Question

What will be the angle of dip if the magnetic field of earth at a certain place has a horizontal component $0.3Gauss$ and the total strength is $0.5Gauss$ ?
A) ${\tan }^{-1}{\displaystyle \frac{3}{4}}$
B) ${\sin }^{-1}{\displaystyle \frac{3}{4}}$
C) ${\tan }^{-1}{\displaystyle \frac{4}{3}}$
D) ${\sin }^{-1}{\displaystyle \frac{3}{5}}$

Answer 1

Hint: -First of all, we have to use the formula of calculating the total strength of the field to find the vertical component using the horizontal component and total strength.
The angle of dip is the angle that is made by the earth’s magnetic field lines with the horizontal.

Complete Step by Step Solution: -
Let the horizontal component be $H$, vertical component be $V$ and total strength be $B$
According to the question, it is given that –
Horizontal component, $\therefore H=0.3G$
where, $G$ stands for Gauss
Total strength of the field, $\therefore B=0.5G$
Now, we have to find the vertical component, so we will use the relationship between the total strength of the field, horizontal component and vertical component which is –
$\therefore B^2=H^2+V^2$
The above equation can also be made as –
$$\begin{gathered} V^2=B^2-H^2 \\ \Rightarrow V=\sqrt{B^2-H^2}\cdots \left(1\right) \end{gathered}$$
Putting the value of total strength of the field and horizontal component in the equation $\left(1\right)$ -
$$\begin{gathered} V=\sqrt{{\left(0.5\right)}^2-{\left(0.3\right)}^2} \\ \Rightarrow V=\sqrt{0.25-0.09} \\ \Rightarrow V=\sqrt{0.16} \\ \therefore V=0.4G \end{gathered}$$
Hence, we got the value of the vertical component which is $0.4Gauss$.
Now, we have to find the angle of dip
We know that, angle of dip can be defined as the angle that is made by the earth’s magnetic field lines with the horizontal. It is denoted by $\varphi$. The formula for calculating angle of dip is –
$$\tan \varphi ={\displaystyle \frac{V}{H}}$$
Putting the values of vertical component and horizontal component in the above formula for calculating the angle of dip –
$$\begin{gathered} \therefore \tan \varphi ={\displaystyle \frac{0.4}{0.3}}={\displaystyle \frac{4}{3}} \\ \Rightarrow \varphi ={\tan }^{-1}\left({\displaystyle \frac{4}{3}}\right) \end{gathered}$$
Hence, the value of angle of dip is ${\tan }^{-1}{\displaystyle \frac{4}{3}}$.

So, the correct option is (C).

Note: -At the equator, magnetic lines of forces are perfectly horizontal such that even the magnetic needle is also horizontal at the equator. So, the angle of dip at the equator is zero.