Question:
A dip needle vibrates in a vertical plane perpendicular to the magnetic meridian. The time period of vibration is
A dip needle vibrates in a vertical plane perpendicular to the magnetic meridian. The time period of vibration is
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An "infinite time period" physically means the object never completes a single swing. It just stays where it is because there is no magnetic "restoring force" to pull it back to a center position.
Updated On: Apr 24, 2026
- zero
- infinite
- same as in the magnetic meridian
- more than in the magnetic meridian
- less than in the magnetic meridian
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The Correct Option is B
Solution and Explanation
Concept:
Physics - Magnetism and Earth's Magnetic Field.
Step 1: Understand the Magnetic Meridian.
The magnetic meridian is the vertical plane passing through the magnetic axis of a freely suspended magnet. The Earth's total magnetic field ($B$) lies entirely within this plane.
Step 2: Analyze the components of the field.
In a plane perpendicular to the magnetic meridian, there is no component of the Earth's horizontal magnetic field ($B_{H}$).
Step 3: Recall the Time Period formula.
The time period of vibration for a magnet is: $$ T = 2\pi\sqrt{\frac{I}{MB_{H}}} $$
Step 4: Apply the condition.
In a plane perpendicular to the meridian, the effective horizontal component becomes zero ($B_{H} = 0$). $$ T = 2\pi\sqrt{\frac{I}{M \times 0}} = \infty $$ The needle will not vibrate; it will remain stationary in any position it is placed.
Physics - Magnetism and Earth's Magnetic Field.
Step 1: Understand the Magnetic Meridian.
The magnetic meridian is the vertical plane passing through the magnetic axis of a freely suspended magnet. The Earth's total magnetic field ($B$) lies entirely within this plane.
Step 2: Analyze the components of the field.
In a plane perpendicular to the magnetic meridian, there is no component of the Earth's horizontal magnetic field ($B_{H}$).
Step 3: Recall the Time Period formula.
The time period of vibration for a magnet is: $$ T = 2\pi\sqrt{\frac{I}{MB_{H}}} $$
Step 4: Apply the condition.
In a plane perpendicular to the meridian, the effective horizontal component becomes zero ($B_{H} = 0$). $$ T = 2\pi\sqrt{\frac{I}{M \times 0}} = \infty $$ The needle will not vibrate; it will remain stationary in any position it is placed.
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