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Chapter 32: Q31P (page 969)

In New Hampshire the average horizontal component of Earth’s magnetic field in 1912 was 16μT, and the average inclination or “dip” was 73°. What was the corresponding magnitude of Earth’s magnetic field?

Short Answer

Expert verified

The magnitude of the earth’s magnetic field is 55μT

Step by step solution

01

Listing the given quantities

Horizontal component of magnetic field,Bh=16μT

Inclination,ϕi=73°

02

Understanding the concepts of magnetic field

We use the horizontal component of magnetic field to calculate the magnitude of the earth’s magnetic field.

B=Bhcoscosϕi

03

Calculations of the magnitude of the earth’s magnetic field

The horizontal component of earth’s magnetic field is given by

Bh=Bcosϕi

ϕiis the inclination or dip.

ϕi=73°Bh=16μT

So,

B=Bhcoscosϕi=16×10-6Tcoscos73°=55μT

The magnitude of the earth’s magnetic field is55μT

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Most popular questions from this chapter

A sample of the paramagnetic salt to which the magnetization curve of Fig. 32-14 applies is immersed in a uniform magnetic field of 2.0T. At what temperature will the degree of magnetic saturation of the sample be (a)50%and (b)90%
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Figure 32-30

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