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Unformatted text preview: 55. (a) The magnitude of the magnetic ﬁeld on the axis of a circular loop, a distance z from the loop center,
is given by Eq. 3028:
N µ0 iR2
B=
,
2(R2 + z 2 )3/2
where R is the radius of the loop, N is the number of turns, and i is the current. Both of the loops
in the problem have the same radius, the same number of turns, and carry the same current. The
currents are in the same sense, and the ﬁelds they produce are in the same direction in the region
between them. We place the origin at the center of the lefthand loop and let x be the coordinate
of a point on the axis between the loops. To calculate the ﬁeld of the lefthand loop, we set z = x
in the equation above. The chosen point on the axis is a distance s − x from the center of the
righthand loop. To calculate the ﬁeld it produces, we put z = s − x in the equation above. The
total ﬁeld at the point is therefore
B= N µ0 iR2
2 (R 2 1
1
+2
2 ) 3/ 2
2 − 2sx + s2 )3/2
+x
(R + x . Its derivative with respect to x is
dB
N µ0 iR2
=−
dx
2 (R 2 3(x − s)
3x
+2
2 ) 5/ 2
2 − 2sx + s2 )5/2
+x
(R + x . When this is evaluated for x = s/2 (the midpoint between the loops) the result is
dB
dx =−
s /2 N µ0 iR2
2 3s/2
3s/2
−2
=0
(R2 + s2 /4)5/2
( R + s 2 / 4 − s 2 + s 2 ) 5/ 2 independently of the value of s.
(b) The second derivative is
d2 B
dx2 = N µ0 iR2
2
− − (R 2 15x2
3
+
2 ) 5/ 2
2 + x2 )7/2
+x
(R 3
15(x − s)2
+2
(R2 + x2 − 2sx + s2 )5/2
(R + x2 − 2sx + s2 )7/2 . At x = s/2,
d2 B
dx2 = N µ0 iR2
2 = s2 − R 2
N µ0 R2 −6(R2 + s2 /4) + 30s2 /4
= 3N µ0 iR2 2
.
2
(R2 + s2 /4)7/2
(R + s2 /4)7/2 s /2 Clearly, this is zero if s = R. − (R 2 30s2 /4
6
+2
2 /4)5/2
+s
(R + s2 /4)7/2 ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Current

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