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work done in moving a magnetic dipole from an angle 1 to an angle 2 w.r.t. the direction of magnetic field = [ 1 ] [ 2 ] .d ... where is the torque reqd. to produce the given angular disp. = [ 1 ] [ 2 ] M.B.sin .d = - MB ( cos 2 - cos 1 ) = MB (cos 1 - cos 2 ) thus this work done is stored as the potential enrgy of the field-magnet system. Thus .. U( 2 ) - U( 1 ) = MB (cos 1 - cos 2 ) Now in HCV .... it is written that let us assume that the potential energy at = 90 o as zero....and so the potential energy at an angle is U( ) - U(90 o ) = MB ( cos90 o - cos ) = - MBcos Now why shuold we suddenly consider potential energy at = 90 o as zero ... ?????????


work done in moving a magnetic dipole from an angle 1 to an angle 2 w.r.t. the direction of magnetic field =
[1 ][2 ] .d ... where  is the torque reqd. to produce the given angular disp.
 
= [1 ][2 ] M.B.sin.d
= - MB ( cos2 - cos1 )
= MB (cos1 - cos2 )
thus this work done is stored as the potential enrgy of the field-magnet system.
Thus .. U(2) - U(1) = MB (cos1 - cos2 )
 
Now in HCV .... it is written that let us assume that the potential energy at  = 90o as zero....and so the potential energy at an angle  is
U() - U(90o) = MB ( cos90o - cos )
                   = - MBcos
                   
 
Now why shuold we suddenly consider potential energy at  = 90o as zero ... ?????????
 


Grade:upto college level

1 Answers

ROSHAN MUJEEB
askIITians Faculty 833 Points
3 years ago
because cos 90 is zero and the formula defines it to be so

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