} \, \text{T} $

The initial time period of oscillation for the magnet is:

$ T_1 = 0.1 \, \text{s} $

The current flowing through the vertical wire is:

$ I = 18 \, \text{A} $

The distance between the wire and the magnet is:

$ r = 20 \, \text{cm} = 0.2 \, \text{m} $

To find the magnetic field due to the current-carrying wire, we use the formula for the magnetic field around a long straight conductor:

$ B_{\text{wire}} = \frac{\mu_0}{2\pi} \cdot \frac{I}{r} $

Substituting the known values:

$ B_{\text{wire}} = \frac{10^{-7} \times 2 \times 18}{0.2} = 18 \times 10^{-6} \, \text{T} $

The time period of a magnetic dipole oscillating in a magnetic field is given by:

$ T = 2 \pi \sqrt{\frac{I}{MB_H}} $

where:

$ I $ is the moment of inertia,

$ M $ is the magnetic moment.

For the initial and new conditions, we have:

$ T_1 = 2 \pi \sqrt{\frac{I}{M B_H}} \quad \text{(i)} $

$ T_2 = 2 \pi \sqrt{\frac{I}{M (B_H + B_{\text{wire}})}} \quad \text{(ii)} $

By dividing equation (i) by equation (ii), we obtain:

$ \frac{T_1}{T_2} = \sqrt{\frac{B_H + B_{\text{wire}}}{B_H}} $

Substitute the given values:

$ \frac{T_1}{T_2} = \sqrt{\frac{24 \times 10^{-6} + 18 \times 10^{-6}}{24 \times 10^{-6}}} $

$ \frac{0.1}{T_2} = \sqrt{\frac{42 \times 10^{-6}}{24 \times 10^{-6}}} = \frac{\sqrt{7}}{2} $

Solving further:

$ T_2 = \frac{0.2}{\sqrt{7}} $

Thus, the new time period of oscillation of the magnet is approximately:

$ T_2 = 0.076 \, \text{s} $

Moving Charges And Magnetism

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