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Question: A solenoid of length 500m, having 100 turns carries a current of 2.5A find the magnetic field (A) In...

A solenoid of length 500m, having 100 turns carries a current of 2.5A find the magnetic field (A) In the interior of the solenoid.
(B) At one end of the solenoid.
Give, μ0=4π×10−7wbA−1m−1{\mu _0} = 4\pi \times {10^{ - 7}}wb{A^{ - 1}}{m^{ - 1}}

Explanation

Solution

We need to know the concept of magnetic field through solenoid.
We have to use Binside=μ0nI{B_{inside}} = {\mu _0}nI and Boutside=Binside2=μ0nI2{B_{outside}} = \dfrac{{{B_{inside}}}}{2} = \dfrac{{{\mu _0}nI}}{2}

Complete step by step answer:
A solenoid is a generic wire used as an electromagnet. It also refers to any device that converts electrical energy to mechanical energy using a solenoid. When electrical current is introduced a magnetic field forms around the coil which the plunger in solenoid has two magnetic fields in surface and out ward surface so we have to solve in both cases.

Interior Magnetite Field.
Where B is the magnetite field n is number terms peps length I is current field through it
Given:

l=50  cm=0.5  ml = 50\;cm = 0.5\;m
i=2.5  Ai = 2.5\;A
N=100  turns                      N = 100\;turns\;\;\;\;\;\;\;\;\;\;\;
So,
Binside=μ0nI n=Nl n=1000.5=200turns B=4π×10−7×200×2.5 =4π×10−7×200×2.5=2π×10−4 =6.28×10−4T  {B_{inside}} = {\mu _0}nI \\\ n = \dfrac{N}{l} \\\ n = \dfrac{{100}}{{0.5}} = 200 turns \\\ B = 4\pi \times {10^{ - 7}} \times 200 \times 2.5 \\\ = 4\pi \times {10^{ - 7}} \times 200 \times 2.5 = 2\pi \times {10^{ - 4}} \\\ = 6.28 \times {10^{ - 4}}T \\\

For B)
As the magnetic field will face discontinuity at the surface of the end of the solenoid the value of the field at the end will be the average of the field lust inside and outside of the sleeve.

In inside B=6.28×10−4TB = 6.28 \times {10^{ - 4}}T
And in outside B=0B = 0
So on surface the average magnetic field will be
B=12μ0nI Binside2=6.28×10−42 =3.14×10−4T  B = \dfrac{1}{2}{\mu _0}nI \\\ \dfrac{{{B_{inside}}}}{2} = \dfrac{{6.28 \times {{10}^{ - 4}}}}{2} \\\ = 3.14 \times {10^{ - 4}}T \\\

Note:
Always remember that solenoids have magnetic fields on the inside and also on the outside.
Also the magnetic field in the inside is contours but in the outside magnetic field is dishonoured. So we have to find both of them by formula for the case inside as it is confined but for outside it is discontinuous so the magnetic field is zero.