Part A Constants | Periodic Table A 2.4 mm -diameter copper wire carries a 36 A

Question

Part A Constants | Periodic Table A 2.4 mm -diameter copper wire carries a 36 A current (uniform across its cross section). Determine the magnetic field at the surface of the wire. Express your answer using two significant figures. 1M ALp 5 - 0 2 ? B-T Submit Request Answer Part B Determine the magnetic field inside the wire, 0,50 mm below the surface. Express your answer using two significant figures. 16 A2+ 6 + O 2 ? B- Submit Request Answer Part C Determine the magnetic field outside the wire 2.5 mm from the surface.

Explanation / Answer

radius of wire R = D/2 = 2.4/2 = 1.2 mm = 1.2*10^-3 m

total current I = 36 A

current density J = I/A = I/(pi*R^2)

part(A)

at the surface

consider a circular loop radius R

current enclosed in loop Ienclosed = I

from Ampere's law

B*2*pi*R = uo*Ienclosed

B = uo*I/(2*pi*R)

B = 4*pi*10^-7*36/(2*pi*1.2*10^-3)

B = 0.006 T   <<<-------------ANSWER

==================================

part(B)

at r < R

consider a circular loop radius r ( < R)= 1.2-0.5 = 0.7 mm = 0.7*10^-3 m

current enclosed in loop Ienclosed = J*pi*r^2 = I*r^2/R^2

from Ampere's law

B*2*pi*r = uo*Ienclosed

B*2*pi*r = uo*I*r^2/R^2

B = uo*I*r/(2*pi*R^2)

B = 4*pi*10^-7*36*0.7*10^-3/(2*pi*(1.2*10^-3)^2)

B = 0.0035 T   <<<-------------ANSWER

==============================

part(C)

at r > R

consider a circular loop radius r ( > R)= 1.2+2.5 = 3.7 mm = 3.7*10^-3 m

current enclosed in loop Ienclosed = J*pi*R^2 = I

from Ampere's law

B*2*pi*r = uo*Ienclosed

B*2*pi*r = uo*I

B = uo*I/(2*pi*r)

B = 4*pi*10^-7*36/(2*pi*(3.7*10^-3))

B = 0.0019 T   <<<-------------ANSWER

DONE please check the answer. any doubts post in comment box

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