Question

You may recognize the figure below from an earlier homework. It, however, represents a very different physical situation: this is now a very long (infinite for our purposes) conducting wire in the form of a cylindrical shell. The inner radius of the shell a=2.9 mm and outer radius b=4.1 mm. The electric current I=13.5 A runs along this wire and is distributed uniformly throughout the shell cross-section. This current is the source of the magnetic field, which you will be evaluating.

Find the magnitude of the magnetic field B(r₁) inside the shell body at distance r₁=3.14 mm from the cylindrical axis:
B(r₁)= ______ G.

Find the magnitude of the magnetic field B(r₂) outside the shell body at distance r₂=6.56 mm from the cylindrical axis:
B(r₂)= ______ G.

You may recognize the figure below from an earlier homework. It, however, represents a very different physical situation: this is now a very long (infinite for our purposes) conducting wire in the form of a cylindrical shell. The inner radius of the shell a=2.9 mm and outer radius b=4.1 mm. The electric current 1=13.5 A runs along this wire and is distributed uniformly throughout the shell cross-section. This current is the source of the magnetic field, which you will be evaluating. I 12 Find the magnitude of the magnetic field B(ru) inside the shell body at distance r1=3.14 mm from the cylindrical axis: B(r1)= G. Find the magnitude of the magnetic field B(ra) outside the shell body at distance r2=6.56 mm from the cylindrical axis: B(r2)= G.

Answer 1

a According to the amperes lirmital law magnetic field around the closed loop is equal to the lo times the current enclosed by this loop $5.de - llo Tenclesed for a cr <b lets take a closed loop of radius r, I enclosed JA J = current density I 17 (6%a2) A area 느 I enline = I (ru -2²) (6%-a2) rang (re- a2). * Crir-a2) I y (67a2) ampens ens law - llo. I ( ra) 67-a2 B.ar llo I ( a ) (brua) llo - B - (2-a2) 217 r (6%-a & B.de - =) I lo = 41 x 10-7 Tim A-1 da, = 3.14 mm = 3.14810-3 m. a=2.9mm = 2.9x10-3 m. ba 4.1mm = 4.1 8103m 5 = current - 13.5A

(34X153) - 6-9X16) (4.14163) (2.9 x17?) B - 211 YHX10-7 (13.5) 3-14x103 = 1.48 810-4T = 1.48x15 4x10 xio4 G. [IT= 104 a = 1.48 G (N) = 1.48 G b I lets take a closed rin loop of radius ry outside the shell. Mb using amperes law, & side llo Tenclosed. [ Ienlosed - I I B.215r2 = lo I [loop outside the shall B = lo I enclosed the original 21TR2 curren II llo=4H X 107 T-MAT, I 2 13.5A r2 = 6.56mm = 6.56*103m x 13.5 B2 4.12 x 10-4 2 ITX 6.56410-3 T - 4.12 G. 4TX10 B (22) = 4.12 G.