no tagsLecture 018: The Magnetic Dipole MomentYourName, 1 April 2011 (created 1 April 2011)The behavior of current loops in externalmagnetic fieldsThere is one more important phenomenon to consider: what happens whena current loop is itself exposed to a magnetic field? This is importantbecause this problem is the basis of electric motors, which are ubiquitous inthe world: from fans that cool a room, to hybrid gas-electric cars. The basisof such a motor is the force exerted on a loop of a loop of current when it isimmersed in an external magnetic field.Consider a simple geometry of a square loop. Consider also the simple caseof a uniform magnetic field into which the loop can be placed. Rememberthat the force exerted by an external magnetic field on a current is:What happens if we place the loop in the field so that originally theplane of the loop is perpendicular to the field?General Physics - E&M (PHY 1308) LectureNotesGeneral Physics - E&M (PHY 1308) LectureNotesF L ~= I~Â B~\oint ds = 2 \piGeneral Physics - E&M (PHY 1308) - Lecture Notes file:///home/sekula/Dropbox/Documents/Notebook...1 of 8 04/05/2011 08:42 AMIn the drawing above, we can think about what happens by breaking theproblem into 4 currents, each one moving in a different direction but all thesame magnitude.The top current is moving from right to left, and is anti-parallel to themagnetic field. Thus its cross-product magnitude isTherefore, there is no force on the top segment of the current.The bottom current leads us to the same conclusion: it is parallel to themagnetic field, and so the magnitude of its cross-product is also zero.Let's consider the left, downward-pointing current. Now we have acurrent moving at a right-angle to the magnetic field (actually, at). We then expect that the force will point OUT of the page (outof the plane of the blackboard). We can also apply the right-hand rule tosee this: stick our fingers in the direction of the current, curl the fingerstoward the magnetic field, and let our thumb tell us the direction of thejI L j LB LB : top~Â B~= I sin Ò = I sin Ù = 0Ò Ù=2 = ÀRGeneral Physics - E&M (PHY 1308) - Lecture Notes file:///home/sekula/Dropbox/Documents/Notebook...2 of 8 04/05/2011 08:42 AMforce - then we see that it points outward. Thus the left-hand currentexperiences an outward-pointing force due to the magnetic field.Finally, we consider the right-hand, upward-pointing current. The onlydifference between it and the left-hand current is the direction of thecurrent. Since the direction is up, the sign of the resulting forcereverses (we now have instead of in the cross-productmagnitude), and the right-hand current experiences a magnetic forcepointing into the page/blackboard.So what have we learned? We have learned that a loop of current thatbegins fully parallel to the magnetic field will experience equal but oppositeforces on the left and right side. This will cause a rotation - and thus there isa torque resulting from the force exerted by the magnetic field.Recall that the definition of a torque is:where is a vector that points from the axis of rotation to the point wherethe force is exerted. In the case above, this is a vector from the verticalbisecting axis of the loop to the right or left-hand sides of the loop.In our example above, the torque vector points vertically upward (again, usethe right-hand rule for the cross-product to verify this: point your fingers inthe direction of , curl your fingers toward , and your thumb indicates thedirection of the torque vector.)More loopsWhat if we then wrap the wire twice, so that there are two overlappingloops of wire that start out in the plane of the paper? Now we have twicethe current on each side of the loop (each wire carries current I, but thereare two wires one each side now). As a result, we expect the magnetic forceto double.If we wind 3 times, we expect the force to triple. And so on. For loops, thetotal current is .sin =2 Ù sin =2 ÙÜ ~ = r~ Â F~r ~r ~ F ~N NI General Physics - E&M (PHY 1308) - Lecture Notes file:///home/sekula/Dropbox/Documents/Notebook...3 of 8 04/05/2011 08:42 AMThe Magnetic Dipole MomentSo the magnetic force on a loop of wire is proportional to the number ofwindings in the loop. It is convenient to define a new vector for a currentloop, just like we defined a special vector for the electric dipole that helpedus to calculate the torque on a dipole in an external electric field. Recallthat we defined the electric dipole moment as:where is the magnitude of either charge in the dipole and is a vectorpointing from the positive to the negative charge. We found that the torqueon an electric dipole could be expressed as:where was the external electric field into which the dipole was placed.Since the magnetic field of a current loop looks like a dipole field, we canalso define the magnetic dipole moment of a current loop:The vector ois a new concept, one we will need going forward into thenext chapter. Here are its properties: has a magnitude equal to the area of the loop. has a direction that is perpendicular to the plane of the loop.How do we determine the direction of ? The convention is ANOTHERright-hand rule:p d electric~ = q~q d ~Ü electric~ = p~ Â E~E ~Ö IA ~ Ñ N~A ~A ~A ~A ~General Physics - E&M (PHY 1308) - Lecture Notes file:///home/sekula/Dropbox/Documents/Notebook...4 of 8 04/05/2011 08:42 AMCurl your fingers on your right hand in the direction of current aroundthe loop. Your thumb points in the direction of .(Yes, you'll have to try to memorize all of these conventions...!)Why Define the Magnetic Dipole Moment?As we saw above, calculating the forces on a loop of current due to anexternal magnetic field is not trivial - you have to think hard about whichparts of the loop are parallel or perpendicular to the field. The MagneticDipole Moment is a convenient short-hand that let's us more quicklydetermine the behavior of current loops in external magnetic fields, just likethe electric dipole moment let us more quickly calculate the electric forceon a dipole.Let's see why.A ~General Physics - E&M (PHY 1308) - Lecture Notes file:///home/sekula/Dropbox/Documents/Notebook...5 of 8 04/05/2011 08:42 AMConsider a square current loop with N windings, like the one depictedabove but with equal-length sides. The area of the loop is just:The magnitude of a force on either the left or right side of the loop isalways:since the current on the
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