Question

1. A toroidal sample of magnetic material of susceptibility x = 2 × 10^-2 is wound with 100 turns of wire carrying a current of 2 A. The toroid is 0.10 m long.

(a) Find the solenoidal current density.(3)

(b) Determine the magnetic field intensity H produced by the current.(3)

(c) Calculate μ, the magnetic permeability of the material.(3)

(d) Calculate the induced magnetization M in the material.(3)

(e) Calculate the magnetic field B resulting from the current and the magnetization of the material.(3)

Answer 1

(a) The solenoidal current density depends on the turn density

$n=\frac{N}{l} \\ n = \frac{100}{0.10}=1000$

(b) The magnetic field intensity H produced by the current

$H= \frac{NI}{l} \\ H= \frac{100 \times 2}{0.10}=2000\;A/m$

(c) The magnetic permeability of the material

$\mu=\mu_0(1+χ)$

χ=susceptibility of the materials

$\mu=4 \pi \times 10^{-7}(1+2 \times 10^{-2}) \;H/m \\ \mu = 1.28 \times 10^{-6} \;H/m$

(d) The induced magnetization

$M= χH \\ M= 2 \times 10^{-2} \times 2000 = 40 \;A/m$

(e) The magnetic field B resulting from the current and the magnetization of the material

$B=\mu_0 nI \\ B= 4 \pi \times 10^{-7} \times 1000 \times 2 \\ B=2.51 \times 10^{-3} \; T$