The horizontal component of the Earth's magnetic field is 3.6 × 10 - 5 T where the dip angle is 60 ° . The magnitude of the Earth's magnetic field is

Horizontal component of Earth’s field, H = B cos θ , since, θ = 60 ° 3.6 × 10 − 5 = B × 1 2 ⇒ B = 7.2 × 10 − 5 tesla

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PhysicsMagnetismProperties of magnetEasy2 minPYQ_2020

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The horizontal component of the Earth's magnetic field is $3.6 \times 10^{- 5} T$ where the dip angle is $60 °$ . The magnitude of the Earth's magnetic field is

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Answer:

Solution:

Horizontal component of Earth’s field,

$H = B \cos θ,$ since, $θ = 60 °$

$3.6 \times 10^{- 5} = B \times \frac{1}{2} \Rightarrow B = 7.2 \times 10^{- 5}$ $tesla$

Stream:JEESubject:PhysicsTopic:MagnetismSubtopic:Properties of magnet

⏱ 2mℹ️ Source: PYQ_2020

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