Home
Class 12
PHYSICS
What is the angle of dip at a place wher...

What is the angle of dip at a place where horizontal and vertical components of earth's field are equal?

Text Solution

AI Generated Solution

To find the angle of dip at a place where the horizontal and vertical components of the Earth's magnetic field are equal, we can follow these steps: ### Step 1: Understand the relationship between the components The angle of dip (δ) is defined as the angle made by the Earth's magnetic field with the horizontal. It can be calculated using the formula: \[ \tan(\delta) = \frac{V}{H} \] where \(V\) is the vertical component and \(H\) is the horizontal component of the Earth's magnetic field. ...
Promotional Banner

Topper's Solved these Questions

  • MAGNETIC EFFECT OF CURRENT AND MAGNETISM

    PRADEEP|Exercise Conceptual Problems (d)|2 Videos
  • MAGNETIC EFFECT OF CURRENT AND MAGNETISM

    PRADEEP|Exercise Very Short Answer Questions (a)|2 Videos
  • ELECTROSTATICS

    PRADEEP|Exercise ASSERTION-REASON TYPE QUESTIONS|2 Videos
  • OPTICS

    PRADEEP|Exercise Multiple choice questions|1 Videos

Similar Questions

Explore conceptually related problems

Calculate the angle of dip at a place where the horizontal and vertical components of Earth's magnetic field are 0.24 gauss and 0.36 gauss respectively. Also, calculate the resultant intensity of Earth's field.

The angle of dip at a place where horizontal and vertical components of earth magnetic field are same, is

At a particular place, horizontal and vertical components of earth's magnetic field are equal. The angle of dip at that place is…………… .

What is the horizontal component of Earth's magnetic field? If the horizontal and vertical components of Earth's magnetic field are equal then what is the angle of dip?

If the earth's field induction at a place is 0.36 G and the angle of dip is 30^(@) , the horizontal and vertical component of the field will be respectively

At a place the earth's horizontal component of magnetic field is 0.36xx10^(-4) "Weber"//m^(2) . If the angle of dip at that place is 60^(@) , then the vertical component of earth's field at that place in "Weber"//m^(2) will be approxmately