Solutions Homework #2

8/3/2019 Solutions Homework #2

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SOLUTIONS HOMEWORK #2

1. This is a simple write down. The rod is moving through a perpendicular magneticfield at a constant velocity v, so there is at any point r from the wire, an electric field

E = vB(r). Integrating this field along a distance L gives a potential difference

between the ends of the rod = vB(r)L = V0(r)

Thus:

() = ()2

2. Here we use the boundary conditions that the tangential components of E areconserved across a boundary, while the

normal components of D(=E) are conserved

across the boundary

From the figure:

= =

Thus: =

3. Derivation of the force on an electric dipole, , in an arbitrary electric field (r).Method #1.

Represent the dipole a two charges of equal magnitude,q, but opposite sign,

separated by a distance d. That is:

= lim qwhere the limit is taken consistent with the charge value such that the product qd

always equals p. The force of the two charges in the field (r) is = q( ( + ) ())


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For d0, the [ ( + ) ()] term is adequately represented as ( ())So

= ( ())Method #2

The energy of a dipole in an external field is obtained by noting that in an external

field of () = is obtained using the same expansion as above:[( + ) ()] = q = () = ()

Then noting that = W(r); we have = ( ())

4. Circular plane with radius R of uniform surface charge rotates at a constantangular frequency around the z axis. At a radius r, in radial element of dr, there is

a current loop or radius r with current I(r) = 2rdr. The dipole momentassociated with this current loop is the (current) x (area of the loop), yielding

m(r) = (r)I(r) () = (r) 2 2rdr. Yielding

= r

dr Or

= 4

5. A sphere of radius R has a surface charge distribution = cos. We can calculatethe overall dipole of this distribution by summing the dipoles from charge rings

(using polar coordinates) (analogous argument to #4)

() = (2[]cos)


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This yields a total dipole moment of

(integrating from from 0; and doubling to account for both hemispheres):

= 2(2R sin[dsin]

)

= 2R 6. For the sphere (radius R) with a surface charge density = cos, rotating about

the z axis with angular frequency , the procedure is essentially the same as in #4:

here (using polar coordinates) the current ring at is

2Rsin(Rd)cos

2

then using p = area x current,m() = Rsin()d(sin())

So, integrating from from 0; and doubling to account for both hemispheres:

= 12 R

7. This is an exercise in converting lab units into MKS for insertion into formulae,then expressing the results in something you as an experimentalist can relate to.

This is the kind of quick calculation you want to be able to do in order to check your

sanity as to the important terms, and those that can be ignored. The question here

is whether the force from the electric dipole of the charge distribution is dominant

over, less than or equal to the magnetic forces arising from the spinning motion of

the charges.

As we discussed in class, we expect that since only in the very special case of an

ultra-intense laser field, where the electron is accelerated to essentially the speed oflight, is the effect of the magnetic force on the electron comparable to the electric

force for any given situatq=1.6 x 10-19C) is moving with 150keV of kinetic energyalong the y (dipole pointed along +z) axis. Convert (non-relativistically-this makes a

10% over-estimation of the velocity, increasing the apparent value of the magnetic

forces): this gives = .77 or v=2.3x108 m/s.


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1mm=0.001m; = 10 ; = 3.810 ;100cm=0.1m;The electric dipole of a sphere, radius R, with a constant charge density of is givenby

=

= (3.14)(10)(10) The magnetic dipole is given in the question from Brau (see # 9)

= 13

= (. 33)(1.610)(10)(3.810)

The electric field due to the electric dipole of the charge distribution along an axis

that is oriented perpendicular to the direction of the dipole (which it is here)

= 14

so evaluating F=eE

= (1.610)(910)(9.910)

(10)

Yields Fe (pointed along the positive z axisopposite to the direction of the e field)

Fe=1.4x10-24N;

The B field at distance 100 cm along the y axis (perpendicular to the dipole

orientation) is given by

= 4

so evaluating F=evB

= (1.610)(2.310)(10)(210)10 Fm = 7.4x10-31N


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As we suspected, the magnetic forces are down on the order of c

from the electric forces.

8. This problem is included solely to get you to read and understand 3.3.3 and 3.3.4of Brau (pp 187-198); and/or 5.6 and 5.7 of Jackson (pp184-191). The required

derivation is contained in this material, along with additional physical insights. You

are responsible for this material.

9. Both of these proofs are best simply done by putting m along the z axis, thenwriting out the expression by B in , . #10 is simply using the fact that thetotal field outside of the superconducting sphere is the sum of the original field plus

the field of a perfect dipole (in this case induced by B0)centered atr=0

Here we are emphasizing that for a magnetic dipolem (MEMORIZE THIS)

() = 4 (3 )

r r

In this case, the spinning sphere of total charge q has a dipole moment calculated

using the procedure of #6 with an uniform charge density = . (You may needthe result () = + ). For the induced dipole for thesuperconducting sphere, this is easily done again following the procedure of #5

once you have noticed that since the B field outside the sphere is tangent to the

surface everywhere along the outside of the spherical surface and that it is exactly

zero-by supposition-inside the sphere, then the induced surface current is simply

given by the change in the tangential B field (divided by ). (You may need

() =

)

Solutions Homework #2

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