The horizontal component of the earth's magnetic field is 3.6 × 1 0 − 5 tesla where the dip angle is 6 0 ∘ . The magnitude of the earth's magnetic field is

Horizontal component of earth's field, H = B cos θ , since, θ = 6 0 ∘ 3.6 × 1 0 − 5 = B × 2 1 ⇒ B = 7.2 × 1 0 − 5 Tesla

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Physics - Magnetism Question with Solution | TestHub

PhysicsMagnetismProperties of magnetEasy2 minPYQ_2018

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The horizontal component of the earth's magnetic field is $3.6 \times 1 0^{- 5}$ tesla where the dip angle is $6 0^{\circ} .$ The magnitude of the earth's magnetic field is

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Solution:

Horizontal component of earth's field, $H = B cos θ$ , since, $θ = 6 0^{\circ}$
$3.6 \times 1 0^{- 5} = B \times \frac{1}{2} \Rightarrow B = 7.2 \times 1 0^{- 5} Tesla$

Stream:VITEEESubject:PhysicsTopic:MagnetismSubtopic:Properties of magnet

⏱ 2mℹ️ Source: PYQ_2018

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