Biot-Savart Law. 1. Given a flat circular coil with N turns (a large number) of wire carrying a current I. The turns are

Business, Finance, Economics, Accounting, Operations Management, Computer Science, Electrical Engineering, Mechanical Engineering, Civil Engineering, Chemical Engineering, Algebra, Precalculus, Statistics and Probabilty, Advanced Math, Physics, Chemistry, Biology, Nursing, Psychology, Certifications, Tests, Prep, and more.
Post Reply
answerhappygod
Site Admin
Posts: 899603
Joined: Mon Aug 02, 2021 8:13 am

Biot-Savart Law. 1. Given a flat circular coil with N turns (a large number) of wire carrying a current I. The turns are

Post by answerhappygod »

Biot Savart Law 1 Given A Flat Circular Coil With N Turns A Large Number Of Wire Carrying A Current I The Turns Are 1
Biot Savart Law 1 Given A Flat Circular Coil With N Turns A Large Number Of Wire Carrying A Current I The Turns Are 1 (184.95 KiB) Viewed 60 times
Please do it by yourself. Do not copy another answer already
existing on the site. It's very important.
Please do it by yourself. Do not copy another answer already
existing on the site. It's very important.
Please do it by yourself. Do not copy another answer already
existing on the site. It's very important.
Biot-Savart Law. 1. Given a flat circular coil with N turns (a large number) of wire carrying a current I. The turns are spread uniformly over a flat surface from the inner radius a to the outer radius b. The same setup as in assignment #2, problem No.1. p²dr NI From B-S Law, BA(z)= where K = 2 b-a IdZ UK ģ (r2 a + z2)3/2 > p=a b (i) In this problem you are asked to find the approximate B (2) at any point z where (z> b) but not (z >> b). There will be lots of terms in the series expansion. It is understandable that you may even not get the correct answer. Be very patient, neat, and try your best. It turns out that 0 is zero, so keep () as the first correction term and ignore terms smaller than (0) (ii) Determine B, (z <a) where a is finite and not approaching zero. Once again keeps only two largest terms as in ☺ and leave out smaller terms. Note that here z <r. rédr z+r? r r The following is useful. Siz +r?)? = ln (** + Z Z n(n-1) (1+x)" =1+nx + n(n-1)(n-2) .x² + x +... n can be a fraction, positive or negative, 2! 3! 1 1 1 1 and ln(1+x)= x- -X2 + x- rt +-X |x|<1 2 3 4 5 (a+b+c)2 = a² +b+c? + 2ab + 2ac + 2bc a = x, b=x?, c=x+. 1 2 4 1 I 1 throw out terms with power 2 6. Power of 2 4 6 3 5 6 .. keep only 4 terms. (a+b+c)} = a + b3 +c? +3(ab+ ab? +a’c+abe)+3(abc + acé +b+c+bc) 1 2 4 I 1 1 1 I 个 1 1 个 个 1 throw out power 6 power of 36 12 4 5 6 7 7 9 8 10 keep only 3 terms
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!
Post Reply