A bar magnet of magnetic moment `2.0` `A-m^2` is free to rotate about a vertical axis through its centre. The magnet is released from rest from the east west position. Find the kinetic energy of the magnet as it takes the north south position. The horizontal component of the earth's magnetic field as `B=25muT`. Earth's magnetic field is from south to north.
A
`40muJ`
B
`50muJ`
C
`60muJ`
D
`70muJ`
Text Solution
Verified by Experts
The correct Answer is:
B
`"Gain in kinetic energy" ="loss in potential energy"` Thus, `KE=U_i-U_f` `U=Mbcostheta` As, `KE=-Mbcos(pi/2)-(-MBcostheta^@)` `=MB` Susbtituting the values we have `KE=(2.0)(25xx10^I-6)J` `=50muJ`
Topper's Solved these Questions
MAGNETICS
DC PANDEY|Exercise Exercise|160 Videos
MAGNETICS
DC PANDEY|Exercise INTRODUCTORY EXERCISE|1 Videos
MAGNETIC FIELD AND FORCES
DC PANDEY|Exercise Medical entrance s gallery|59 Videos
MAGNETISM AND MATTER
DC PANDEY|Exercise Medical gallery|1 Videos
Similar Questions
Explore conceptually related problems
A bar magnet of magnetic moment 2.0 A m^2 is free to rotate about a vertical axis through its centre. The magnet is released from rest from the east-west position. Find the kinetic energy of the magnet as it takes the north-south position. The horizontal component of the earth's magnetic field is B = 25 muT .
A bar of magnetic moment 2.5 Am^(2) is free to rotate about a vertical axis through its centre . The magnet is released from the rest from the east - west direction . Find the kinetic energy of the magnet as it aligns itself in the north - south direction . The horizontal component of earth's magnetic field is 0.3 G
A bar magnet of moment 40 A-m^(2) is free to rotate about a vetical axis passing through its centre. The magnet is released from rest from east west direction. The kinetic energy of the magnet as it takes north-south direction is (B_(H)=30muT)
A bar magnet of magnetic moment 2Am^(2) is free to rotate about a vertical axis passing through its centre. The magnet is released from rest from east-west position. Then the KE of the magnet as it takes N-S position is (B_(H)=25muT)
A bar magnet of magnetic moment 4Am^(2) is free to rotate about a vertical axis passing through it centre.The magnet is released from rest from east-west position.Then K.E of the magnet as it takes north-south position is B_(H)=20 mu T
49.A bar magnet of magnetic moment 2Am^(2) is free to rotate about a vertical axis passing through its centre.The magnet is released from rest east-west position.Then the K.E of the magnet as it takes north-south position is (2n+40)mu J then "n" is ( B_(H)=25 mu T )
Explain the horizontal and vertical components of earth's magnetic field.
At a certain place, the angle of dip is 30^(@) and the horizontal component of earth's magnetic field is 50muT . The total magnetic field (in muT ) of the earth at this place, is
At the magnetic north pole of the earth, the value of horizontal component of earth's magnetic field and angle of dip are, respectively
