Consider the plane $S$ formed by the dipole axis and the axis of earth. Let $P$ be point on the magnetic equator and in $S$. Let $Q$ be the point of intersection of the geographical and magnetic equators. Obtain the declination and dip angles at $P$ and $Q$.
$P$ is in the plane $S$, needle is in north, so the declination is zero.
$P$ is also on the magnetic equator, so the angle of dip $=0$, because the value of angle of dip at equator is zero. $Q$ is also on the magnetic equator, thus the angle of dip is zero. As earth tilted on its axis by $11.3^{\circ}$, thus the declination at $Q$ is $11.3^{\circ}$.
There are two current carrying planar coil made each from identical wires of length L. $C_1$ is circular (radius $R$ ) and $C_2$ is square (side $a$ ). They are so constructed that they have same frequency of oscillation when they are placed in the same uniform B and carry the same current. Find $a$ in terms of $R$.
$C_1=$ circular coil of radius $R$, length $L$, number of turns per unit length
$$n_1=\frac{L}{2 \pi R}$$
$C_2=$ square of side $a$ and perimeter $L$, number of turns per unit length $n_2=\frac{L}{4 a}$
Magnetic moment of $C_1$
$$\Rightarrow \quad m_1=n_1 I A_1$$
Magnetic moment of $C_2$
$$\Rightarrow \quad m_2=n_2 I A_2$$
$$ \begin{aligned} m_1 & =\frac{L \cdot I \cdot \pi R^2}{2 \pi R} \\ m_2 & =\frac{L}{4 a} \cdot I \cdot a^2 \\ m_1 & =\frac{L I R}{2} \quad\text{.... (i)}\\ m_2 & =\frac{L I a}{4}\quad\text{.... (ii)} \end{aligned}$$
Moment of inertia of $C_1 \Rightarrow I_1=\frac{M R^2}{2}\quad\text{.... (iii)}$
Moment of inertia of $\mathrm{C}_2 \Rightarrow I_2=\frac{\mathrm{Ma}^2}{12}\quad\text{.... (iv)}$
Frequency of $C_1 \Rightarrow f_1=2 \pi \sqrt{\frac{I_1}{m_1 B}}$
Frequency of $C_2 \Rightarrow f_2=2 \pi \sqrt{\frac{I_2}{m_2 B}}$
According to question, $f_1=f_2$
$$\begin{aligned} 2 \pi \sqrt{\frac{I_1}{m_1 B}} & =2 \pi \sqrt{\frac{I_2}{m_2 B}} \\ \frac{I_1}{m_1} & =\frac{I_2}{m_2} \text { or } \frac{m_2}{m_1}=\frac{I_2}{I_1} \end{aligned}$$
Plugging the values by Eqs. (i), (ii), (iii) and (iv)
$$\begin{aligned} \frac{L I a \cdot 2}{4 \times L I R} & =\frac{M a^2 \cdot 2}{12 \cdot M R^2} \\ \frac{a}{2 R} & =\frac{a^2}{6 R^2} \\ 3 R & =a \end{aligned}$$
Thus, the value of $a$ is $3 R$.