Question 7.3: A circular loop located on x^2 + y^2 = 9, z = 0 carries a di...

Consider the circular loop shown in Figure 7.8(a). The magnetic field intensity dH at point P(0,0,h) contributed by current element I dl is given by Biot–Savart’s law:

dH=\frac{Idl\times R}{4\pi R^{3}}

where

dl=\rho d\phi a_{\phi},       R=(0,0,h)-(x,y,0)=-\rho a_{\rho}+ha_{z}

and

dl\times R=\left | \begin{matrix} a_{\rho} & a_{\phi} & a_{z} \\ 0 & \rho d\phi & 0 \\ -\rho & 0 & h \end{matrix} \right | =\rho hd\phi a_{\rho}+\rho ^{2}d\phi a_{z}

Hence

dH=\frac{I}{4\pi[\rho^{2}+h^{2}]^{3/2}}(\rho hd\phi a_{\rho}+\rho^{2}d\phi a_{z})=dH_{\rho}a_{\rho}+dH_{z}a_{z}

By symmetry, the contributions along a_{\rho} add up to zero because the radial components produced by current element pairs 180^{\circ} apart cancel.

This may also be shown mathematically by writing a_{\rho} in rectangular coordinate systems (i.e., a_{\rho}=\cos\phi a_{x}+\sin\phi a_{y}). Integrating

\cos\phi   or   \sin\phi   over   0\leq \phi\leq 2\pi

gives zero, thereby showing that H_{\rho}=0. Thus

H=\int dH_{z}a_{z}=\int_{0}^{2\pi}\frac{I\rho ^{2}d\phi a_{z}}{4\pi [\rho^{2}+h^{2}]^{3/2}}=\frac{I\rho ^{2}2\pi a_{z}}{4\pi [\rho^{2}+h^{2}]^{3/2}}

or

H=\frac{I\rho ^{2}a_{z}}{2[\rho^{2}+h^{2}]^{3/2}}

(a) Substituting I=10A,  \rho=3,  h=4 gives

H(0,0,4)=\frac{10(3)^{2}a_{z}}{2[9+16]^{3/2}}=0.36a_{z}A/m

(b) Notice from dl\times R in the Biot–Savart law that if h is replaced by -h, the z component of dH remains the same while the \rho-component still adds up to zero due to the axial symmetry of the loop. Hence

H(0,0,-4)=H(0,0,4)=0.36a_{z}A/m

The flux lines due to the circular current loop are sketched in Figure 7.8(b).