Question

A solenoid has 1000 turns per metre length. If a current of 5A is flowing through it, then magnetic field inside the solenoid is

Answer 1

Answer:

answer in attachment

Answer 2

Concept

The magnetic field of the solenoid is proportional to both the applied current and the number of turns per unit length. The solenoid type magnetic field  is given as $B=u_{0} nI$ where, 'B' represents the magnetic flux density, '$u_{0}$ 'represents the magnetic constant of the value $4\pi\times10^-^7 T / mA$, and 'n' represents the number of turns per unit length and 'I' the current flowing through the solenoid,

Given

We have given a solenoid of 1000 turns per unit length having current 5A

Find

We are asked to determine the magnetic field inside the solenoid .

Solution

Magnetic field inside the solenoid is given by

$B=u_{0} nI$  ...(1)

It is given that  $n=1000m \ , I=5A$

Putting values of n and I in equation (1) , we get

$B=4\pi \times10^-^7\times1000\times 5\\=20000\pi \times 10^-^7\\=2\pi \times10^-^3 T$

Therefore, the magnetic field inside the solenoid is $2\pi \times10^-^3T$ .

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