A circular coil consisting of 200 turns has a radius of 10 cm and carries a current of 2.0 A . At what distance from the centre along the axis of the coil will the magnetic field drop to half of its value at the centre?

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 7.66 cm

The magnetic field at the centre of a circular coil is given by

B_{c} = \dfrac{\mu_{0} N I}{2 R}

The magnetic field at a distance

x

from the centre on its axis is given by

B_{x} = \dfrac{\mu_{0} N I R^{2}}{2 (R^{2} + x^{2})^{3/2}}

Given that

B_{x} = \dfrac{B_{c}}{2}

\dfrac{\mu_{0} N I R^{2}}{2 (R^{2} + x^{2})^{3/2}} = \dfrac{\mu_{0} N I}{4 R}

\dfrac{R^{2}}{(R^{2} + x^{2})^{3/2}} = \dfrac{1}{2 R}

(R^{2} + x^{2})^{3/2} = 2 R^{3}

Taking the power of

2/3

on both sides:

R^{2} + x^{2} = 2^{2/3} R^{2}

x^{2} = R^{2} (4^{1/3} - 1)

x = R \sqrt{4^{1/3} - 1}

Substituting

R = 10 \text{ cm}

and

4^{1/3} = 1.5874

:

x = 10 \sqrt{1.5874 - 1}

x = 10 \sqrt{0.5874}

x \approx 10 \times 0.7664 = 7.664 \text{ cm}

A circular coil consisting of 200 turns has a radius of 10 cm and...

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