Question

The figure shows a cross section across a long cylindrical conductor of radius a = 2.97 cm carrying uniform current 27.5 A. What is the magnitude of the current's magnetic field at radial distance (a) 0, (b) 2.10 cm, (c) 2.97 cm (wire's surface), (d)3.84 cm?

Answer 1

given
a = 2.97 cm = 0.0297 m
I = 27.5 A

current density in the wire, J = I/A

= I/(pi*a^2)

a) at r = 0

B = 0 <<<<<<<<---------------Answer

b) at r = 2.10 cm = 0.021 m
imagine a circular amperian loop with radius r.

current enclosed by the amperian loop, I_enclosed = J*Area of the amprian loop

= I/(pi*a)^2*(pi*r^2)

= I*r^2/a^2

now use Ampere's law

integral B.dL = mue*I_enclosed

B*2*pi*r = mue*I*r^2/a^2

B = mue*I*r/(2*pi*a^2)

= 4*pi*10^-7*27.5*0.021/(2*pi*0.0297^2)

= 1.31*10^-4 T <<<<<<<---------------Answer

c) at wires surface, a = 2.97 cm = 0.0297 m

B = mue*I/(2*pi*a)

= 4*pi*10^-7*27.5/(2*pi*0.0297)

= 1.85*10^-4 T <<<<<<<---------------Answer

d) at r = 3.84 cm = 0.0384 m
imagine a circular amperian loop with radius r.

current enclosed by the amperian loop, I_enclosed =I

now use Ampere's law

integral B.dL = mue*I_enclosed

B*2*pi*r = mue*I

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

= 4*pi*10^-7*27.5/(2*pi*0.0384)

= 1.43*10^-4 T <<<<<<<---------------Answer