Question

Question no 01) Calculate the magnetic induction B and flux Φ at the center of a toroidal 

solenoid with mean circumference of 50 cm and a cross-sectional area of 200 

mm2

, wound with 800 turns of wire carrying 1.0 A, (1) when the solenoid has 

an air core and (2) when the solenoid has a soft iron core of relative 

permeability 1000.

Answer 1

The magnetic induction

$\int{Bdl}=\mu_0I\to B\cdot2\pi r=\mu_0NI\to B=\frac{\mu_0NI}{2\pi r}$

$L=2\pi r=50\to r\approx8\ (cm)$

$B=\frac{\mu_0NI}{2\pi r}=\frac{4\pi\cdot10^{-7}\cdot 800\cdot1}{2\pi\cdot0.08}=0.002\ (T)$ with air

$B=\frac{\mu\mu_0NI}{2\pi r}=\frac{1000\cdot4\pi\cdot10^{-7}\cdot 800\cdot1}{2\pi\cdot0.08}=2\ (T)$ with iron

The magnetic flux

$\Phi=BA\cos\alpha$

$\Phi=BA\cos\alpha=0.002\cdot 200\cdot10^{-6}=0.4\cdot10^{-6}\ (Wb)$ with air

$\Phi=BA\cos\alpha=2\cdot 200\cdot10^{-6}=400\cdot10^{-6}\ (Wb)$ with iron