ELEG3203 2015-2016 First Term
Assignment 2
1155047511
Li Man Hin
2. Divide the wire into three parts: two long straight wire and one semicircular coil.
Using right hand grip rule, all three parts provides an out-of-paper magnetic flux
density at point P.
By symmetry, the two long wire can be considered as one long wire extending to
both infinities. By Biot-Savart Law,
Z L
Z
b dz 0
L
µ0 I
µ0 I
µ0 I
lim
lim √
= aφ
= aφ
B = dB = aφ
02
2
3/2
2
2
4π L→∞ −L (z + b )
2πb L→∞ L + b
2πb
The magnetic flux density due to the half-coil is given by substituting z = 0 into
(6-38). Thus we have
µ0 I
Bhc =
4b
Total magnetic flux density is
µ0 I µ0 I
+
â
B=
2πb
4b
where the direction of â is given by right-hand grip rule, i,e, out of paper.
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3. Suppose we divide each wire into n each segments. Define
dl =
w
n
Consider an arbitrary point P (x, y, z) inside the coil where −
w
w
< x, y <
2
2
Calculate the distance between P and
By Biot-Savart Law, each line segment contributes a magnetic flux density
dBi =
where
µ0 I dl × sin θi
4π
Ri2
q
Ri = ∆x2i + ∆yi2 + ∆zi2
and
p 2
2

 ∆yi + ∆zi , for two horizontal wires

i
sin θi = p R
2

∆x
+
∆zi2

i

, for two vertical wires
Ri
The direction dBi is determined by right hand grip rule.
Decompose it into the x, y, z direction, then sum up the dBix , dBiy , dBiz separately
for :
(a) x = −w/2, y = −w/2 to w/2
(b) x = w/2, y = −w/2 to w/2
(c) y = −w/2, x = −w/2 to w/2
(d) y = w/2, y = −w/2 to w/2
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