Question

In New Hampshire the average horizontal component of Earth’s magnetic field in 1912 was $\mathbf{16}\mathbf{}\mathit{\mu }\mathbf{T}$, and the average inclination or “dip” was $\mathbf{73}\mathbf{°}$. What was the corresponding magnitude of Earth’s magnetic field?

Short answer

The magnitude of the earth’s magnetic field is $55\mu \mathrm{T}$

Step-by-step solution

Listing the given quantities

Horizontal component of magnetic field,$B_h=16\mu \mathrm{T}$

Inclination,${\varphi }_i=73°$

Understanding the concepts of magnetic field

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

$B=\frac{B_h}{coscos{\varphi }_i}$

Calculations of the magnitude of the earth’s magnetic field

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

$B_h=Bcos{\varphi }_i$

${\varphi }_i$is the inclination or dip.

$\begin{array}{rcl}{\varphi }_i& =& 73°\\ B_h& =& 16\mu \mathrm{T}\end{array}$

So,

$\begin{array}{rcl}B& =& \frac{B_h}{coscos{\varphi }_i}\\ & =& \frac{16\times {10}^{-6}\mathrm{T}}{coscos73°}\\ & =& 55\mu \mathrm{T}\end{array}$

The magnitude of the earth’s magnetic field is$55\mu \mathrm{T}$