A magnetized sewing needle has a magnetic moment of 14.42 mA · m2. At its location, the Earth's magnetic field is 56.5 µT northward at 48.0° below the horizontal.

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A magnetized sewing needle has a magnetic moment of 14.42 mA · m2. At its location, the Earth's magnetic field is 56.5 µT northward at 48.0° below the horizontal.

(a) Identify the orientation of the needle that represent the minimum potential energy.
    

(b) Identify the orientation of the needle that represent the maximum potential energy of the needle–field system.
    

(c) How much work must be done on the system to move the needle from the minimum to the maximum potential energy orientation?
µJ 

 

asked Feb 19, 2012 in Physics by apush ~Expert~ (552 points)
    

1 Answer

+4 votes

 

The field exerts torque on the needle tending to align it with the field, so the minimum
energy orientation of the needle is: 
 
pointing north at 48.0° below the horizontal
 
where its energy is
 
Potential Energy min = -magnetic dipole moment * electric field * cos(0)
 
where:
magnetic dipole moment = 14.42 mA · m= 14.42*10^-3 mA*m^2
electric field =  56.5 µT = 56.5*10^-6 T
 
plugging back into the equation we get that:
 

Potential Energy min = -(14.42*10^-3 mA*m^2) * ( 56.5*10^-6 T)* cos(0) =

Potential Energy min = -8.1473*10^-7 J

 

 

It has maximum energy when pointing in the opposite direction,

south at 48.0° above the horizontal

 

 

where its energy is 
 
Potential Energy max = -magnetic dipole moment * electric field * cos(180)
 
Potential Energy max = -(14.42*10^-3 mA*m^2) * ( 56.5*10^-6 T)* cos(180)
 
Potential Energy max = 8.1473*10^-7 J
 
 
The work done is simply (Potential Energy max)-(Potential Energy min)
 
work =   (8.1473*10^-7 J) - (-8.1473*10^-7 J) = 1.62946*10^-6 J
 
work =  1.62946*10^-6 J  = 1.62946 µJ

 

 

answered Feb 23, 2012 by Roar3215 ~Expert~ (1,318 points)

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