Question

A short magnet makes 40 oscillations per minute when used in an oscillation magnetometer at a place where the earth's horizontal magnetic field is 25 μT. Another short magnet of magnetic moment 1.6 A m² is placed 20 cm east of the oscillating magnet. Find the new frequency of oscillation if the magnet has its north pole (a) towards north and (b) towards south.

Answer 1

Here,
Frequency of oscillations, $\nu$₁ = 40 oscillations/min
Earth's horizontal magnetic field, B_H = 25 μT
Magnetic moment of the second magnet, M = 1.6 A-m²
Distance at which another short magnet is placed, d = 20 cm = 0.2 m

(a) For the north pole of the short magnet facing the north, frequency $\left(v_1\right)$ is given by
$v_1=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{M{B}_H}{I}}$
Here,
M = Magnetic moment of the magnet
I = Moment of inertia
B_H = Horizontal component of the magnetic field
Now, let B be the magnetic field due to the short magnet.
When the north pole of the second magnet faces the north pole of the first magnet, the effective magnetic field $\left(B_{eff}\right)$ is given by
B_effective = B_H $-$ B
The new frequency of oscillations $\left(v_2\right)$ on placing the second magnet is given by
$v_{\mathit{2}}\mathit{=}\frac{1}{2\mathrm{\pi }}\mathit{}\sqrt{\frac{M\left(B_H\mathit{-}B\right)}{\mathit{1}}}$
The magnetic field produced by the short magnet $\left(B\right)$ is given by
$$\begin{gathered} B=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{m}{d^{\mathit{3}}} \\ \Rightarrow B=\frac{{10}^{-7}\times 1.6}{8\times {10}^{-3}}=20\mathrm{\mu T} \end{gathered}$$
Since the frequency is proportional to the magnetic field,
$$\begin{gathered} \frac{v_{\mathit{1}}}{v_{\mathit{2}}}=\sqrt{\frac{B_H}{B_H\mathit{-}B}} \\ \Rightarrow \frac{40}{v_2}=\sqrt{\frac{25}{5}} \\ \Rightarrow \frac{40}{v_2}=\sqrt{5} \\ \Rightarrow v_2=\frac{40}{\sqrt{5}}=17.88 \\ =18\mathrm{oscillations}/\min \end{gathered}$$

(b) For the north pole facing the south,
$$\begin{gathered} v_1=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{M{B}_H}{I}} \\ \Rightarrow v_2=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{M\left(B\mathit{+}{B}_H\right)}{\mathit{1}}} \\ \Rightarrow \frac{v_1}{v_2}=\sqrt{\frac{B_H}{B\mathit{+}{B}_H}} \\ \Rightarrow \frac{40}{v_2}=\sqrt{\frac{25}{45}} \\ \Rightarrow v_2=\frac{40}{\sqrt{25/45}}=54\mathrm{oscillations}/\min \end{gathered}$$