A small loop of wire (radius a) is held a distance z above the center of a large loop (radius b), as shown in Fig. 7.37. The planes of the two loops are parallel, and perpendicular to the common axis. (a) Suppose current I flows in the big loop. Find the flux through the little loop. (The little

Question

A small loop of wire (radius a) is held a distance z above the center of a large loop (radius b), as shown in Fig. 7.37. The planes of the two loops are parallel, and perpendicular to the common axis.

(a) Suppose current I flows in the big loop. Find the flux through the little loop. (The little loop is so small that you may consider the field of the big loop to be essentially constant.)

(b) Suppose current I flows in the little loop. Find the flux through the big loop. (The little loop is so small that you may treat it as a magnetic dipole.)

(c) Find the mutual inductances, and confirm that [latex]M_{12}=M_{21}[/latex] .

Accepted Answer

(a) From Eq. 5.41, the field (on the axis) is [latex]B =\frac{\mu_{0} I}{2} \frac{b^{2}}{\left(b^{2}+z^{2} ight)^{3 / 2}} \hat{z }[/latex] , so the flux through the little loop [latex]\left(\operatorname{area} \pi a^{2} ight)[/latex] is [latex]\Phi=\frac{\mu_{0} \pi I a^{2} b^{2}}{2\left(b^{2}+z^{2} ight)^{3 / 2}}[/latex] .

[latex]B(z)=\frac{\mu_{0} I}{4 \pi}\left(\frac{\cos heta}{ᴫ^{2}} ight) 2 \pi R=\frac{\mu_{0} I}{2} \frac{R^{2}}{\left(R^{2}+z^{2} ight)^{3 / 2}} [/latex] (5.41)

(b) The field (Eq. 5.88) is [latex]B =\frac{\mu_{0}}{4 \pi} \frac{m}{r^{3}}(2 \cos heta \hat{ r }+\sin heta \hat{ heta }), ext { where } m=I \pi a^{2}[/latex] . Integrating over the spherical “cap” (bounded by the big loop and centered at the little loop):

[latex]B _{ dip }( r )= abla imes A =\frac{\mu_{0} m}{4 \pi r^{3}}(2 \cos heta \hat{ r }+\sin heta \hat{ heta })[/latex] (5.88)

[latex]\Phi=\int B \cdot d a =\frac{\mu_{0}}{4 \pi} \frac{I \pi a^{2}}{r^{3}} \int(2 \cos heta)\left(r^{2} \sin heta d heta d \phi ight)=\frac{\mu_{0} I a^{2}}{2 r} 2 \pi \int_{0}^{\bar{ heta}} \cos heta \sin heta d heta[/latex]

where [latex]r=\sqrt{b^{2}+z^{2}} ext { and } \sin \bar{ heta}=b / r . \quad ext { Evidently } \Phi=\left.\frac{\mu_{0} I \pi a^{2}}{r} \frac{\sin ^{2} heta}{2} ight|_{0} ^{\bar{ heta}}=\frac{\mu_{0} \pi I a^{2} b^{2}}{2\left(b^{2}+z^{2} ight)^{3 / 2}}[/latex] , the same as in (a)!!

(c) Dividing off [latex]I\left(\Phi_{1}=M_{12} I_{2}, \Phi_{2}=M_{21} I_{1} ight): M_{12}=M_{21}=\frac{\mu_{0} \pi a^{2} b^{2}}{2\left(b^{2}+z^{2} ight)^{3 / 2}}[/latex] .

Question 7.22: A small loop of wire (radius a) is held a distance z above t...

Related Answered Questions