Question

A current of 2 A flows in an air core solenoid of length
1 m and number of turns 1000. What is the magnetic flux
density inside the solenoid ?

Please no one liners
Detailed explanation

Answer 1

Answer:

PLEASE SEE THE ATTACHMENT FOR CALCULATION PART..

IN A SOLENOID ....

WE KNOW B ( MAGNETIC FIELD OR MAGNETIC FIELD DENSITY.) = MAGNETIC PERMEABILITY × n ( turn density) × i ( current).

@AMINUL HOQUE.

Answer 2

Answer :

Current flow = 2A

Length of solenoid = 1m

No. of turns = 100

We have to find magnetic flux density (magnetic field) inside the solenoid$.$

⧪ Magnetic field due to straight solenoid :

At a point inside the solenoid

• B = μ｡× n × I

At either end of the solenoid

• B' = μ｡× n × I / 2

where n denotes number of turns per unit length and μ｡denotes magnetic permeability of free space.

⭆ B = μ｡× n × I

⭆ B = 4π × 10‾⁷ × 100 × 2

⭆ B = 8π × 10‾⁵

⭆ B = 2.51 × 10‾⁴ T

Cheers!