Question

Use Biot Savart’s law and obtain the value of the magnetic field along the axis of an ideal solenoid?

Answer 1

Answer:

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Answer 2

Answer:

Explanation:

I → current  

R → Radii

X → Axis  

x → Distance of OP

dl → Conducting element of the loop

According to Biot-Savart's law, the magnetic field at P is  

∣dl×r∣=rdl(they are perpendicular)

  cancelled out and only the x-component remains ∴dB  

Summation of dl over the loop is given by 2πR

(b) Toriod is a hollow circular ring on which a large number of turns of wire are closely wound.Three circular Amperian loops 1,2 and 3 are shown by dashed lines.

By symmetry, the magnetic field should be tangential to each of them and constant in magnitude for a given loop.

Let the magnetic field inside the toroid be B. We shall now consider the magnetic field at S.By Ampere's Law,

Where L is the length of the loop for which B is tangential I be the current enclosed by the loop and N be the number of turns.

We find, L=2πr

The current enclosed I is NL

B(2πr)=μ  

For a loop inside the toroid, no current exists thus, I=0 Hences, B=0

Exterior to the toroid :

Each turn of current coming out of the plane of the paper is cancelled exactly by the current going into it. Thus I=0,and,B=0

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