A vibration magnetometer is used at two different places \(A\) and \(B\) on the earth. The time period of a magnet suspended freely in the magnetometer at \(A\) is twice that at \(B\). If the horizontal component of the earth's magnetic field at \(B\) is \(32\times 10^{-6}\ \text{T}\), then its value at \(A\) is
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For a vibration magnetometer,
\[
T=2\pi\sqrt{\frac{I}{MH}}
\]
For the same magnet,
\[
T\propto \frac{1}{\sqrt{H}}
\]
So, if time period increases, the horizontal magnetic field decreases.
Step 1: Write the time period formula of a vibration magnetometer. A magnet oscillating in the horizontal component of Earth's field $ H $ has time period \[ T = 2\pi\sqrt{\frac{I}{MH}} \] where $ I $ is the moment of inertia and $ M $ is the magnetic moment. For the same magnet, $ I $ and $ M $ are constant, so: \[ T \propto \frac{1}{\sqrt{H}} \] Step 2: Form the ratio of time periods at A and B. \[ \frac{T_A}{T_B} = \sqrt{\frac{H_B}{H_A}} \] This is derived by dividing the formula at A by the formula at B and simplifying. Step 3: Substitute the given condition. We are told $ T_A = 2T_B $, so $ \frac{T_A}{T_B} = 2 $. Substituting: \[ 2 = \sqrt{\frac{H_B}{H_A}} \] Step 4: Square both sides and solve for H_A. Squaring: \[ 4 = \frac{H_B}{H_A} \implies H_A = \frac{H_B}{4} \] Step 5: Substitute the value of H_B. Given $ H_B = 32 \times 10^{-6}\text{ T} $: \[ H_A = \frac{32 \times 10^{-6}}{4} = 8 \times 10^{-6}\text{ T} \] Step 6: Physical interpretation and final answer. Place A has a weaker horizontal field than B, which is why the magnet oscillates more slowly there (longer period). The result is consistent: weaker field at A gives longer time period. \[ \boxed{H_A = 8 \times 10^{-6}\text{ T}} \]