A compass needle whose magnetic moment is `60Am^2` pointing geograhic north at a certain place, where the horizontal component of earth's magnetic field is `40muWbm^-2` experiences a torque `1*2xx10^-3Nm`. What is the declination of the place?
A
`alpha=45^(@)`
B
`alpha=60^(@)`
C
`alpha=25^(@)`
D
`alpha=30^(@)`
Text Solution
Verified by Experts
The correct Answer is:
D
A compass needle in stable equilibrium position points towards nroth i.e. alon the horizontal components H of earth's magnetic field. When it is turned through the angle of declination `alpha`, so as as to point geographical north, then it experience a torque of magnitude MH sin `alpha`. `Here, M=60A-m^(2)` `" " H=40xx10^(-6)Wb//m^(2)` `therefore " " sin alpha=(1.2xx10^(-3))/(60xx40xx10)=0.5 therefore alpha=30^(@)`
Topper's Solved these Questions
MAGNETISM AND MATTER
DC PANDEY|Exercise 5.1|14 Videos
MAGNETISM AND MATTER
DC PANDEY|Exercise 5.2|14 Videos
MAGNETICS
DC PANDEY|Exercise MCQ_TYPE|1 Videos
MODERN PHYSICS
DC PANDEY|Exercise Integer Type Questions|17 Videos
Similar Questions
Explore conceptually related problems
A compass needle whose magnetic moment is 60Am^2 pointing geographic north at a certain place, where the horizontal component of earth's magnetic field is 40mu Wb//m^2 , experience a torque 1*2xx10^-3Nm . What is the declination of the place?
A compass needle whose magnetic moment is 60Am^2 pointing geographic north at a certain place where horizontal component of earth's magnetic field is 40muWb//m^2 experiences a torque of 1*2xx10^-3Nm . What is the declination of the place?
A compass needle of magnetic moment 50 Am^2 experiences a torque of 1.3 xx 10^(-3) Nm when pointing towards geographical north at a certain place where horizontal component of Earth's magnetic field is 30 mu Wb//m^2 . What is the declination of the place?
A compass needle whose magnetic moment is 60Am^2 points geographic north at a certain place where H=40muT . The compass needle experience a torque of 1*2xx10^-3N-m . What is the declination at the place?
The horizontal component of earth's magnetic field is sqrt3 times the vertical component . What is the value of angle of dip at this place ?
At a certain place, the horizontal component of earth's magnetic field is sqrt(3) times the vertical component. The angle of dip at that place is
Horizontal component of earth’s magnetic field at a place is sqrt()3 times the vertical component. What is the value of inclination at that place?
In the magnetic meridian of a certain place, the horizontal component of the earth's magnetic field is 0*26G and dip angle is 60^@ . What is the magnetic field of earth at this location?
