The wire loop PQRSP formed by joining two semicircular wires of radii [tex]\sf R_1[/tex] and [tex]\sf R_2[/tex] carries a current [tex]\sf I[/tex] as shown. the magnitude of the magnetic induction at the centre C :
[tex]\sf (A) \quad\dfrac{\mu_{0}I}{2}\bigg( \dfrac{R_2 - R_1}{R_2R_1}\bigg)[/tex]
[tex]\sf (B) \quad \dfrac{\mu_{0}I}{2\pi}\bigg( \dfrac{R_2 - R_1}{R_2R_1}\bigg)[/tex]
[tex]\sf (C) \quad \dfrac{\mu_{0}I}{4}\bigg( \dfrac{1}{R_2} - \dfrac{1}{R_1}\bigg) [/tex]
[tex]\sf (D) \quad \dfrac{\mu_{0}I}{4\pi}\bigg(\dfrac{1}{R_2} - \dfrac{1}{R_1}\bigg)[/tex]
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Answer:
B
(
4
μ
0
)I[
R
1
1
−
R
2
1
]
Magnetic field at the centre of a coil B
c
=
2r
μ
o
IN
where N is the number of turn.
For semi-circular coil N=0.5
So, magnetic field at C due to semicircular coil of radius R
1
,
B
1
=
2R
1
μ
o
I(0.5)
=
4R
1
μ
o
I
out of plane of paper
So, magnetic field at C due to semicircular coil of radius R
2
,
B
2
=
2R
2
μ
o
I(0.5)
=
4R
2
μ
o
I
into the plane of paper
Magnetic field at C due to section SR and PQ is zero.
So net magnetic field at C B
net
=B
1
−B
2
=
4
μ
o
I
[
R
1
1
−
R
2
1
]
Explanation:
hope it is helpful to you
Solution:
First of all, wires PQ and SR will not contribute any field intensity at point O, because their axes through point O.
So, field intensity will only be given by the semi-circular part.
Hope It Helps.