Question

A solenoid has a core of a material with a relative permeability of 400. The solenoid windings are insulated from the core and carry a current of 2A. If the number of turns is 1000 per meter, then the value of magnetic intensity will be 

Hint:

{Magnetic intensity} (\( H \)) inside a solenoid depends on the {number of turns per unit length} and {current}. The presence of a magnetic core modifies the intensity due to the {relative permeability} of the material.

Answer 1

Step 1: Understanding Magnetic Intensity (\(H\)) 
Magnetic intensity (\( H \)) in a solenoid is given by the formula: \[ H = n I \] where:
\( n \) is the number of turns per meter, 
\( I \) is the current in amperes. 
Step 2: Substituting the given values 
Given: \[ n = 1000 \, {turns/m}, \quad I = 2 \, {A} \] Calculating: \[ H = (1000) \times (2) = 2000 \, {Am}^{-1} \] 
Step 3: Effect of Relative Permeability 
The presence of a material with relative permeability (\( \mu_r \)) modifies the magnetic field. 
The permeability factor is given as: \[ B = \mu_0 \mu_r H \] 
Given \( \mu_r = 400 \), we calculate: \[ B = (4\pi \times 10^{-7}) \times (400) \times (2000) \] \[ B = 1.0053 { Am}^{-1} \] .Thus, the final value of magnetic intensity is \( 1.0053 { Am}^{-1} \).