Home
Class 12
PHYSICS
The earth's magnetic field at a certain ...

The earth's magnetic field at a certain place has a horizontal component 0.3 Gauss and the total strength 0.5 Gauss. The angle of dip is

A

`tan^(-1),(3)/(4)`

B

`sin^(-1),(3)/(4)`

C

`tan^(-1),(4)/(3)`

D

`sin^(-1),(4)/(3)`

Text Solution

Verified by Experts

The correct Answer is:
C
Promotional Banner

Topper's Solved these Questions

  • MAGNETISM

    ERRORLESS |Exercise Magnet and it s Properties|15 Videos

  • MAGNETISM

    ERRORLESS |Exercise Earth Magnetism|8 Videos

  • MAGNETIC EFFECT OF CURRENT

    ERRORLESS |Exercise Exercise|394 Videos

  • RAY OPTICS

    ERRORLESS |Exercise SET|25 Videos

Similar Questions

Explore conceptually related problems

The vertical component of earth's magnetic field at a place is sqrt3 times the horizontal component the value of angle of dip at this place is

Vertical component of earth's magnetic field at a place is sqrt3 times its horizontal component. If total intensity of earth's magnetic field at the place is 0*4Gauss , find (i) angle of dip (ii) horizontal component of earth's magnetic field.

Our earth behaves as it has a powerful magnet within it. The value of magnetic field on the surface of earth is a few tenth of gauss (1G = 10^(-4) T) There are three elements of Earth’s magnetism (i) Angle of declination (ii) Angle of dip (iii) Horizontal component of Earth’s magnetic field. In the magnetic meridian of a certain place, the horizontal component of Earth’s magnetic field is 0.6 G and the dip angle is 53^@ . The value of vertical component of earth's magnetic field at this place is

In the magnetic meridian of a certain place, the horizontal component of earth's magnetic fied is 0.26 G and the dip angle is 60^@ . Find Vertical component of earth's magnetic field b. the net magnetic field at this place

Calculate earth's magnetic field at a place where angle of dip delta is 60^(@) and horizontal component of earth's magnetic field is 0.3 G