A hollow, thick-walled, conducting cylinder carries a current of 11.2 A and has an inner radius ri = r and outer radius ro = 3r/2, where r = 4.90 mm. Determine the magnitude of the magnetic field at the following distances from the center of the cylinder. (a) ra = r/2 T

(b) rb = 5r/4 T

(c) rc = 2r T

a. 0 T b. 1.03 × 10⁻⁴ T c. 2.29 × 10⁻⁴ T

Explanation:

a. Using ampere's law ∫B.ds = μ₀i

for ra = r/2, i = 0 (since no current is enclosed) and

∫B.ds = B∫ds = B(2πr/2) = Bπr

So, Bπr = 0

B = 0 T

b. rb = 5r/4

WE find the current enclosed between r = r and r = 5r/4. The total current density in the holow thick-walled conducting cylinder is J = i/[π(3r/2)² - πr²] = i/[9πr²/4 - πr²] = i/8πr²/4 = i/2πr².

Since the current density is constant, we find the current, i' enclosed between  r = r and r = 5r/4. J = i'//[π(5r/4)² - πr²] = i'/[25πr²/16 - πr²] = i'/9πr²/16 = 16i'/9πr²

So, i/2πr² = 16i'/9πr²

i' = 9i/32

Using ampere's law ∫B.ds = μ₀i'

∫B.ds = = B∫ds = B × 2π(5r/4) = 5Bπr/2

5Bπr/2 = μ₀(9i/32)

B = 9μ₀i/32πr × 2/5

B = 9μ₀i/80πr

Substituting r = 4.90 mm = 4.90 × 10⁻³ m and i = 11.2 A, we have

B = 9μ₀i/80πr

= 9 × 4π × 10⁻⁷ H/m × 11.2 A/80π(4.90 × 10⁻³ m)

= 403.2/392 × 10⁻⁴ T

= 1.029 × 10⁻⁴ T

≅ 1.03 × 10⁻⁴ T

c. rc = 2r

Using ampere's law ∫B.ds = μ₀i

∫B.ds = = B∫ds = B × 2π(2r) = 4Bπr

4Bπr = μ₀i  (since i = current enclosed = 11.2 A)

B =  μ₀i/4πr

= 4π × 10⁻⁷ H/m × 11.2 A/4π(4.90 × 10⁻³ m)

= 2.286 × 10⁻⁴ T

≅ 2.29 × 10⁻⁴ T

what type of charge does it have? negative

cant without the picture! ?