Question

Consider a circular current-carrying loop of radius R in the x-y plane with centre at origin. Consider the line integral ζ(L) = | integral_-L ^+L B • dl | taken along z axis.

a) show that ζ(L) monotonically increases with L.
b) use an appropriate loop to show that ζ(∞) = μ_0I.
c) verify directly the above result.
​

Answer 1

Magnetic field due to a circular current-carrying loop lying in the xy- plane acts along z-axis as shown in figure.

a) $\displaystyle\sf \zeta(L) = \left| \int\limits_{-L}^{+L} \vec{B} \cdot \vec{d \ell}\right|$

$\displaystyle\sf = \int\limits_{-L}^{+L} Bd\ell\:\cos 0^{\circ}$

$\displaystyle\sf = \int\limits_{-L}^{+L} Bd\ell = 2BL$

$\displaystyle\sf \therefore$ ζ(L) is a monotonic function of L.

b) Now consider an Amperean loop around the circular coil of such a large radius that L → ∞, since this loop encloses a current I,

• now using ampere's law,

$\displaystyle\sf \zeta(\infty) = \oint\limits_{-\infty}^{+\infty} \vec{B}\cdot\vec{d\ell} = \mu_0I$

c) the magnetic field at the axis (z-axis) of circular coil is given by

$\displaystyle\sf B_z = \dfrac{\mu_0IR^2}{2(z^2+R^2)^{3/2}}$

Now on integrating,

$\displaystyle\sf \int\limits_{-\infty}^{+\infty} B_z d_z = \int\limits_{-\infty}^{+\infty} \dfrac{\mu_0IR^2}{2(z^2+R^2)^{3/2}}dz$

Let z = R tanθ so that dz = R sec²θ dθ

and $\displaystyle\sf (z^2+R^2)^{3/2} = (R^2\:\tan^2\theta+R^2)^{3/2}$

$\displaystyle\sf = R^3\:\sec^3\theta\:\;\;\;(\bf \because1+tan^2\boldsymbol\theta = sec^2\boldsymbol\theta)$

Thus, $\displaystyle\sf \int\limits_{-\infty}^{+\infty} B_zd_z = \dfrac{\mu_0I}{2} = \int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{R^2(R\sec^2\theta\:d\theta)}{R^3\:sec^3\theta}$

$\displaystyle\sf = \dfrac{\mu_0I}{2}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \:\cos\theta\:d\theta$

$\displaystyle\boxed{\bf \therefore\int\limits_{-\boldsymbol\infty}^{+\boldsymbol\infty} B_zd_z = \boldsymbol\mu_0I}$