The Magnetic Force Exerted Upon a Magnetic Dipole

We now start our study of magnetism, and also, analogous to the method in which we began our study of electrical energy, we start by discussing the result of a given magnetic area without initially explaining how such a magnetic area might be led to to exist. We delve into the reasons of magnetic fields in subsequent chapters.

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A magnetic field is a vector field. That is, it is an boundless collection of vectors, one at each suggest in the region of room where the magnetic area exists. We usage the expression “magnetic field” to designate both the boundless collection of vectors, and also, as soon as one is talking about the magnetic area at a suggest in area, the one magnetic field vector at that point in room. We usage the symbol (vecB) to represent the magnetic field. The a lot of fundamental effect of a magnetic area is to exert a torque on an object that has a property known as magnetic dipole minute, and also, that finds itself in the magnetic field. A pshort article or object that has a non-zero value of magnetic dipole minute is referred to as a magnetic dipole. A magnetic dipole is a bar magnet. The worth of the magnitude of the magnetic dipole moment of a things is a meacertain of exactly how strong a bar magnet it is. A magnetic dipole has actually 2 ends, recognized as poles—a north pole and a south pole. Magnetic dipole minute is a residential or commercial property of matter which has actually direction. We deserve to specify the direction, of the magnetic dipole minute of a things, by considering the object to be an arrow whose north pole is the arrowhead and whose south pole is the tail. The direction in which the arrow 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 selection of magnetic dipole moments, a representative worth for the magnetic dipole moment of a compass needle is (.1Acdot m^2).

Again, the most basic effect of a magnetic field is to exert a torque on a magnetic dipole that finds itself in the magnetic field. The magnetic area vector, at a provided suggest in area, is the maximum feasible torque-per-magnetic-dipole-moment-of-would-be-victim that the magnetic area would/will certainly exert on any magnetic dipole (victim) that might find itself at the suggest in question. I need to say “maximum possible” bereason the torque exerted on the magnetic dipole relies not just on the magnitude of the magnetic area at the suggest in room and the magnitude of the magnetic dipole moment of the victim, however it also depends on the orientation of the magnetic dipole relative to the direction of the magnetic field vector. In fact:

where:

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

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

(vecB) is the magnetic area vector at the place in space at which the magnetic dipole is.

For the cross product of any type of two vectors, the magnitude of the cross product is the product of the magnitudes of the 2 vectors, times the sine of the angle the two vectors develop as soon as put tail to tail. In the instance of (vec au=vecmu imes vecB), this means:

< au=mu oom Bcos heta>

In the SI device of units, torque has units of (Ncdot m) (newton-meters). For the units on the ideal side of ( au=muspace Bcos heta) to work-related out to be (Ncdot m), what with (mu)having actually devices of electrical dipole moment ((Acdot m^2) ) and (sin heta) having actually no devices at all, (B) need to have actually devices of torque-per-magnetic-dipole-moment, namely, (fracNcdot mAcdot m^2). That combicountry unit is given a name. It is called the tesla, abbreviated (T).

<1T= 1fracNcdot mAcdot m^2>


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

*

Recall 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 cause the magnetic dipole to point in the direction of the magnetic field. For the instance depicted over, that would be clockwise as regarded from the vantage suggest of the creator of the diagram. The magnitude of the torque for such a case 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>

Hence, 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 pwrite-up having 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 allude in space 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 unicreate magnetic area exerts no pressure on a bar magnet that is in the magnetic area. You need to more than likely pausage here for a moment and let that sink in. A uniform magnetic area exerts no force on a bar magnet that is in that magnetic field.

You have actually more than likely had some suffer with bar magnets. You understand that favor poles repel and also unchoose poles tempt. And, from your study of the electrical area, you have more than likely (correctly) hypothesized that in the area allude of check out, the method we see this is that one bar magnet (contact it the resource magnet) creates a magnetic area in the area of room approximately itself, and also, that if tbelow is an additional bar magnet in that area of space, it will certainly be affected by the magnetic area it is in. We have currently discussed the truth that the victim bar magnet will certainly endure a torque. But you understand, from your suffer with bar magnets, that it will certainly likewise experience a pressure. How can that be as soon as I just declared that a unicreate magnetic field exerts no force on a bar magnet? Yes, of course. The magnetic area of the resource magnet need to be non-unicreate. Enough about the nature of the magnetic area of a bar magnet, I’m intended to save that for an upcoming chapter. Suffice it to say that it is non-uniform and 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-unidevelop magnetic area will certainly exert a torque on a magnetic dipole (a bar magnet) just as prior to ((vec au=vecmu imes vecB)). But, a non-unicreate magnetic field (one for which the magnitude, and/or direction, depends on position) additionally exerts a pressure on a magnetic dipole. The pressure is given by:

where

(vecF_B) is the force exerted by the magnetic area (vecB) on a particle having actually a magnetic dipole moment (vec au) (vecmu) is the magnetic dipole of the "victim", and also, (vecB) is the magnetic area at the place in room wright here the victim finds itself. To evaluate the pressure, once must recognize (vecB) as a function of (x,y) and also (z) (wherea (vecmu) is a constant) .

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

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Just to make certain that you know exactly how to use this equation, please note that if (vecmu) and also (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 force we seek) is given 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 (utilizing (vecmucdotvecB=mu_x B_x+mu_y B_y+mu_z B_z) from just above) have the right to be expressed as:

wbelow we have actually taken benefit of the fact that the components of the magnetic dipole minute of the victim are not functions of position. Also note that the derivatives are all partial derivatives. Partial derivatives are the simple kind in the sense that, once, for circumstances, you take the derivative through respect to (x), you are to treat (y) and (z) as if they were constants. Finally, it is vital to realize that, after you take the derivatives, you have to plug the worths of (x,y) and (z) matching to the area of the magnetic dipole (the victim), right into the given expression for the pressure.