Answer:
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
