makes an angle of

6 0^{\circ}

with the horizontal. The actual value of the angle of dip is

(A) $tan^{- 1} (\frac{3}{2})$ ✓ Correct
(B) $tan^{- 1} (6)$
(C) $tan^{- 1} (\frac{2}{3})$
(D) $tan^{- 1} (\frac{1}{2})$

Answer: (A)

Solution

When a magnet is hung at an angle

α

with the magnetic meridian and makes an angle

θ^{'}

with the horizontal, the actual angle of dip

θ

is given by

tan θ^{'} = \frac{t a n θ}{c o s α}

. Here

θ^{'} = 6 0^{\circ}

and

α = 4 5^{\circ}

. Substituting:

3 = \frac{t a n θ}{1/ 2}

, which gives

tan θ = 3/2

. Therefore

θ = tan^{- 1} 3/2

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Single correct option

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