The Magnetic Force Exerted Upon a Magnetic Dipole

We now begin our research of magnetism, and, analogous to the means in which we began our research of electrical energy, we begin by mentioning the result of a given magnetic area without initially explaining just how such a magnetic area might be caused to exist. We delve into the reasons of magnetic areas in succeeding chapters.

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A magnetic area is a vector field. That is, it is an boundless set of vectors, one at each point in the region of room wright here the magnetic area exists. We use the expression “magnetic field” to designate both the boundless collection of vectors, and, once one is talking about the magnetic area at a point in space, the one magnetic field vector at that allude in room. We use the symbol (vecB) to represent the magnetic area. The many standard result of a magnetic field is to exert a torque on an object that has a residential property known as magnetic dipole minute, and, that finds itself in the magnetic area. A pshort article or object that has a non-zero worth of magnetic dipole minute is dubbed a magnetic dipole. A magnetic dipole is a bar magnet. The value of the magnitude of the magnetic dipole minute of an object is a meacertain of exactly how strong a bar magnet it is. A magnetic dipole has actually 2 ends, known as poles—a north pole and a southern pole. Magnetic dipole moment is a home of issue which has direction. We have the right to define the direction, of the magnetic dipole minute of an item, by considering the object to be an arrowhead whose north pole is the arrowhead and also whose south pole is the tail. The direction in which the arrowhead is pointing is the direction of the magnetic dipole moment of the object. The unit of magnetic dipole moment is the (Acdot m^2) (ampere meter-squared). While magnetic compass needles come in a range of magnetic dipole moments, a representative worth for the magnetic dipole minute of a compass needle is (.1Acdot m^2).

Again, the most standard result of a magnetic field is to exert a torque on a magnetic dipole that finds itself in the magnetic area. The magnetic area vector, at a provided suggest in space, is the maximum feasible torque-per-magnetic-dipole-moment-of-would-be-victim that the magnetic field would/will exert on any kind of magnetic dipole (victim) that could discover itself at the allude in question. I have to say “maximum possible” bereason the torque exerted on the magnetic dipole relies not only on the magnitude of the magnetic area at the allude in area and also the magnitude of the magnetic dipole minute of the victim, yet it likewise counts on the orientation of the magnetic dipole relative to the direction of the magnetic area vector. In fact:

where:

(vec au) is the torque exerted on the magnetic dipole (the bar magnet) by the magnetic field,

(vecmu) is the magnetic dipole moment of the magnetic dipole (the bar magnet, the victim), and

(vecB) is the magnetic field vector at the area in room at which the magnetic dipole is.

For the cross product of any kind of two vectors, the magnitude of the cross product is the product of the magnitudes of the two vectors, times the sine of the angle the two vectors develop when put tail to tail. In the situation of (vec au=vecmu imes vecB), this means:

< au=muarea Bcos heta>

In the SI device of devices, torque has units of (Ncdot m) (newton-meters). For the devices on the appropriate side of ( au=muarea Bcos heta) to occupational out to be (Ncdot m), what through (mu)having devices of electric dipole minute ((Acdot m^2) ) and (sin heta) having no devices at all, (B) need to have units of torque-per-magnetic-dipole-moment, namely, (fracNcdot mAcdot m^2). That combination unit is offered a name. It is called the tesla, abbreviated (T).

<1T= 1fracNcdot mAcdot m^2>


Consider a magnetic dipole having a magnetic dipole minute (mu=0.045 , Acdot m^2), oriented so that it makes an angle of (23^circ) via the direction of a unidevelop magnetic area of magnitude (5.0 imes10^-5 T) as depicted listed below. Find the torque exerted on the magnetic dipole, by the magnetic area.

*

Respeak to that the arrowhead represents the north pole of the bar magnet that a magnetic dipole is. The direction of the torque is such that it often tends to reason the magnetic dipole to allude in the direction of the magnetic field. For the instance illustrated over, that would be clockwise as perceived from the vantage point of the creator of the diagram. The magnitude of the torque for such a instance deserve to be calculated as follows:

< au=mu Bsin heta>

< au=(.045Acdot m^2)(5.0 imes10^-5T)sin 23^circ>

< au=8.8 imes 10^-7 Acdot m^2 cdot T>

Recalling that a tesla is a (fracNcdot mAcdot m^2) we have:

< au=8.8 imes 10^-7 Acdot m^2 cdot fracNcdot mAcdot m^2>

< au=8.8 imes 10^-7 Ncdot m>

Therefore, the torque on the magnetic dipole is ( au=8.8 imes 10^-7 Ncdot m) clockwise, as perceived from the vantage point of the creator of the diagram.


A pshort article having actually a magnetic dipole moment (vecmu=0.025 Acdot m^2 hati-0.035 Acdot m^2 hatj+0.015 Acdot m^2 hatk) is at a suggest in room wbelow the magnetic field (vecB=2.3 mT hati+5.3mThatj-3.6mThatk). Find the torque exerted on the pshort article by the magnetic field

< vec au=eginvmatrix hati&hatj&hatk\ 0.025Acdot m^2&-0.035 Acdot m^2&0.015 Acdot m^2 \ 0.0023 fracNmAm^2 &0.0053 fracNmAm^2 & -0.0036 fracNmAm^2endvmatrix>

>

<+hatj Big< (0.015Am^2)(0.0023 fracNmAm^2)-(0.025Am^2)(-0.0036 fracNmAm^2) Big>>

<+hatk Big< (0.025Am^2)(0.0053 fracNmAm^2)-(-0.035Am^2)(0.0023 fracNmAm^2) Big>>



The Magnetic Force Exerted Upon a Magnetic Dipole

A uniform magnetic area exerts no force on a bar magnet that is in the magnetic area. You need to most likely pausage here for a minute and also let that sink in. A unidevelop magnetic area exerts no pressure on a bar magnet that is in that magnetic area.

You have more than likely had actually some endure via bar magnets. You know that favor poles repel and also unprefer poles lure. And, from your research of the electric area, you have actually more than likely (correctly) hypothesized that in the field point of view, the method we see this is that one bar magnet (speak to it the source magnet) creates a magnetic area in the region of space around itself, and, that if there is an additional bar magnet in that area of area, it will certainly be affected by the magnetic field it is in. We have actually currently questioned the reality that the victim bar magnet will certainly experience a torque. But you understand, from your experience via bar magnets, that it will likewise suffer a force. How can that be as soon as I just proclaimed that a unidevelop magnetic field exerts no force on a bar magnet? Yes, of course. The magnetic area of the source magnet have to be non-unicreate. Enough around the nature of the magnetic field of a bar magnet, I’m expected to conserve that for an upcoming chapter. Suffice it to say that it is non-unicreate and also to focus our attention on the result of a non-unidevelop area on a bar magnet that finds itself in that magnetic area.

First of all, a non-uniform magnetic area will exert a torque on a magnetic dipole (a bar magnet) simply as before ((vec au=vecmu imes vecB)). But, a non-unicreate magnetic area (one for which the magnitude, and/or direction, counts on position) likewise exerts a force on a magnetic dipole. The force is given by:

where

(vecF_B) is the force exerted by the magnetic field (vecB) on a ppost having a magnetic dipole minute (vec au) (vecmu) is the magnetic dipole of the "victim", and also, (vecB) is the magnetic field at the position in area wright here the victim finds itself. To evaluate the pressure, once have to know (vecB) as a duty of (x,y) and also (z) (wherea (vecmu) is a constant) .

Note that after you take the gradient of (vecmucdot vecB), you have to evaluate the outcome at the worths of (x,y) and also (z) equivalent to the area of the victim.

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Just to make certain that you know how to usage this equation, please note that if (vecmu) and (vecB) are expressed in (hati,hatj,hatk) notation, so that they show up as (vecmu=mu_x hati+mu_y hatj+mu_z hatk) and (vecB=B_x hati+B_y hatj+B_z hatk) respectively, then:

And the gradient of (vecmucdotvecB) (which by equation ( ef15-2) is the pressure we seek) is offered by

< abla (vecmucdot vecB)=fracpartial (vecmucdot vecB)partial x hati +fracpartial (vecmucdot vecB)partial y hatj+fracpartial (vecmucdot vecB)partial z hatk>

wbelow derivatives in this equation deserve to (making use of (vecmucdotvecB=mu_x B_x+mu_y B_y+mu_z B_z) from simply above) can be expressed as:

where we have taken benefit of the truth that the components of the magnetic dipole minute of the victim are not features of position. Also note that the derivatives are all partial derivatives. Partial derivatives are the simple type in the sense that, as soon as, for circumstances, you take the derivative through respect to (x), you are to treat (y) and also (z) as if they were constants. Finally, it is vital to realize that, after you take the derivatives, you need to plug the values of (x,y) and (z) corresponding to the location of the magnetic dipole (the victim), right into the offered expression for the pressure.