Question

An ideal solenoid has a core of relative permeability 5000 and it's winding has 1000 turns per meter.if a steady current of 1.6A is passed through its winding.what is the magnetic field intensity within the solenoid

Answer 1

Answer:

I = 79.48 ×105 Am−1

Explanation:

According to given question statement;

relative permeability = μr =5000,

I′ = 1.6A

numer of turns =n = 1000 per metre

H = nI ′

= 1000×1.6

= 1.6×103 Am−1

B= μ H = μ0μrH

= 4π×10−7 × 5000 x (1.6×103)

=10 T

Magnetic field intensity is given by the formula,

I = B/μ0 − H

= 10 / 4π×10−7  − 1.6×103

=79 . 5×105 −0⋅016×105

=79.48 ×105 Am−1