A short bar magnet with its axis inclined at 30 degree to the external magnetic field of 800G acting horizontally experiences a torque of 0.016Nm. calculate the
a) the magnetic moment of the magnet.
b) the work done by an external force in moving it from most stable to most unstable position.
c) what is the work done by the force due to the external magnetic field in the process mentioned in (b).

2

Answers

2015-01-01T04:41:38+05:30

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Torque = M X B
    =  cross product of magnetic moment of the bar magnet and Magnetic field
Magnitude of Torque = M * B Sin Ф
  0.016 = M * 800 G Sin 30
 
      M = 0.016 / (400 G)   A-m²

Work done = Integral of M B Sin Ф   from Ф = 0 to Ф = 180 deg
    W =  M B ( Cos 0 - Cos 180)  = 2 M B = 2 * 0.016 /(400G) * 800 G  Joules
         = 0.064 Joules

c)     same as in (b) but negative in sign.   so  W = - 0.064 J

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thx a lot
can u please give me the reason for "C"
thx
The Brainliest Answer!
2015-01-06T06:46:26+05:30
Magnitude of Torque = M * B Sin Ф
  0.016 = M * 800 G Sin 30
 
      M = 0.016 / (400 G)   A-m²
also 
Work done = Integral of M B Sin Ф   from Ф
that is 0.064 J
we already got .064 but we have to consider a negative sign here 
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