by applying boit sawart's law . derive an expression for magnetic field due to long staright conductor along with special cases.
Answers
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
db=
4πr
3
μ
0
I∣dl×r∣
r
2
=x
2
+R
2
∣dl×r∣=rdl(they are perpendicular)
∴dB=
4π
μ
0
(x
2
+R
2
)
Idl
dB has two components −dB
x
and dB
⊥
.dB
⊥
is cancelled out and only the x-component remains ∴dB
x
=dBcosθ cosθ=
(x
2
+R
2
)
1/2
R
dB
x
=
4π
μ
0
Idl
fr
(x
2
+R
2
)
1/2
R
Summation of dl over the loop is given by 2πR
∴B=Bxi=
2(x
2
+R
2
)
3/2
μ
0
IR
2
i
(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,
∫
B
ˉ
.
dI
ˉ
=μ
0
IBL=μ
0
NI
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
0
NI,therefore,B= 2πr NI
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
Answer: I will answer it later is it oksy
Explanation:(-_-)