Question

a solenoid has a core of a material with relative permeability 400. the windings of the solenoid are insulated from the core and carry a current of 2a. if the number of turns is 1000 per metre calculate (a) h (b) m (c) b and (d) the magnetising current im.

Answer 1

Answer:

First we need to calculate the magnetic field of Solenoid.

It is given by the formula,

B=μnI

where, μ is the permeability of the material, n is the number of turns per unit length and I is the current.

μ=μr μ₀

It is given that, μr=400

⇒μ=400 × 4π×10⁻⁷T.m/A

I= 2A

n=1000/m

Substitute the values:

B=1600π×10⁻⁷T.m/A×1000/m×2A=1.005 T

B=1.005 T

Magnetic field strength is given by:

H= B/μ=nI = 1000/m ×2A=2000 A/m

H=2000 A/m

Magnetisation, M =(μr-1)H

⇒M=(400-1)(2000 A/m) =798000 A/m

M =798000 A/m

Magnetisation current,

$I_M=\frac{M}{n}=798000/1000 A=798 A$

$I_M=798 A$          

Answer 2

Explanation:

First we need to calculate the magnetic field of Solenoid.

It is given by the formula,

B=μnI

where, μ is the permeability of the material, n is the number of turns per unit length and I is the current.

μ=μr μ₀

It is given that, μr=400

⇒μ=400 × 4π×10⁻⁷T.m/A

I= 2A

n=1000/m

Substitute the values:

B=1600π×10⁻⁷T.m/A×1000/m×2A=1.005 T

B=1.005 T

Magnetic field strength is given by:

H= B/μ=nI = 1000/m ×2A=2000 A/m

H=2000 A/m

Magnetisation, M =(μr-1)H

⇒M=(400-1)(2000 A/m) =798000 A/m

M =798000 A/m

Magnetisation current,

I_M=\frac{M}{n}=798000/1000 A=798 A I

M

=

n

M

=798000/1000A=798A

I_M=798 AI

M

=798A