Question

If the intensity of magnetic field at a point on the axis of current coil is half of that the centre of the coil then the distance of that point from the centre of the coil will be

Answer 1

Answer:

Let the distance of the axis of the current-carrying coil from the centre be “x”.

The magnetic field induction, at the centre of circular coil of “N” no. of turns,  radius “r”, and current “I” is given by,

Bo = [µo*N*I] / [2*r] ….. (i)

The magnetic field induction, at a point on the axis of the coil of “N” no. of turns,  radius “r”, and current “I” is given by,

B = [µo*2π*N*I*r²] / [4π(x² + r²)^(3/2)] ….. (ii)

We are given that magnetic field at a point on the axis of current coil is half of that the centre of the coil i.e.,

B = ½ * Bo

Substituting the values from (i) & (ii), we get

[µo*2π*N*I*r²] / [4π(x² + r²)^(3/2)] = ½ * [{µo*N*I} / {2*r}]

Cancelling the similar terms

⇒ [r² / (x² + r²)^(3/2)] = 1/(2r)

⇒ 2r³= (x² + r²)^(3/2)

Squaring on both sides

⇒ 4r⁶ = (x² + r²)³

Taking cube roots on both sides

⇒ r² * ∛4 = x² + r²

⇒ x² = r² (∛4 - 1)

⇒ x = r * √[∛4 - 1]

Thus, the distance of the axis of the current-carrying coil from the centre is [r * √{∛4 – 1}].