source of magnetic field

2010-04-26 7:13 pm
1.A square loop of wire with side length
圖片參考:http://session.masteringphysics.com/render?var=I_2
. The infinite wire and loop are in the same
plane; two sides of the square loop are parallel to the wire and two are
perpendicular as shown.

圖片參考:http://imgcld.yimg.com/8/n/HA00021001/o/701004260035113873387730.jpg

part A
What is the magnitude,
圖片參考:http://session.masteringphysics.com/render?var=F
, of the net force on the loop?
Express the force in terms of
圖片參考:http://session.masteringphysics.com/render?var=mu_0
.

ans:


圖片參考:http://session.masteringphysics.com/render?var=F

=
圖片參考:http://session.masteringphysics.com/render?tex=%7B%5Cmu%7D_%7B0%7DI_%7B1%7DI_%7B2%7D%5Cfrac%7Ba%5E%7B2%7D%7D%7B2%7B%5Cpi%7D%5Cleft%28d%5E%7B2%7D-%5Cfrac%7Ba%5E%7B2%7D%7D%7B4%7D%5Cright%29%7D

part b
The magnetic moment
圖片參考:http://session.masteringphysics.com/render?var=m_vec
of a current loop is defined as the vector
whose magnitude equals the area of the loop times the magnitude of the
current flowing in it (
圖片參考:http://session.masteringphysics.com/render?tex=m%3DI+A
), and whose direction is perpendicular to
the plane in which the current flows. Find the magnitude,
圖片參考:http://session.masteringphysics.com/render?var=F
, of the force on the loop from Part A in terms of the
magnitude of its magnetic moment.
Express
圖片參考:http://session.masteringphysics.com/render?var=mu_0
.


ans:
圖片參考:http://session.masteringphysics.com/render?var=F

=
圖片參考:http://session.masteringphysics.com/render?tex=%7B%5Cmu%7D_%7B0%7DI_%7B2%7D%5Cfrac%7Bm%7D%7B2%7B%5Cpi%7D%5Cleft%28d%5E%7B2%7D-%5Cfrac%7Ba%5E%7B2%7D%7D%7B4%7D%5Cright%29%7D

can you also explain why some forces each up will be zero.. and how do you determine it?

2.Find the magnetic field a distance
圖片參考:http://imgcld.yimg.com/8/n/HA00021001/o/701004260035113873387741.jpg

part a
First find the magnetic field,
圖片參考:http://session.masteringphysics.com/render?tex=%5Cvec%7BB%7D_%7B%5Crm+out%7D%28%5Cvec%7Br%7D%29
, outside
the wire (i.e., when the distance
圖片參考:http://session.masteringphysics.com/render?var=a
).
(Part A figure)


Express
圖片參考:http://session.masteringphysics.com/render?tex=%5Cvec%7BB%7D_%7B%5Crm+out%7D%28%5Cvec%7Br%7D%29
in terms of
the given parameters, the permeability constant
圖片參考:http://session.masteringphysics.com/render?var=z_unit
. You may not need all these
in your answer.
ans:




圖片參考:http://session.masteringphysics.com/render?tex=%5Cvec%7BB%7D_%7B%5Crm+out%7D%28%5Cvec%7Br%7D%29

=
圖片參考:http://session.masteringphysics.com/render?tex=%7B%5Cmu%7D_%7B0%7D%5Cfrac%7Bja%5E%7B2%7D%7D%7B2r%7D%5Chat%7B%7B%5Ctheta%7D%7D

part b

圖片參考:http://imgcld.yimg.com/8/n/HA00021001/o/701004260035113873387742.jpg


Now find the magnetic field
圖片參考:http://session.masteringphysics.com/render?tex=%5Cvec%7BB%7D_%7B%5Crm+in%7D%28%5Cvec%7Br%7D%29

inside the wire (i.e., when the distance
圖片參考:http://session.masteringphysics.com/render?var=a
).
(Part B figure)


Express
圖片參考:http://session.masteringphysics.com/render?tex=%5Cvec%7BB%7D_%7B%5Crm+in%7D%28%5Cvec%7Br%7D%29
in terms of
the given parameters, the permeability constant
圖片參考:http://session.masteringphysics.com/render?var=z_unit
. You may not need all these
in your answer.



ans:


圖片參考:http://session.masteringphysics.com/render?tex=%5Cvec%7BB%7D_%7B%5Crm+in%7D%28%5Cvec%7Br%7D%29

=
圖片參考:http://session.masteringphysics.com/render?tex=%5Cfrac%7Brj%7B%5Cmu%7D_%7B0%7D%7D%7B2%7D%5Chat%7B%7B%5Ctheta%7D%7D




回答 (1)

2010-04-26 7:55 pm
✔ 最佳答案
1a) By symmetry, the net force of the magnetic forces acting on the 2 horizontal wire segments of the loop is zero.

Hence we only need to find out the net force of the magnetic forces acting on the 2 vertical wire segments of the loop.

The magnetic force on the closer vertical wire segment is:

μ0I1I2a/[2π(d - a/2)] (attractive)

The magnetic force on the farther vertical wire segment is:

μ0I1I2a/[2π(d + a/2)] (repulsive)

So the net force will be:

μ0I1I2a/[2π(d - a/2)] - μ0I1I2a/[2π(d + a/2)]

= [μ0I1I2a/(2π)] x [1/(d - a/2) - 1/(d + a/2)]

= [μ0I1I2a/(2π)] x a/[(d - a/2)(d + a/2)]

= μ0I1I2a2/[2π(d2 - a2/4)]

b) Magnetic moment of the loop = I1a, hence:

F = μ0I1I2a2/[2π(d2 - a2/4)]

= μ0I2ma/[2π(d2 - a2/4)]

2a) Using Ampere's circuital law: (Refer to http://en.wikipedia.org/wiki/Ampere%27s_law)

When r > a, current enclosed = j(πa2)

Hence:

B(2πr) = μ0j(πa2)

B = μ0ja2/(2r)

where B is the magnitude of the B-field.

Hence by right hand grip rule, Bout (vector) = μ0ja2θ/(2r) where θ is the unit vector as indicated in the question.

b) Applying the Ampere's circuital law again:

When r <= a, current enclosed = j(πr2)

Hence:

B(2πr) = μ0j(πr2)

B = μ0jr/2

where B is the magnitude of the B-field.

Hence by right hand grip rule, Bin (vector) = μ0jrθ/2 where θ is the unit vector as indicated in the question.
參考: Myself


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