Question

b) 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. (uo = 471 x 10-7Hm-1).
(5​

Answer 1

The magnitudes of magnetic intensity H and the magnetic field B are as follows :

Explanation:

The magnetic field induction at the center of solenoid is given by :

$B=\dfrac{\mu _o IN}{l}$

Here,

I is current, I = 1.5 A

l is length of the solenoid, l = 0.22 m

Number of turns, N = 1500

So,

$B=\dfrac{4\pi \times 10^{-7}\times 1.5\times 1500}{0.22}\\\\B=1.28\times 10^{-3}\ T\\\\B=1.28\ mT$

The relation between magnetic field intensity and magnetic induction are given by :

$H=\dfrac{B}{\mu_o}\\\\H=\dfrac{12.8\times 10^{-3}}{4\pi \times 10^{-7}}\\\\H=10185.91\ A/m\\\\H=1.01\times 10^{4}\ A/m$

Learn more,

Solenoid

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