Question

Class Management | Help Homework 10 Begin Date: 3/20/2020 12:00:00 PM -- Due Date: 3/27/2020 11:59:00 PM End Date: 3/27/2020 11:59:00 PM (8%) Problem 7: A wire of circular cross-section carries current density that is not uniform but varies with distance from the center j(r)=A(1-r/R), for radius r in the range (<r<R. Here, A is a constant with units amperes per square meter, and the radius of the wire is R = 0.24 mm. 11% Part (a) Find an expression for the current enclosed in a cylinder with a radius of r <R. i(r)=((2 1 Ar)2)-((2nAr(3R) Correct! 11% Part (b) If the total current in the wire is 1, find an expression for the constant A, in terms of the other variables in the problem. A =(31)(R) Correct! 11% Part (c) If the total current is 7.5 A, what is the constant A, in amperes per square meter? A = 124339799.3 A = 1.243E+8 ✓ Correct! 11% Part (d) Find an expression for the magnetic field inside the wire, r<R, in terms of the current I. B(r) =(( 3 I Mor (2 TR-))-((3 I Hor)(3 R )) ✓ Correct! 11% Part(e) Find an expression for the magnetic field outside of the wire, for r>R. Bír) = (H1) (2 nr) ✓ Correct! 11% Part (1) For what r, in meters, is the current enclosed maximum? 10,1 = 0.00024 Correct! 11% Part (g) What is the maximum value of the enclosed current, in amperes? imax = 7.5 Correct! 11% Part (h) For what r, in meters, is the magnetic field maximized? POB = 1.8" 101.4) 0.-0.00018 Correct! > * 11% Part (i) What is the maximum value of the magnetic field, in tesla? Bmax = Grade Summary Deductions Potential 100%

Part i

Answer 1

Since magni magnetic field i nside the wehe rLR(R=0.2470m) Bln - ( ) - ) for, magnetic field is to be maximum ano - Bug? - 1341 ) 27 = 0 BAR² 3 AR² where - T R - 0.18 mm = 0.00018 ) Bray = Bute! )*0.015 ) (0-0000) = 5x4mxolx I1 70-0018 - 50B4 < (0.00018)2 ZAR? r = 21Age - 24A 23 [A= 1.24 x 108] | 7 , 80 1.x10 ( xiờ) Kỳ 1.24 x1 , ( 19 218) 8-D** 5872x10x130 - 1861-53x1062 **** x 10-2 fr=ress 631.67 X 10²4 Ir = 6.32A So, Bmax = 3 x 4 x 10 x 6 32 x 18 x 105 - 3x44x10 x 6 32 x (18810) Xxxx (24 x 10 ) = 6x107x 6.32 x 18 810* - 4X 6. 32 x 10 x 324x lote 13824 x 10-05 3x4x10 1 8 0.108 = 3.14 1.24 y 3 2K (241105) 5768 10-005

Bmax = (1.185x10 2 - 0.592 x 10-2) T -Bmax = 0.593x102T or Bmax = 0.00593 T