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How do I find the mass of a magnet using the trendline from this graph? In our physics class we
did a lab where we hung a magnet from a spring and let it oscillate inside of a solenoid. The
attached photo is what we recorded when the magnet oscillated for a few seconds. Im supposed
to take the info from the trendline of the graph and calculate the mass of the magnet how do i do
that? Thanks.
Solution
One of Maxwell's equations is
?×E? =?dB? dt.?×E?=?dB?dt.
Consider an imaginary disk whose normal vector is parallel to the axis of the coil and which is
inside the coil. If you integrate this equation over the area of that disk you get [a][a]
E=???E=???
where ?? is the flux threading the disk and EE is the EMF drop around the loop. This is called
Faraday's law.
So, for each imaginary disk inside your coil we get some EMF as the flux through that disk
changes in time.
Now think about the bar magnet's descent. Suppose we drop it starting way above the entrance
to the coil. It's far away, so there's no flux and no EMF. As it descends and gets close to the
entrance of the coil, some of the magnetic field from the magnet threads the top few imaginary
disks of the coil. The time changing flux induces some EMF. This is the initial rise in the red part
of the diagram. As the magnet continues to fall, and enters the coil, more of its magnetic field is
threading imaginary disks in the coil, so as it moves the time rate of change of total flux
increases, so the EMF goes up. Note that the field lines above and below the bar magnet point in
the same direction.
At some point, the bar reaches the middle of the coil. At this point, the amount of flux added to
the top half of the coil by a small motion of the magnet is equal to the amount of flux removed
from the bottom half. Therefore, at this point the EMF is zero. This is the midopint of the
diagram where the EMF crosses the horizontal axis. The falling part of the red section is just the
approach to the mid section of the coil.
As the bar magnet exits the coil, more flux is leaving than is entering, so the EMF versus time in
the blue section is just the opposite (except for the stretching which you already understand) of
the red section.
[a]: On the right hand side, the area integral of the magnetic field is the flux ?? by definition, and
the time derivative just goes along for the ride. On the left hand side you are doing an area
integral of a curl of a vector, which by Stokes's theorem is equivalent to the line integral of the
vector itself around the boundary of the area. The line integral of the electric field vector is the
EMF by definition.

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How do I find the mass of a magnet using the trendline from this graph.docx

  • 1. How do I find the mass of a magnet using the trendline from this graph? In our physics class we did a lab where we hung a magnet from a spring and let it oscillate inside of a solenoid. The attached photo is what we recorded when the magnet oscillated for a few seconds. Im supposed to take the info from the trendline of the graph and calculate the mass of the magnet how do i do that? Thanks. Solution One of Maxwell's equations is ?×E? =?dB? dt.?×E?=?dB?dt. Consider an imaginary disk whose normal vector is parallel to the axis of the coil and which is inside the coil. If you integrate this equation over the area of that disk you get [a][a] E=???E=??? where ?? is the flux threading the disk and EE is the EMF drop around the loop. This is called Faraday's law. So, for each imaginary disk inside your coil we get some EMF as the flux through that disk changes in time. Now think about the bar magnet's descent. Suppose we drop it starting way above the entrance to the coil. It's far away, so there's no flux and no EMF. As it descends and gets close to the entrance of the coil, some of the magnetic field from the magnet threads the top few imaginary disks of the coil. The time changing flux induces some EMF. This is the initial rise in the red part of the diagram. As the magnet continues to fall, and enters the coil, more of its magnetic field is threading imaginary disks in the coil, so as it moves the time rate of change of total flux increases, so the EMF goes up. Note that the field lines above and below the bar magnet point in the same direction. At some point, the bar reaches the middle of the coil. At this point, the amount of flux added to the top half of the coil by a small motion of the magnet is equal to the amount of flux removed from the bottom half. Therefore, at this point the EMF is zero. This is the midopint of the
  • 2. diagram where the EMF crosses the horizontal axis. The falling part of the red section is just the approach to the mid section of the coil. As the bar magnet exits the coil, more flux is leaving than is entering, so the EMF versus time in the blue section is just the opposite (except for the stretching which you already understand) of the red section. [a]: On the right hand side, the area integral of the magnetic field is the flux ?? by definition, and the time derivative just goes along for the ride. On the left hand side you are doing an area integral of a curl of a vector, which by Stokes's theorem is equivalent to the line integral of the vector itself around the boundary of the area. The line integral of the electric field vector is the EMF by definition.