Question

A short magnet oscillates in an oscillation magnetometer with a time period of 0.10 s where the earth's horizontal magnetic field is 24 μT. A downward current of 18 A is established in a vertical wire placed 20 cm east of the magnet. Find the new time period.

Answer 1

Here,
Horizontal component of Earth's magnetic field, B_H = 24 × 10⁻⁶ T
Time period of oscillation, T₁ = 0.1 s
Downward current in the vertical wire, I = 18 A
Distance of wire from the magnet, r = 20 cm = 0.2 m
In the absence of the wire net magnetic field, B_H = 24 × 10⁻⁶ T
When a current-carrying wire is placed near the magnet, the effective magnetic field gets changed.
Now the net magnetic field can be obtained by subtracting the magnetic field due to the wire from Earth's magnetic field.
B = B_H − B_wire
Thus, the magnetic field due to the current-carrying wire $\left(B_{wire}\right)$ is given by
B = $\frac{{\mu }_{0}I}{2\pi r}$
The net magnetic field $\left(B\right)$ is given by
$$\begin{gathered} B=24\times {10}^{-6}-\frac{{\mathrm{\mu }}_0\mathit{}I}{2\mathrm{\pi }r} \\ B=24\times {10}^{-6}-\frac{2\times {10}^{-7}\times 18}{0.2} \\ B=\left(24-10\right)\times {10}^{-6} \\ B=14\times {10}^{-6} \end{gathered}$$

Time period of the coil $\left(T\right)$ is given by
$T=2\mathrm{\pi }\sqrt{\frac{I}{M{B}_{\mathrm{H}}}}$
Let T₁ and T₂ be the time periods of the coil in the absence of the wire and in the presence the wire respectively.
As time period is inversely proportional to magnetic field,

$$\begin{gathered} \frac{T_1}{T_2}=\sqrt{\frac{B}{B_{\mathrm{H}}}} \\ \Rightarrow \frac{0.1}{T_2}=\sqrt{\frac{14\times {10}^{-6}}{24\times {10}^{-6}}} \\ \Rightarrow {\left(\frac{0.1}{T_2}\right)}^2=\frac{14}{24} \\ \Rightarrow T_2^2=\frac{0.01\times 14}{24} \\ \Rightarrow T_2=0.076\mathrm{s} \end{gathered}$$