JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Magnetic Properties of Matter

At an angle of

3 0^{\circ}

to the magnetic meridian, the apparent dip is

4 5^{\circ}

. Find the true dip:

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

Answer: (A)

Solution

In a vertical plane turned through an angle

θ

from the magnetic meridian, only the component

H cos θ

of the horizontal field lies in that plane, while the vertical component

V

is unchanged. The apparent dip

δ^{'}

measured there satisfies

tan δ^{'} = \frac{V}{H cos θ} = \frac{tan δ}{cos θ}

With

δ^{'} = 4 5^{\circ}

and

θ = 3 0^{\circ}

tan δ = tan 4 5^{\circ} cos 3 0^{\circ} = \frac{3}{2}

The true dip is

tan^{- 1} \frac{3}{2}

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Single correct option

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