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HW8solutions

# HW8solutions - Physics 213 28.60 Homework 8 Solutions...

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Physics 213 Homework 8 Solutions Spring 2009 z r ˆ r a dl P 28.60. We must find the vector sum of the magnetic fields due to each wire. For a long straight wire B 0 I 2 r . The direction of B is given by the right-hand rule and is perpendicular to the line from the wire to the point where then field is calculated. (a) The magnetic field vectors are shown in Figure (a) below. (b) At a position on the x -axis 0 0 0 net 22 2 2 2 2 2 sin , 2 ( ) I I a Ia B r x a x a x a   in the positive x -direction. (c) The graph of B versus x / a is given below (Figure (b)). (d) The magnetic field is a max at the origin, x = 0. (e) When 0 2 ,. Ia x a B x  28.69. According to the Biot-Savart Law, 0 2 ˆ 4 μ Id πr lr B= . If the current loop can be broken up into segments i=0,1,…,n, then the integral can also be broken up: 0 2 0 ˆ 4 n i i μ Id . In our case, there are 3 segments: the straight parts and the quarter circle. The contribution from the straight segments is zero since 0. d  This leaves only the curved segment contributing to B . For each interval on the curved segment, ˆ is perpendicular to dr l and so the Biot-Savart Law gives 0 2 4 Idl dB R in an upward direction. (Note that the contributions from all intervals around the curved segment are the same magnitude and direction: no cancellations, etc). To calculate the total B , we integrate over all dB for ¼ of a circle. In the expression for dB, everything is a constant except dl. So the integral becomes just 2 / 4 dl R , giving 00 1 4 28 μ I μ I B RR     directed out of the page. Note that in this expression, the B field from one quarter of a circle is just ¼ of the B field from a whole circle. (Because of symmetries, it is very simple to calculate B at point P but it would be much more difficult to calculate B at other points.) 28.70. The horizontal wire yields zero magnetic field since 0. d For the vertical current, the setup is as shown on the right. The magnetic field contribution d B = o I d r ^ r 2 at point P due to every current element I d along this segment is directed out of the plane of the page, and the magnitude dB = |d B | = o I dz r 2 cos . So the total magnetic field due to this straight segment is out of the page, and its magnitude is 0 2 0 cos 4 I dz B dB r  . B 1x =B 1 sin

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Physics 213 Homework 8 Solutions Spring 2009 We can evaluate the integral by making the usual change of variable to the angle : z = a tan , dz = a sec 2 , r 2 = a 2 + z 2 = a 2 (1 + tan 2 ) = a 2 sec 2 , to get: /2 0 0 0 0 0 cos sin 4 4 4 I I I Bd a a a   . So we end up with a magnetic field which is half that an infinite wire. Actually, we could have predicted this ahead of time.
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HW8solutions - Physics 213 28.60 Homework 8 Solutions...

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