Question:medium

Two bar magnets having same geometry with magnetic moments m and 2m, are firstly placed such that their similar poles are on the same side. The time period of oscillation is $T₁$. When the polarity of one is reversed, the time period is $T₂$. Then:

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Time period is inversely proportional to square root of magnetic moment.
Updated On: May 24, 2026
  • $T₁ < T₂$
  • $T₁ = T₂$
  • $T₁ > T₂$
  • $T₂ = ∞$
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The Correct Option is A

Solution and Explanation

The question involves the oscillation of two bar magnets, which are placed under different conditions that affect their time periods of oscillation. Let's analyze the problem step-by-step:

  1. We have two bar magnets with magnetic moments \(m\) and \(2m\). They are placed in such a way that either similar poles are on the same side or one of them is reversed, causing a differential interaction between them.
  2. The time period of oscillation of a magnet in a magnetic field is given by the formula: \(T = 2\pi \sqrt{\frac{I}{m_{eff} \cdot B}}\) where \(I\) is the moment of inertia of the magnet, \(m_{eff}\) is the effective magnetic moment, and \(B\) is the magnetic field.
  3. When the similar poles are on the same side (scenario for \(T₁\)), the effective magnetic moment is \(m_{eff} = m + 2m = 3m\).
  4. When one magnet is reversed (scenario for \(T₂\)), the effective magnetic moment becomes \(m_{eff} = 2m - m = m\) (since they oppose each other).
  5. The time periods for these scenarios are:
    • For \(T₁\)\(T₁ = 2\pi \sqrt{\frac{I}{3m \cdot B}}\)
    • For \(T₂\)\(T₂ = 2\pi \sqrt{\frac{I}{m \cdot B}}\)
  6. From the above expressions, \(T₁\) has a larger denominator due to the larger effective magnetic moment. Thus, \(T₁\) is less than \(T₂\), given that an increase in the denominator for a fixed numerator will decrease the overall expression.

Thus, the correct answer is \(T₁ < T₂\), which makes option \(T₁ < T₂\) the right choice.

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