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Bar magnet and a current carrying wire

  • Thread starter Jahnavi
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  • #1
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Homework Statement


Bar Magnet.jpg


Homework Equations




The Attempt at a Solution



The magnetic field lines due to the infinite wire will be clockwise .When P is the midpoint of the magnet , Magnetic field at the South Pole will have a component upwards and a component towards right .This means S will experience a force downwards and towards left .

Similarly N will experience a force downwards and towards right .

The net force on the magnet will be downwards . But since equal forces act on both S and N downwards , there will be zero torque on the magnet .

I think both 1) and 3) should be right but the given answer is option 3) .

What is the mistake in my reasoning ?
 

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  • #2
Charles Link
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@Jahnavi I agree. I think (1) should also be correct. There will be torques in case 1 on the different parts of the magnet, but the net torque is zero for this case. i.e. the net torque measured about the center of mass. If the torque is measured about some other origin, you do get a different result, i.e. a non-zero torque.
 
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  • #3
kuruman
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@Jahnavi I agree. I think (1) should also be correct. There will be torques in case 1 on the different parts of the magnet, but the net torque is zero for this case.
I think the question is ambiguous. As you say, there are torques acting on the bar, but the net torque is zero. Consider though that when a book is at rest on a table top, we say that there are two forces acting on the book, not zero forces.
 
  • #4
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we say that there are two forces acting on the book, not zero forces.
But we do say , net force is zero .

Question is asking about net torque and net force on the magnet .

What is the ambiguity ?
 
  • #5
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There will be torques in case 1 on the different parts of the magnet
With respect to P , how ?
 
  • #6
kuruman
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But we do say , net force is zero .

Question is asking about net torque and net force on the magnet .

What is the ambiguity ?
I don't see the word "net" in any of the four choices.
 
  • #7
Charles Link
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With respect to P , how ?
The torque per unit volume is ## \tau=\vec{M} \times \vec{B} ##. The torque on the left side will cancel the torque on the right. Alternatively, the left side and right side will both experience a force that pulls the magnet towards the wire. This could actually result in a "torque" if the "torque" is not computed from the center of mass. I do think @kuruman is correct though=the question is rather ambiguous.
 
  • #8
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torque per unit volume
Interesting !

What's that ? I thought a bar magnet consists of two poles . Is there more to it ?
 
  • #9
Charles Link
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Interesting !

What's that ? I thought a bar magnet consists of two poles . Is there more to it ?
@Jahnavi You need to read my Insights article about Permanent Magnets and Magnetic Surface Currents. In a year or two, you may see a lot more ferromagnetism in an advanced E&M course. https://www.physicsforums.com/insights/permanent-magnets-ferromagnetism-magnetic-surface-currents/ ## \\ ## Edit: Meanwhile though, torque is a slightly tricky subject, because the origin needs to be specified. My statement of torque per unit volume ## \tau= \vec{M} \times {B} ## I believe is not completely correct. A magnetic moment ## \vec{m} ## on a magnetic field experiences a torque about its center given by ## \tau= \vec{m} \times \vec{B} ##. I don't know that that can be generalized to a large body composed of microscopic magnetic moments throughout its volume. In some cases, the net effect of these microscopic magnetic moments simplifies to having a ## "+" ## pole and a ## "-" ## pole on the ends.## \\ ## One equation that does apply though is the energy per unit volume ## U=-\vec{M} \cdot \vec{B} ##. The force that the object experiences can be found by considering the total energy ## E=\int U \, d^3r ## as a function of its location. ## \vec{F}=-\nabla E ## will give the force that the magnet experiences. ## \\ ## The generation of the magnetic field ## B ## from the permanent magnet can be done by using Biot-Savart on the magnetic surface currents. Alternatively, the magnetic field ## B ## can also be computed by treating the magnet with magnetic pole calculations. ## \\ ##
 
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  • #10
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Thank you very much :smile:
 
  • #11
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@kuruman , @Charles Link The textbook states that the work done by magnetic force is always zero . But if we place two bar magnets close to each other with their poles facing each other , they do exert a force and accelerate .This means work is done by the magnetic force .

So , does this mean work done by magnetic force can be non zero and zero work by magnetic force qv × B applies only to point charges ?

Please let me know what do you think .

Thanks
 
  • #12
kuruman
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The textbook states that the work done by magnetic force is always zero.
Read the textbook carefully. "Always" is too strong a word. As you pointed out, there are cases where magnetic fields actually exert forces and do work. It's the magnetic force on moving charges that does no work because the force is "always" instantaneously perpendicular to the displacement. More generally, uniform magnetic fields exert no net force on magnetic dipoles. The key word here is "uniform". The magnetic field of bar magnets is most certainly non-uniform, therefore bar magnets do exert forces on one another and on magnetic dipoles in general.

This does not only apply to point charges. A current loop in a uniform magnetic field may experience a torque depending on its orientation but not a net force. This is an extension to what I wrote above if you consider that a current loop has a magnetic dipole.
 
  • #13
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Very nice explanation :smile:

Thanks !
 
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  • #14
Charles Link
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