The value of the horizontal component of the earth's magnetic field and and angle of dip are `1.8xx10^(-5) Weder//m^(2)` and `30^(@)` respectively at some place. The total intensity of earth's magnetic field at that place will be
A
`2.08xx10^(-5)Weber//m^(2)`
B
`3.67xx10^(-5)Weber//m^(2)`
C
`3.18xx10^(-5)Weber//m^(2)`
D
`5.0xx10^(-5)Weber//m^(2)`
Text Solution
Verified by Experts
The correct Answer is:
A
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 horizontal compound of the earth's magnetic field at a given place is 0.4xx10^(-4) Wb//m^(2) and angle of dip =30^(@) . What is the total intensity of the earth's magnetic field at that place?
The vartical component of the earth's magnetic field is zero at a place where the angle of dip is
If the angle of dip at two places are 30^(@) and 45^(@) respectively, then the ratio of horizontal components of earth's magnetic field at the two places will be
At the magnetic north pole of the earth, the value of horizontal component of earth's magnetic field and angle of dip are, respectively
The horizontal component of the earth's magnetic field at any place is 0.36 xx 10^(-4) Wb m^(-2) If the angle of dip at that place is 60 ^@ then the value of the vertical component of earth's magnetic field will be ( in Wb m^(-2))
The value of horizontal component of earth's magnetic field at a place is -.35xx10^(-4)T . If the angle of dip is 60^(@) , the value of vertical component of earth's magnetic field is about:
