Question

Calculate the magnitudes of magnetic intensity
H and the magnetic field B at the centre of a 1500-turn solenoid which is 0.22 m long and carries a current of
1.5 A.

Answer 1

We have

$N=1500$ $l=0.22 m$ $I=1.5A$

Magnetic intensity and magnetic induction are related by the formula

$H=\frac{B}{\mu_0}$

Magnetic induction at the center of the solinoid is

$B=\frac{\mu_0 \cdot I \cdot N}{l}=\frac{4\pi \cdot 10^{-7}\cdot 1.5 \cdot 1500}{0.22}=4\pi \cdot 10^{-7}\cdot1.023 \cdot10^{4}=12.85mT$

The magnetic intensity is

$H=\frac{B}{\mu_0}=\frac{4\pi \cdot 10^{-7}\cdot1.023 \cdot10^{4}}{4\pi \cdot 10^{-7}}=1.023 \cdot10^{4} A/m$