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- Homework Statement
- A superconducting solenoid of length ##l##, which we can nonetheless asumme as of infinite length has radius ##r_1,\ n\ \text{turns}/m## and is coaxial with a metallic ring of radius ##r_2<r_1##, of trascurable self-inductance and resistance ##R##. The current through the solenoid and the ring at ##t=0s## is, respectively, ##i_{01}=0\ A## and ##i_{02}>0## in the direction shown in the figure.

Find:

(a) Inductance of the solenoid and the coefficient of mutual inductance between the solenoid and the ring;

(b) The electric current going around the ring at ##t_0>0##;

(c) The total magnetic field at the center of the ring at time ##t_0##;

(d) The magnetic energy of the solenoid at time ##t_0##:

(e) The electric charge which has gone around the ring between times ##t=0## and ##t_0##.

- Relevant Equations
- ##\phi=LI;\ \mathcal{E}_2=-\frac{d\phi_2}{dt}=-M\frac{dI_1}{dt}; M_{21}=M_{12}.##

The following is my solution to this problem; I would appreciate some feedback, especially on part (b), which I have found the most challenging. Thanks.

(a) Using Ampere's Law I get ##B=\mu_0 n i_1## where ##i_1## is the current through the solenoid, and since ##\phi=Li_1##, where ##L## is the self-inductance, it follows that ##L=\pi r_1^2 n^2 \mu_0 l##.

For the mutual inductance, since ##M_{12}=M_{21}=M## we can consider the situation where the current ##i## is flowing through the solenoid giving a magnetic flux through the ring ##\phi=B_1 \pi r_2^2=\mu_0 ni\pi r_2^2=M i## so ##M=\mu_0n\pi r_2^2##.

(b) From Faraday's Law we have that ##\mathcal{E}_1=-M\frac{di_2}{dt}-L\frac{di_1}{dt}## (1) and ##\mathcal{E}_2=-M\frac{di_1}{dt}## (2). From (1) we get ##i_1 R_{sol}=M\frac{di_2}{dt}-L\frac{di_1}{dt}## which implies (since ##R_{sol}=0##, the solenoid being superconducting) ##0=M\frac{di_2}{dt}-L\frac{di_1}{dt}## so ##\frac{di_1}{dt}=-\frac{M}{L}\frac{di_2}{dt}## which, substituted in (2) gives: ##\mathcal{E}_2=-M\left( -\frac{M}{L}\frac{di_2}{dt} \right) \Rightarrow i_2 R=\frac{M^2}{L}\frac{di_2}{dt}\Rightarrow \frac{LR}{M^2}dt=\frac{di_2}{i_2}\Rightarrow \int_{t=0}^{t=t_0}\frac{LR}{M^2}d\bar{t}=\int_{i_2(0)}^{i_2(t_0)}\frac{d\bar{i_2}}{\bar{i_2}}\Rightarrow i_2(t_0)=i_2(0)e^{\frac{LR}{M^2}t_0}##

(c) ##\vec{B}=\left(\frac{\mu_0 i_2(t_0)}{2r_2} + \mu_0 ni_1(t_0)\right)\hat{k}## where the first term is the magnetic field at the center of the ring due to the current through the ring which I got using Biot-Savart's law; also, we can get ##i_1(t_0)## from (b)(1) since ##\frac{di_1}{dt}=-\frac{M}{L}\frac{di_2}{dt}=-\frac{M}{L}i_2(0)\frac{LR}{M^2}e^{\frac{LR}{M^2}t}=-\frac{i_2(0)R}{M}e^{\frac{LR}{M^2}t}\Rightarrow i_1(t_0)=-\frac{i_2(0)M}{L}e^{\frac{LR}{M^2}t_0}##

(d) ##U(t_0)=\frac{1}{2}Li_1^2(t_0)##

(e) ##q_2(t_0)=\int_{0}^{t_0}i_2(t) dt##

(a) Using Ampere's Law I get ##B=\mu_0 n i_1## where ##i_1## is the current through the solenoid, and since ##\phi=Li_1##, where ##L## is the self-inductance, it follows that ##L=\pi r_1^2 n^2 \mu_0 l##.

For the mutual inductance, since ##M_{12}=M_{21}=M## we can consider the situation where the current ##i## is flowing through the solenoid giving a magnetic flux through the ring ##\phi=B_1 \pi r_2^2=\mu_0 ni\pi r_2^2=M i## so ##M=\mu_0n\pi r_2^2##.

(b) From Faraday's Law we have that ##\mathcal{E}_1=-M\frac{di_2}{dt}-L\frac{di_1}{dt}## (1) and ##\mathcal{E}_2=-M\frac{di_1}{dt}## (2). From (1) we get ##i_1 R_{sol}=M\frac{di_2}{dt}-L\frac{di_1}{dt}## which implies (since ##R_{sol}=0##, the solenoid being superconducting) ##0=M\frac{di_2}{dt}-L\frac{di_1}{dt}## so ##\frac{di_1}{dt}=-\frac{M}{L}\frac{di_2}{dt}## which, substituted in (2) gives: ##\mathcal{E}_2=-M\left( -\frac{M}{L}\frac{di_2}{dt} \right) \Rightarrow i_2 R=\frac{M^2}{L}\frac{di_2}{dt}\Rightarrow \frac{LR}{M^2}dt=\frac{di_2}{i_2}\Rightarrow \int_{t=0}^{t=t_0}\frac{LR}{M^2}d\bar{t}=\int_{i_2(0)}^{i_2(t_0)}\frac{d\bar{i_2}}{\bar{i_2}}\Rightarrow i_2(t_0)=i_2(0)e^{\frac{LR}{M^2}t_0}##

(c) ##\vec{B}=\left(\frac{\mu_0 i_2(t_0)}{2r_2} + \mu_0 ni_1(t_0)\right)\hat{k}## where the first term is the magnetic field at the center of the ring due to the current through the ring which I got using Biot-Savart's law; also, we can get ##i_1(t_0)## from (b)(1) since ##\frac{di_1}{dt}=-\frac{M}{L}\frac{di_2}{dt}=-\frac{M}{L}i_2(0)\frac{LR}{M^2}e^{\frac{LR}{M^2}t}=-\frac{i_2(0)R}{M}e^{\frac{LR}{M^2}t}\Rightarrow i_1(t_0)=-\frac{i_2(0)M}{L}e^{\frac{LR}{M^2}t_0}##

(d) ##U(t_0)=\frac{1}{2}Li_1^2(t_0)##

(e) ##q_2(t_0)=\int_{0}^{t_0}i_2(t) dt##