Question

Fig. 28-9 shows the cross-section of a hollow cylinder of inner radius a = 5 cm and outer radius b = 7 cm. A uniform current density j = 1 A/cm2 flows through the cylinder. Calculate the magnitude of the magnetic field at a distance d = 10 cm from the axis of the cylinder.

Answer 1

Answer:

1.507×10⁻⁴ T

Explanation:

a = 5 cm = 0.05 m

b = 7 cm = 0.07 m

j = 1 A/cm²

Distance from magnetic field = d = 10 cm = 0.1 m

μ₀ = Vacuum permeability = 4π×10⁻⁷ H/m

Magnetic of hollow cylinder

$\oint B.ds=\mu_0 I\\\Rightarrow B2\pi d=\mu_0 I\\\Rightarrow B2\pi d=\mu_0J\pi(b^2-a^2)\\\Rightarrow B=\frac{\mu_0J\pi(b^2-a^2)}{2\pi d}\\\Rightarrow B=\frac{4\pi\times 10^{-7}\times 1\times 10^{-4}\pi(0.07^2-0.05^2)}{2\pi 0.1}\\\Rightarrow B=1.507\times 10^{-4}\ T$

∴ Magnitude of the magnetic field at a distance d = 10 cm from the axis of the cylinder is 1.507×10⁻⁴ T