^16 A very long solenoid of radius a, with n turns per unit length, carries a current Is. Coaxial with the solenoid, at radius b ≪ a, is a circular ring of wire, with resistance R. When the current in the solenoid is (gradually) decreased, a current Ir is induced in the ring. (a) Calculate Ir, in

Question

[latex]^{16}[/latex] A very long solenoid of radius a, with n turns per unit length, carries a current [latex]I_{s}[/latex]. Coaxial with the solenoid, at radius b ≪ a, is a circular ring of wire, with resistance R. When the current in the solenoid is (gradually) decreased, a current [latex]I_{r}[/latex] is induced in the ring.

(a) Calculate [latex]I_{r} ext {, in terms of } d I_{s} / d t[/latex] .

(b) The power [latex]\left(I_{r}^{2} R ight)[/latex] delivered to the ring must have come from the solenoid. Confirm this by calculating the Poynting vector just outside the solenoid (the electric field is due to the changing flux in the solenoid; the magnetic field is due to the current in the ring). Integrate over the entire surface of the solenoid, and check that you recover the correct total power

Accepted Answer

[latex] ext { (a) } ε =-\frac{d \Phi}{d t} ; \Phi=\pi a^{2} B ; B=\mu_{0} n I_{s} ; ε =I_{r} R . ext { So } I_{r}=-\frac{1}{R}\left(\mu_{0} \pi a^{2} n ight) \frac{d I_{s}}{d t}[/latex].

[latex] ext { (b) } \oint E \cdot d l =-\frac{d \Phi}{d t} \Rightarrow E(2 \pi a)=-\mu_{0} \pi a^{2} n \frac{d I_{s}}{d t} \Rightarrow E =-\frac{1}{2} \mu_{0} a n \frac{d I_{s}}{d t} \hat{\phi} . B =\frac{\mu_{0} I_{r}}{2} \frac{b^{2}}{\left(b^{2}+z^{2} ight)^{3 / 2}} \hat{ z }[/latex] (Eq. 5.41).

[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)

[latex]S =\frac{1}{\mu_{0}}( E imes B )=\frac{1}{\mu_{0}}\left(-\frac{\mu_{0} a n}{2} \frac{d I_{s}}{d t} ight)\left(\frac{\mu_{0} I_{r}}{2} \frac{b^{2}}{\left(b^{2}+z^{2} ight)^{3 / 2}} ight)(\hat{ \phi } imes \hat{ z })=-\frac{1}{4} \mu_{0} I_{r} \frac{d I_{s}}{d t} \frac{a b^{2} n}{\left(b^{2}+z^{2} ight)^{3 / 2}} \hat{ s }[/latex] .

Power:

[latex]P=\int S \cdot d a =\int_{-\infty}^{\infty}(S)(2 \pi a) d z=-\frac{1}{2} \pi \mu_{0} a^{2} b^{2} n I_{r} \frac{d I_{s}}{d t} \int_{-\infty}^{\infty} \frac{1}{\left(b^{2}+z^{2} ight)^{3 / 2}} d z [/latex]

The integral is [latex]\left.\frac{z}{b^{2} \sqrt{z^{2}+b^{2}}} ight|_{-\infty} ^{\infty}=\frac{1}{b^{2}}-\left(-\frac{1}{b^{2}} ight)=\frac{2}{b^{2}}[/latex] .

[latex]=-\left(\pi \mu_{0} a^{2} n \frac{d I_{s}}{d t} ight) I_{r}=\left(R I_{r} ight) I_{r}=I_{r}{ }^{2} R . \quad ext { qed } [/latex]

Question 8.13: ^16 A very long solenoid of radius a, with n turns per unit ...

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