The Question is from neet-physics magnetic-effects-current-and-magnetism terrestrial-magnetism, and it was created by the author VENKATESWARARAO

Question

A dip circle is placed such that its magnetic needle aligns with the magnetic meridian.  It is then rotated through an angle $	heta$ about a vertical axis. The apparent dip $\delta'$ is related to the true dip $\delta$ by:

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
 **Primary Concept**: Dip circle and its relation to Earth's magnetic field.
 **Related Topics**: Earth's magnetism, magnetic inclination, components of magnetic field.
 **Prerequisites**: Understanding of trigonometry and vector resolution.
## SOLUTION PATHWAY
**Given Information**:
  * True dip angle: $\delta$
  * Apparent dip angle after rotation: $\delta'$
  * Angle of rotation about vertical axis: $	heta$
**Method & Steps**:
 1. Resolve the horizontal component of Earth's magnetic field into components parallel and perpendicular to the plane of the dip circle after rotation.
  Reasoning: When the dip circle is rotated, the horizontal component of Earth's magnetic field gets resolved into two components. One component lies in the plane of the dip circle, and the other is perpendicular to it.
  Formula/Rule:  Let $H$ be the horizontal component of Earth's magnetic field. After rotation, the component in the plane of the dip circle is $H \cos	heta$.
 2. Relate the apparent dip ($\delta'$) to the true dip ($\delta$) using the resolved horizontal component and the vertical component of Earth's magnetic field.
  Working: The vertical component of Earth's magnetic field remains unchanged ($V$).  The tangent of the apparent dip is given by the ratio of the vertical component to the horizontal component in the plane of the dip circle:
  $	an \delta' = \frac{V}{H \cos 	heta}$
  Since $	an \delta = \frac{V}{H}$, we can substitute this into the equation above:
  $	an \delta' = \frac{H 	an \delta}{H \cos 	heta} = \frac{	an \delta}{\cos 	heta} = 	an \delta \sec 	heta$
 3. Final Result:
  Result: $	an \delta' = 	an \delta \sec 	heta$
  Verification: When $	heta = 0$, $	an \delta' = 	an \delta$, which is consistent with the true dip.
 ## OPTION ANALYSIS
A ✓: $	an \delta' = 	an \delta \sec 	heta$: This is the correct relationship derived from the resolution of the horizontal component of the Earth's magnetic field.
B :x: $	an \delta' = 	an \delta \cos 	heta$: This equation incorrectly suggests that the apparent dip decreases with increasing rotation angle.
C :x: $\cot \delta' = \cot \delta \sec 	heta$: This is not the correct relationship between the true and apparent dip angles.
D :x: $\cot \delta' = \cot \delta \cos 	heta$: This is also an incorrect relationship.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Incorrectly resolving the horizontal component of the Earth's magnetic field:  Remember to resolve the horizontal component only into components parallel and perpendicular to the plane of the rotated dip circle.
  2.Confusing true dip and apparent dip:  The true dip is the angle of inclination when the dip circle is aligned with the magnetic meridian. The apparent dip changes with rotation.
 **Quick Tips**:
  1.Visualize the vector resolution of the horizontal component. Draw a diagram to help understand the geometry.
  2.Remember the basic trigonometric relationships between the true and apparent dip angles.

VENKATESWARARAO · Answer

The correct Option is "$	an \delta' = 	an \delta \sec 	heta$"

VENKATESWARARAO · Answer

$	an \delta' = 	an \delta \cos 	heta$

VENKATESWARARAO · Answer

$\cot \delta' = \cot \delta \sec 	heta$

VENKATESWARARAO · Answer

$\cot \delta' = \cot \delta \cos 	heta$

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