Home
Class 12
PHYSICS
The horizontal component of the earth's ...

The horizontal component of the earth's magnetic field is `3.6 xx 10^(-5) T` where the dip is `60^@`. Find the magnitude of the earth's magnetic field.

Text Solution

AI Generated Solution

To find the magnitude of the Earth's magnetic field given the horizontal component and the angle of dip, we can follow these steps: ### Step 1: Write down the given values - Horizontal component of the Earth's magnetic field, \( B_H = 3.6 \times 10^{-5} \, \text{T} \) - Angle of dip, \( \delta = 60^\circ \) ### Step 2: Use the relationship between horizontal component and total magnetic field The relationship between the horizontal component of the magnetic field \( B_H \) and the total magnetic field \( B \) is given by the formula: ...
Promotional Banner

Topper's Solved these Questions

  • PERMANENT MAGNETS

    HC VERMA ENGLISH|Exercise Worker out examples|19 Videos
  • PERMANENT MAGNETS

    HC VERMA ENGLISH|Exercise Objective|13 Videos
  • OPTICAL INSTRUMENTS

    HC VERMA ENGLISH|Exercise Question for short Answer|12 Videos
  • PHOTO ELECTRIC EFFECT AND WAVE PARTICLE DUALITY

    HC VERMA ENGLISH|Exercise question for short answer|13 Videos

Similar Questions

Explore conceptually related problems

The strength of the earth's magnetic field is

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip down at 60^(@) with the horizontal.The horizontal component of the earth's magnetic field at the place is known to be 0.4G .Determine the magnitude of the earth's magnetic field at the place.

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 a. Vertical component of earth's magnetic field b. the net magnetic field at this place

A tangent galvanometer shown a deflection of 45^@ when 10 mA of current is passed through it. If the horizontal component of the earth's magnetic field is B_H = 3.6 xx 10^(-5) T and radius of the coil is 10cm , find the number of turns in the coil.

At a place, the horizontal component of earth's magnetic field is B and angle of dip is 60^@ . What is the value of horizontal component of earth's magnetic field at equator?

At a place, the horizontal component of earth's magnetic field is V and angle of dip is 60^(@) . What is te value of horizontal component of earth's magnetic field at equator ?

A magnetic needle free to rotate about the vertical direction (compass ) point 3.5 west of the geographic north . Another magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 18^(@) with the horizontal . The magnitude of the horizontal component of the earth ' magnetic field at the place is known to be 0.40 G . what is the direction and magnitude of the earth's magnetic field at the place ?

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22^@ with the horizontal. The horizontal component of the earth's magnetic field at the place is known to be 0*35G . Determine the strength of the earth's magnetic field at the place.

At a certain place, the angle of dip is 60^(@) and the horizontal component of the earth's magnetic field (B_(H)) is 0.8xx10^(-4) T. The earth's overall magnetic field is

The angle of dip at a place is 40.6^(@) and the intensity of the vertical component of the earth's magnetic field V=6xx10^(-5) Tesla. The total intensity of the earth's magnetic field (I) at this place is