Question

At two different places, the angles of dip are respectively \[30^\circ\]and \[45^\circ\] . At these two places the ratio of the horizontal component of earth’s magnetic field is:
A. \[\sqrt 3 :\sqrt 2 \]
B. \[1:\sqrt 2 \]
C. \[1:2\]
D. \[1:\sqrt 3 \]

Answer 1

Hint: First of all, we will calculate the horizontal component of the earth’s magnetic field separately for the two different angles. Then we will divide one equation by the other. We will manipulate accordingly and obtain the result.

Complete step by step answer:
In the given question, we are supplied with the following data:
There are two places, for which the angles of dip are given as \[30^\circ\] and \[45^\circ\] respectively.
We are asked to find the ratio of the horizontal component of earth’s magnetic field at these two places.
To begin with, we know that earth’s magnetic field will contain two components, one along the vertical and the other along the horizontal. Since, we are asked to calculate the ratio of the horizontal components, we will take the component along the horizontal plane into the account. For these cases, we will use the trigonometric ratio of cosine to find out the horizontal component, as we already know that the horizontal component is associated to cosine while the vertical component is associated to sine of an angle.
Mathematically, horizontal component of the earth’s magnetic field is given by:
\[{B_{\text{H}}} = B\cos \theta \] …… (1)
Where,
\[{B_{\text{H}}}\] indicates the horizontal component of the earth’s magnetic field.
\[B\] indicates the earth’s magnetic field.
\[\theta \] indicates the angle.
Now, according to the question, we calculate the horizontal components for the two different angles given:
For angle \[30^\circ\], we have:
\[{B_{\text{H}}} = B\cos 30^\circ \] …… (2)
For angle \[45^\circ\], we have:
\[B_{\text{H}}^1 = B\cos 45^\circ \] …… (3)
Now, we divide the equation (2) by equation (3) and we get:
\begin{align*}
\Rightarrow \dfrac{{{B_{\text{H}}}}}{{B_{\text{H}}^1}} &= \dfrac{{B\cos 30^\circ}}{{B\cos 45^\circ }} \\
\Rightarrow \dfrac{{{B_{\text{H}}}}}{{B_{\text{H}}^1}} &= \dfrac{{\cos 30^\circ}}{{\cos 45^\circ}} \\
\Rightarrow \dfrac{{{B_{\text{H}}}}}{{B_{\text{H}}^1}} &= \dfrac{{\dfrac{{\sqrt 3 }}{2}}}{{\dfrac{1}{{\sqrt 2 }}}} \\
\Rightarrow \dfrac{{{B_{\text{H}}}}}{{B_{\text{H}}^1}} &= \dfrac{{\sqrt 3 }}{2} \times \sqrt 2 \\
\end{align*}
Again, we further simplify,
\begin{align*}
\Rightarrow \dfrac{{{B_{\text{H}}}}}{{B_{\text{H}}^1}} &= \dfrac{{\sqrt 3 }}{2} \times \dfrac{{\sqrt 2 \times \sqrt 2 }}{{\sqrt 2 }} \\
\Rightarrow \dfrac{{{B_{\text{H}}}}}{{B_{\text{H}}^1}} &= \dfrac{{\sqrt 3 \times 2}}{{2\sqrt 2 }} \\
\Rightarrow \dfrac{{{B_{\text{H}}}}}{{B_{\text{H}}^1}} &= \dfrac{{\sqrt 3 }}{{\sqrt 2 }} \\
\end{align*}
\[\therefore {B_{\text{H}}}:B_{\text{H}}^1 = \sqrt 3 :\sqrt 2 \]
Hence, the ratio of the horizontal component of earth’s magnetic field at these two places is \[\sqrt 3 :\sqrt 2 \] .

Therefore, the correct option is A.

Note: While solving this problem, it is important to note that the horizontal component is associated with the trigonometric ratio cosine. When the magnetic field points downwards, the dip angle is assumed to be positive. The dip angle is assumed to be negative as the magnetic field points upwards.