Q31P - Fundamentals Of Physics

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?

Step-by-Step Solution

Answer

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

Step 1: Listing the given quantities

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

Inclination, ${\varphi }_i=73°$

Step 2: Understanding the concepts of magnetic field

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

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

Step 3: 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,

$\begin{array}{rcl}B& =& \frac{B_h}{coscos {\varphi }_i }\\ & =& \frac{16\times {10}^{-6} \mathrm{T}}{coscos 73° }\\ & =& 55 \mu \mathrm{T}\end{array}$

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