Question

A solenoid has 800 turns over a length of 0.5m when a current is 8A. what will be tha magnetic field at the ends. Area of cross section is 0.02m sqaure

Answer 1

Answer:

A solenoid has a core of material of relative permeability 4000. The number of turns is 1000 per metre. A current of 2 A flows through the solenoid. Find magnetic intensity, the magnetic field in core, magnetization, magnetic current and susceptibility.

Given: Relative permeability = μr = 4000, Number of turns per metre = n = 1000, Current flowing = i = 2A, μo = 4π x 10-7 Wb/Am.

To Find:magnetic intensity = H = ?, the magnetic field in core = B = ?, magnetization = MZ = ?, magnetic current = Im =?, and susceptibility = χ = ?

Solution:

Magnetic intensity H = n i = 1000 x 2 = 2000 A/m

Intensity of magnetic field B = μ H = μr μ0 H = 4000 x 4π x 10-7 x 2000 = 10 T

Magnetization MZ = (μr – 1)H = (4000 – 1) x 2000 = 3999 x 2000 = 7.998 x 106 A/m

We have B = μr μ0 (i + Im)

10 = 4000 x 4π x 10-7 x (2 + Im)

(2 + Im) = 10 / (4000 x 4 x 3.142 x 10-7)

2 + Im = 1989

Magnetic current Im = 1987 A

We have μr = 1 + χ

∴ Susceptibility = χ = μr – 1 = 4000 – 1 = 3999

Example – 02:

A solenoid has 1000 turns and is 20 cm long. Find the magnetic induction produced at the centre of the solenoid by the current of 2 A. What is the flux at this point if the diameter of solenoid is 4 cm?

Given: Number of turns = N = 1000, Length = l = 20 cm = 0.2 m, Current through solenoid = 2 A, diameter = 4 cm, radius of solenoid = 4/2 = 2 cm = 0.02 m, μo = 4π x 10-7 Wb/Am.