工程电磁场
如何描述线1周围的用来决定对线2作用力的力场？
Note that in the third case (perpendicular currents), I2 is in the same direction as H, so that their cross product (and the resulting force) is zero. The actual force computation involves a different field quantity, B, which is related to H through B = μ0H in free space. This will be taken up in a later lecture. Our immediate concern is how to find H from any given current distribution.
第三种情况，磁场与电流平行，叉乘=0
特别注意与距离的平方成反比, 而且叉矢量指向纸内（右手螺旋法则决定）Note the similarity to Coulomb’s Law
a point charge of magnitude dQ1at Point 1 would generate electric field at Point 2 given by:
The units of H are [A/m]
To determine the total field arising from the closed circuit path,
we sum the contributions from the current elements that make up
the entire loop, or
The contribution to the field at P from any portion of the current will be just the above integral evalated over just that portion.
..and so the differential current quantity that
appears in the Biot-Savart law becomes:
The magnetic field arising from a current
sheet is thus found from the two-dimensional
form of the Biot-Savart law:
of three-dimensional current elements, and so the Biot-Savart
and so..
so that:
Integrate this over the entire wire:
..after carrying out the cross product
Example: concluded
finally:
Current is into the page.
Magnetic field streamlines
are concentric circles, whose magnitudes
decrease as the inverse distance from the 线是如何画？（力线疏密反应强
如何画？
定值
/ρ 2，所以，磁力线的间隔怎么画？
..after a few additional steps (see Problem 7.8), we find:
carry out the cross products to find:
but we must include the angle dependence in the radial
unit vector注意:rho的单位矢量不是常量！！！
with this substitution, the radial component will integrate to zero, meaning that all radial components will cancel on the z axis. rho分量消了，仅仅z分量
Note the form of the numerator: the product of
the current and the loop area. We define this as
the magnetic moment:
环可定义磁矩
似偶极矩的定义
横向涡旋磁场
路径a和b的磁场环流= 总电流I
路径c的磁场环流= 总电流I的一部分
ρ
so that:as before.
solid conductors that carry equal and opposite
currents, I.
The line is assumed to be infinitely long, and the
circular symmetry suggests that
φ-directed, and will vary only with radius
Our objective is to find the magnetic field
for all values of ρ
无限长
内外导体等量异向电流
仅沿rho变化的phi向磁场
求全径向磁场分布
导体间磁场
内导体可看作无限长线电流束
考虑1、2处的电流丝产生的磁场叠加
仅phi向存在
ab间场直接为
But now, the current enclosed is
or finally:
体外的场
As the current is uniformly distributed, and since we
have circular symmetry, the field would have to
be constant over the circular integration path, and so it
must be true that:
+外导体圈内反向电流百分比
..and so finally:
同轴线磁场强度的径向分布
s circuital law to the path we find:
In other words, the magnetic field is discontinuous across the current sheet by the magnitude of the surface current density.
生的磁场（续）
电场的均匀分布
If instead, the upper path is elevated to the line between and , the same current is enclosed and we would have
from which we conclude that
constant in each region(above and below the current plane)
a N is the unit vector that is normal to the
current sheet, and that points into the region in
which the magnetic field is to be evaluated.
引申为等效电流
aN X H = aN X (0.5KXaN)=0.5K
K
H
H
K
We will now use this result as a building block
to construct the magnetic field on the axis of
a solenoid --formed by a stack of identical current
loops, centered on the z axis.
contribution to the total field from a stack of N closely-spaced
loops, each of which carries current I . The length of the stack so therefore the density of turns will be N/d.Now the current in the turns within a differential length, dz , will be
z -d/2
d/2
so that the previous result for H from a single loop:now becomes:
in which z is measured from the center of the coil,
where we wish to evaluate the field.
We consider this as our differential “loop current ”
z
-d/2
d/2
长螺线管的场近似
z
-d/2d/2
We now have the on-axis field at the solenoid midpoint (z = 0):
Note that for long solenoids, for which , the
result simplifies to:
( )
This result is valid at all on-axis positions deep within long coils --at distances from each end of several radii.
d/2
Therefore:
In other words, the on-axis field magnitude near the center of a cylindrical
current sheet, where current circulates around the z axis, and whose length
is much greater than its radius, is just the surface current density.
Solenoid Field --Off-Axis
螺线管非轴线上的场
用安培环路定律来求解:
The illustration below shows the solenoid cross-section, from a lengthwise cut through the
the windings flows in and out of the screen in the circular current path. Each turn carries current
field along the z axis is NI/d as we found earlier.
Where allowance is made for the existence of a radial H component,
The radial integrals will now cancel, because they are oppositely-directed, and because in the long coil,
Conclusion: The magnetic field within a long solenoid is approximately constant throughout the coil cross-section, and is H z = NI/d.