Question:
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
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
Show Hint
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.
Updated On: Jun 25, 2026
- \(H_A=8\times 10^{-6}\ \text{T}\)
- \(H_A=32\times 10^{-6}\ \text{T}\)
- \(H_A=4\times 10^{-6}\ \text{T}\)
- \(H_A=16\times 10^{-6}\ \text{T}\)
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The Correct Option is A
Solution and Explanation
Step 1: Write the time period formula of a vibration magnetometer.
For a magnet suspended freely in a vibration magnetometer, \[ T=2\pi\sqrt{\frac{I}{MH}} \] where \[ T=\text{time period} \] \[ I=\text{moment of inertia of the magnet} \] \[ M=\text{magnetic moment} \] and \[ H=\text{horizontal component of earth's magnetic field} \] For the same magnet, \[ I \text{ and } M \] remain constant.
Therefore, \[ T\propto \frac{1}{\sqrt{H}} \]
Step 2: Relate time periods and magnetic fields at \(A\) and \(B\).
Since \[ T\propto \frac{1}{\sqrt{H}}, \] we have \[ \frac{T_A}{T_B}=\sqrt{\frac{H_B}{H_A}} \] Given: \[ T_A=2T_B \] Therefore, \[ \frac{T_A}{T_B}=2 \] So, \[ 2=\sqrt{\frac{H_B}{H_A}} \]
Step 3: Square both sides.
\[ 4=\frac{H_B}{H_A} \] Thus, \[ H_A=\frac{H_B}{4} \]
Step 4: Substitute the value of \(H_B\).
Given: \[ H_B=32\times 10^{-6}\ \text{T} \] Hence, \[ H_A=\frac{32\times 10^{-6}}{4} \] \[ H_A=8\times 10^{-6}\ \text{T} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{H_A=8\times 10^{-6}\ \text{T}} \]
For a magnet suspended freely in a vibration magnetometer, \[ T=2\pi\sqrt{\frac{I}{MH}} \] where \[ T=\text{time period} \] \[ I=\text{moment of inertia of the magnet} \] \[ M=\text{magnetic moment} \] and \[ H=\text{horizontal component of earth's magnetic field} \] For the same magnet, \[ I \text{ and } M \] remain constant.
Therefore, \[ T\propto \frac{1}{\sqrt{H}} \]
Step 2: Relate time periods and magnetic fields at \(A\) and \(B\).
Since \[ T\propto \frac{1}{\sqrt{H}}, \] we have \[ \frac{T_A}{T_B}=\sqrt{\frac{H_B}{H_A}} \] Given: \[ T_A=2T_B \] Therefore, \[ \frac{T_A}{T_B}=2 \] So, \[ 2=\sqrt{\frac{H_B}{H_A}} \]
Step 3: Square both sides.
\[ 4=\frac{H_B}{H_A} \] Thus, \[ H_A=\frac{H_B}{4} \]
Step 4: Substitute the value of \(H_B\).
Given: \[ H_B=32\times 10^{-6}\ \text{T} \] Hence, \[ H_A=\frac{32\times 10^{-6}}{4} \] \[ H_A=8\times 10^{-6}\ \text{T} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{H_A=8\times 10^{-6}\ \text{T}} \]
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