A current I flows through a ring of radius R. Determine the magnetic field induction at the center of the ring and on its axis at a distance h from the center of the ring.

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    <h3 id="back-link"><a href="/#9.2">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $9.2.10.$ A current $I$ flows through a ring of radius $R$. Determine the magnetic field induction at the center of the ring and on its axis at a distance $h$ from the center of the ring.
    </p>

    <h3>Solution</h3>
    <p>

The magnetic field $B$ will be directed perpendicular to the plane of the ring. Let's consider a section of the ring of length $Rd\theta$ for it the formula from 9.2.8 will be valid then
$$dB = \frac{\mu_0 I R}{4\pi (R^2 + h^2)}\cdot d\theta$$
but since the field will be directed along the axis, its projection onto this axis is needed there
$$dB\cdot \cos\alpha= \frac{\mu_0 I Rd\theta}{4\pi (R^2 + h^2)} \cdot \frac{R}{\sqrt{R^2 + h^2}}$$

$$dB_z = \frac{\mu_0 I R^2d\theta}{4\pi (R^2 + h^2)^{3/2}}$$
When integrating $dB_z$ from $0$ to $2\pi$ we get:
$$B(h)=\int_0^{2\pi}dB_z=\int_0^{2\pi}\frac{\mu_0 I R^2}{4\pi (R^2 + h^2)^{3/2}}~d\theta$$

$$B(h) = \frac{\mu_0 I R^2}{2 (R^2 + h^2)^{3/2}}$$
substituting $h = 0$ we get the formula you know for the inductance in the center of the ring:
$$B(0) = \frac{\mu_0 I}{2R}$$

    </p>

    <h4>Answer</h4>
    <p>
        $$ B=\frac{\mu_0 qv}{4\pi r^2} \sin\alpha; \quad B = \frac{\mu_0 Il}{4\pi r^2}\sin\alpha$$
    </p>
	<p style="text-align: right; font-style: italic; font-size: 14;">
      Andrei Yersh<br>
    </p>


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