The Magnetic force Exerted upon a Magnetic Dipole

We now begin our study of magnetism, and, analogous to the way in i beg your pardon we started our examine of electricity, we begin by pointing out the effect of a given magnetic ar without first explaining how such a magnetic field can be resulted in to exist. Us delve right into the reasons of magnetic fields in subsequent chapters.

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A magnetic ar is a vector field. That is, it is an infinite collection of vectors, one at each point in the an ar of an are where the magnetic ar exists. We usage the expression “magnetic field” come designate both the infinite set of vectors, and, when one is talking around the magnetic ar at a suggest in space, the one magnetic ar vector at that suggest in space. We usage the prize \$$\\vecB\$$ to represent the magnetic field. The most straightforward effect that a magnetic ar is come exert a talk on an object that has a property recognized as magnetic dipole moment, and, the finds chin in the magnetic field. A particle or thing that has actually a non-zero worth of magnetic dipole minute is referred to as a magnetic dipole. A magnetic dipole is a bar magnet. The value of the size of the magnetic dipole minute of an object is a measure up of how strong a bar magnet that is. A magnetic dipole has two ends, well-known as poles—a phibìc pole and a south pole. Magnetic dipole moment is a property of matter which has direction. Us can specify the direction, the the magnetic dipole moment of an object, through considering the object to be an arrow whose phibìc pole is the arrowhead and also whose southern pole is the tail. The direction in i beg your pardon the arrowhead is pointing is the direction that the magnetic dipole minute of the object. The unit the magnetic dipole minute is the \$$A\\cdot m^2\$$ (ampere meter-squared). If magnetic compass needles come in a variety of magnetic dipole moments, a representative worth for the magnetic dipole minute of a compass needle is \$$.1A\\cdot m^2\$$.

Again, the most an easy effect that a magnetic ar is to exert a torque on a magnetic dipole that finds chin in the magnetic field. The magnetic ar vector, in ~ a given point in space, is the maximum feasible torque-per-magnetic-dipole-moment-of-would-be-victim that the magnetic ar would/will exert on any magnetic dipole (victim) that can find itself at the point in question. I have to say “maximum possible” because the torque exerted on the magnetic dipole counts not only on the size of the magnetic field at the suggest in space and the size of the magnetic dipole moment of the victim, however it additionally depends on the orientation of the magnetic dipole family member to the direction the the magnetic field vector. In fact:

\\<\\vec\\tau=\\vec\\mu\\times \\vecB\\label15-1\\>

where:

\$$\\vec\\tau\$$ is the torque exerted ~ above the magnetic dipole (the bar magnet) through the magnetic field,

\$$\\vec\\mu\$$ is the magnetic dipole minute of the magnetic dipole (the bar magnet, the victim), and

\$$\\vecB\$$ is the magnetic field vector in ~ the place in an are at i beg your pardon the magnetic dipole is.

For the overcome product of any type of two vectors, the magnitude of the cross product is the product the the magnitudes that the two vectors, time the sine that the angle the 2 vectors form when inserted tail come tail. In the instance of \$$\\vec\\tau=\\vec\\mu\\times \\vecB\$$, this means:

\\<\\tau=\\mu\\space B\\cos \\theta\\>

In the SI system of units, torque has actually units that \$$N\\cdot m\$$ (newton-meters). For the devices on the best side that \$$\\tau=\\mu\\space B\\cos \\theta\$$ to occupational out to it is in \$$N\\cdot m\$$, what v \$$\\mu\$$having units of electric dipole minute (\$$A\\cdot m^2\$$ ) and \$$\\sin \\theta\$$ having actually no systems at all, \$$B\$$ must have units that torque-per-magnetic-dipole-moment, namely, \$$\\fracN\\cdot mA\\cdot m^2\$$. That mix unit is provided a name. That is dubbed the tesla, abbreviated \$$T\$$.

\\<1T= 1\\fracN\\cdot mA\\cdot m^2\\>

Consider a magnetic dipole having a magnetic dipole moment \$$\\mu=0.045 \\, A\\cdot m^2\$$, oriented so the it makes an angle of \$$23^\\circ\$$ with the direction of a uniform magnetic ar of magnitude \$$5.0\\times10^-5 T\$$ as illustrated below. Discover the talk exerted top top the magnetic dipole, through the magnetic field.

Recall that the arrowhead to represent the north pole of the bar magnet that a magnetic dipole is. The direction that the talk is such that it has tendency to reason the magnetic dipole to point in the direction that the magnetic field. Because that the situation depicted above, that would certainly be clockwise as regarded from the vantage point of the creator the the diagram. The magnitude of the torque for such a situation can it is in calculated as follows:

\\<\\tau=\\mu B\\sin\\theta\\>

\\<\\tau=(.045A\\cdot m^2)(5.0\\times10^-5T)\\sin 23^\\circ\\>

\\<\\tau=8.8\\times 10^-7 A\\cdot m^2 \\cdot T\\>

Recalling that a tesla is a \$$\\fracN\\cdot mA\\cdot m^2\$$ us have:

\\<\\tau=8.8\\times 10^-7 A\\cdot m^2 \\cdot \\fracN\\cdot mA\\cdot m^2\\>

\\<\\tau=8.8\\times 10^-7 N\\cdot m\\>

Thus, the talk on the magnetic dipole is \$$\\tau=8.8\\times 10^-7 N\\cdot m\$$ clockwise, as regarded from the vantage allude of the creator of the diagram.

A particle having a magnetic dipole minute \$$\\vec\\mu=0.025 A\\cdot m^2 \\hati-0.035 A\\cdot m^2 \\hatj+0.015 A\\cdot m^2 \\hatk\$$ is at a suggest in space where the magnetic ar \$$\\vecB=2.3 mT \\hati+5.3mT\\hatj-3.6mT\\hatk\$$. Discover the torque exerted on the bit by the magnetic field

\\<\\vec\\tau=\\vec\\mu\\times \\vecB\\>

\\< \\vec\\tau=\\beginvmatrix \\hati&\\hatj&\\hatk\\\\ 0.025A\\cdot m^2&-0.035 A\\cdot m^2&0.015 A\\cdot m^2 \\\\ 0.0023 \\fracNmAm^2 &0.0053 \\fracNmAm^2 & -0.0036 \\fracNmAm^2\\endvmatrix\\>

\\<\\vec\\tau=\\hati \\Big<(-0.035Am^2)(-0.0036 \\fracNmAm^2)-(0.015Am^2)(0.0053 \\fracNmAm^2) \\Big>\\>

\\<+\\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>\\>

\\<\\vec\\tau=1.2\\times10^-4 Nm \\hati-1.2\\times 10^-4 Nm \\hatj+2.1\\times 10^-4 Nm\\hatk\\>

## The Magnetic pressure Exerted upon a Magnetic Dipole

A uniform magnetic ar exerts no pressure on a bar magnet the is in the magnetic field. Friend should most likely pause here for a moment and let the sink in. A uniform magnetic field exerts no pressure on a bar magnet the is in that magnetic field.

You have probably had actually some endure with bar magnets. You recognize that choose poles repel and unlike poles attract. And, from your examine of the electrical field, you have probably (correctly) hypothesized that in the field allude of view, the method we see this is that one bar magnet (call the the source magnet) creates a magnetic ar in the an ar of room around itself, and, that if over there is another bar magnet in that an ar of space, it will certainly be influenced by the magnetic field it is in. Us have currently discussed the truth that the victim bar magnet will endure a torque. However you know, indigenous your suffer with bar magnets, that it will also experience a force. How can that be as soon as I just proclaimed that a uniform magnetic ar exerts no force on a bar magnet? Yes, that course. The magnetic ar of the source magnet have to be non-uniform. Enough around the nature the the magnetic ar of a bar magnet, I’m an alleged to save that because that an upcoming chapter. Suffice it to say the it is non-uniform and also to focus our attention on the effect of a non-uniform field on a bar magnet the finds chin in that magnetic field.

First the all, a non-uniform magnetic field will exert a talk on a magnetic dipole (a bar magnet) just as prior to (\$$\\vec\\tau=\\vec\\mu\\times \\vecB\$$). But, a non-uniform magnetic field (one because that which the magnitude, and/or direction, relies on position) also exerts a force on a magnetic dipole. The pressure is given by:

\\<\\vecF_B=\\nabla (\\vec\\mu\\cdot \\vecB) \\label15-2\\>

where

\$$\\vecF_B\$$ is the force exerted by the magnetic field \$$\\vecB\$$ top top a particle having actually a magnetic dipole moment \$$\\vec\\tau\$$ \$$\\vec\\mu\$$ is the magnetic dipole of the \"victim\", and, \$$\\vecB\$$ is the magnetic ar at the place in space where the victim finds itself. To advice the force, as soon as must know \$$\\vecB\$$ together a duty of \$$x,y\$$ and \$$z\$$ (wherea \$$\\vec\\mu\$$ is a constant) .

Note the after you take it the gradient the \$$\\vec\\mu\\cdot \\vecB\$$, you need to evaluate the result at the values of \$$x,y\$$ and \$$z\$$ matching to the ar of the victim.

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Just to make sure that friend know how to use this equation, please keep in mind that if \$$\\vec\\mu\$$ and \$$\\vecB\$$ space expressed in \$$\\hati,\\hatj,\\hatk\$$ notation, so that they show up as \$$\\vec\\mu=\\mu_x \\hati+\\mu_y \\hatj+\\mu_z \\hatk\$$ and \$$\\vecB=B_x \\hati+B_y \\hatj+B_z \\hatk\$$ respectively, then:

\\<\\vec\\mu\\cdot\\vecB=(\\vec\\mu=\\mu_x \\hati+\\mu_y \\hatj+\\mu_z \\hatk)(\\vecB=B_x \\hati+B_y \\hatj+B_z \\hatk)\\>

\\<\\vec\\mu\\cdot\\vecB=\\mu_x B_x+\\mu_y B_y+\\mu_z B_z\\>

And the gradient that \$$\\vec\\mu\\cdot\\vecB\$$ (which through equation \$$\\ref15-2\$$ is the pressure we seek) is given by

\\<\\nabla (\\vec\\mu\\cdot \\vecB)=\\frac\\partial (\\vec\\mu\\cdot \\vecB)\\partial x \\hati +\\frac\\partial (\\vec\\mu\\cdot \\vecB)\\partial y \\hatj+\\frac\\partial (\\vec\\mu\\cdot \\vecB)\\partial z \\hatk\\>

where derivatives in this equation have the right to (using \$$\\vec\\mu\\cdot\\vecB=\\mu_x B_x+\\mu_y B_y+\\mu_z B_z\$$ from simply above) have the right to be express as:

\\<\\frac\\partial (\\vec\\mu\\cdot\\vecB)\\partial x=\\mu_x \\frac\\partial B_x\\partial x+\\mu_y \\frac\\partial B_y\\partial x+\\mu_z \\frac\\partial B_z\\partial x\\>

\\<\\frac\\partial (\\vec\\mu\\cdot\\vecB)\\partial y=\\mu_x \\frac\\partial B_x\\partial y+\\mu_y \\frac\\partial B_y\\partial y+\\mu_z \\frac\\partial B_z\\partial y\\>

\\<\\frac\\partial (\\vec\\mu\\cdot\\vecB)\\partial z=\\mu_x \\frac\\partial B_x\\partial z+\\mu_y \\frac\\partial B_y\\partial z+\\mu_z \\frac\\partial B_z\\partial z\\>

where we have actually taken benefit of the fact that the materials of the magnetic dipole minute of the victim space not functions of position. Additionally note that the derivatives room all partial derivatives. Partial derivatives room the simple kind in the sense that, when, because that instance, you take the derivative v respect to \$$x\$$, you space to law \$$y\$$ and also \$$z\$$ as if they to be constants. Finally, it is necessary to realize that, after ~ you take the derivatives, you need to plug the worths of \$$x,y\$$ and \$$z\$$ corresponding to the location of the magnetic dipole (the victim), into the provided expression because that the force.