A magnetometer stationed at a field site records the horizontal component of the Earth's m… — JAMB UTME

A magnetometer stationed at a field site records the horizontal component of the Earth's magnetic field as $36\ \mu\text{T}$ and the total magnetic field intensity as $45\ \mu\text{T}$ . What is the magnitude of the vertical component of the Earth's magnetic field, in $\mu\text{T}$ ?

Answer: 27 μT

Answer

The magnitude of the vertical component of the Earth's magnetic field is

27\ \mu\text{T}

The total magnetic field strength

B

is the hypotenuse of a right-angled triangle formed by the horizontal component

B_h

and vertical component

B_v

. By applying the Pythagorean relation

B^2 = B_h^2 + B_v^2

, substituting

B = 45\ \mu\text{T}

and

B_h = 36\ \mu\text{T}

yields

B_v = \sqrt{45^2 - 36^2} = \sqrt{729} = 27\ \mu\text{T}

Step-by-Step Solution

Relate total magnetic intensity to its orthogonal components

B^2 = B_h^2 + B_v^2

The total magnetic field vector of the Earth is the vector sum of mutually perpendicular horizontal and vertical components.

Isolate the vertical component variable

B_v = \sqrt{B^2 - B_h^2}

Applying the Pythagorean theorem allows direct calculation of the missing perpendicular side.

Substitute given values and compute numerical result

B_v = \sqrt{45^2 - 36^2} = \sqrt{2025 - 1296} = \sqrt{729} = 27\ \mu\text{T}

Evaluates the exact magnitude of the vertical component.

Key Concept

Orthogonal resolution of Earth's magnetic field components

Estimated Time:1m 15s

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