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A circular coil of radius R carries a cu...

A circular coil of radius `R` carries a current `i`. The magnetic field at its centre is `B`. The distance from the centre on the axis of the coil where the magnetic field will be `B//8` is

A

`sqrt2R`

B

`sqrt3R`

C

`2R`

D

`3R`

Text Solution

Verified by Experts

The correct Answer is:
B
`B = (mu_0"nir"^2)/(2(r^2 + x^2)^(3//2) ) = 1/8 xx (mu_0 "ni")/(2r) `
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NARAYNA-MOVING CHARGES AND MAGNETISM-EXERCISE - 1(C.W) (AMPERE.S CIRCUITAL LAW BIOT-SAVART LAW AND ITS APPLICATIONS)
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