A circular coil of one turn in formed by 6.28 m length wire which carried a current of 3.14 A magnetic field at the center of coil
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Answered by
33
Answer:
Explanation:
Given A circular coil of one turn in formed by 6.28 m length wire which carried a current of 3.14 A magnetic field at the center of coil
Given length = l = 6.28 m , I = 3.14 A
= 2 x 3.14
= 2π m
Again for coil we have l = 2πr
2π = 2πr
So r = 1 m
We know that
Magnetic field of coil = μo I / 2r
= 4π x 10^-7 x 3.14 / 2 x 1
= 19.7 x 10 ^-7
= 1.97 x 10^-6
= 1.97 μ T
Answered by
1
Explanation:
When it convert to coil
l=2πr
2π=2π×r
r=Lm
Magnetic field of coil =
2r
μ
0
I
=
2×L
4π×10
−7
×3.14
=19.7×10
−7
=1.97×10
−6
B≈2μT
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