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The strength of magnetic induction at the centre of a current carrying circular coil is B1 and at a point on its Axis at a distance equal to its radius from the centre is B2 find B1 by B2

The strength of magnetic induction at the centre of a current carrying circular coil is B1 and at a point on its Axis at a distance equal to its radius from the centre is B2 find B1 by B2

Grade:12

1 Answers

Khimraj
3007 Points
3 years ago
Magnetic field dueto current carrying coil is given by
B = (u0nI/2)(r2/(x2+r2)3/2)
For B1 x =0
So B1 =(u0nI/2r)
for B2 x= r
So B2 = (u0nI/25/2r)
So B1/B2 = 23/2.
Hope it clears.

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