{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW#6_Solutions

HW#6_Solutions - Problem 5.9(a The straight segments...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Problem 5.9 (a) The straight segments produce no ﬁeld at P. The two quarter-circles give B = “01 the half-circle contributes — (b) The two half-hues are the same as one inﬁmte line: Mr 4R ; SOB: :—% (1-4-13) (intothepage). 3 Problem 5.10 1 1 [2 (a) The forces on the two sides cancel. At the bottom, B: ”a => F: (EL) Ia = #0 a (up). At the 2—775 21m 27m — "°’ — ii“— - _Wiai. top, B — 2"“ + a) => F — 2"“ + a) (down). The net force 13 (up). I (b) The force on the bottom is the same as before, ”012/21 (up). On the left side, B- — %‘ z; 41" = 1(dl x B): [(dzi + dyS' +dzi) x (5%: z) = \$37154— «my + dyx). But the a: component cancels the H012 (a/V§+a/2) 1 corresponding term from the right side, and F, = — f — do. Here 1:: V5.7, so 21! .//3‘ y F---u°I2h'1s/‘ﬁH-a/2 ”— 2V§1r s/V§ force on the triangle is M Problem 5.13 21rlcs3 \$3 Problem 5.14 By the right-hand-rule, the ﬁeld points in the —3" direction for z > 0, and in the +5' direction for z < 0. At z— - 0, B: 0. Use the amperian loop shown. {B -=dl BI: palm: polzJ=}M(—a<z<a). I:z>a,Ien¢=polaJ, _ﬂoJﬂ Y. for z > +a;} am rian loo Problem 5.15 The ﬁeld inside a. solenoid' 1s haul, and outside it' 15 zero. The outer solenoid’s ﬁeld points to the left (— z), whereas the' mner one points to the right (+2). So: (i)— B— - ml (71; n2) 2, (ii) m _(iii) - ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online