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$$
\begin{array}{l}{\text { Two coils are wound around the same cylindrical form, }} \\ {\text { like the coils in Example } 30.1 . \text { When the current in the first coil is }} \\ {\text { decreasing at a rate of }-0.242 \mathrm{A} / \mathrm{s} \text { , the induced emf in the second }} \\ {\text { coil has magnitude } 1.65 \times 10^{-3} \mathrm{V} \text { . (a) What is the mutual induc- }} \\ {\text { tance of the pair of coils? ( b) If the second coil has } 25 \text { turns, what }}\end{array}
$$

$$
\begin{array}{l}{\text { is the flux through each turn when the current in the first coil }} \\ {\text { equals } 1.20 \mathrm{A} ?(\mathrm{c}) \text { If the current in the second coil increases at a }} \\ {\text { rate of } 0.360 \mathrm{A} / \mathrm{s}, \text { what is the magnitude of the induced emf in the }} \\ {\text { first coil? }}\end{array}
$$

$$
\varepsilon_{1}=M\left|\frac{\Delta i_{2}}{\Delta t}\right|
$$

$$
\varepsilon_{2}={M}\left|\frac{\Delta i_{1}}{\Delta t}\right|
$$

$$
M=\left|\frac{N_{2} \Phi_{B 2}}{i_1}\right|
$$

(a) $$
M=\frac{\varepsilon_{2}}{\left|\frac{\Delta  i_{1}}{\Delta t}\right|}=
$$$$
\frac{1.65*10^{-3}}{|-0.242|}
$$$$
=6.82*10^{-3} \mathrm{H}=6.82 \mathrm{mH}
$$

(b) $$
\Phi_{B_{2}}=\frac{M_{i_ 1}}{N_{2}}
$$$$
=\frac{6.82 * 10^{-3} * 1.2}{25}=
$$$$
3.27 * 10^{-4} \mathrm{wb}
$$

(c) $$
\varepsilon_{1}=M\left|\frac{\Delta i_{2}}{\Delta t}\right|
$$$$
=6.82 * 10^{-3} * 0.36=2.46 * 10^{-3} V=2.46 \mathrm{mv}
$$

$$
\begin{array}{l}{\text { A toroidal solenoid with mean radius } r \text { and cross-sectional }} \\ {\text { area } A \text { is wound uniformly with } N_{1} \text { turns. A second toroidal sole- }} \\ {\text { noid with } N_{2} \text { turns is wound uniformly on top of the first, so that }} \\ {\text { the two solenoids have the same cross-sectional area and mean }} \\ {\text { radius. (a) What is the mutual inductance of the two solenoids? }}\end{array}
$$

$$
\begin{array}{l}{\text { Assume that the magnetic field of the first solenoid is uniform }} \\ {\text { across the cross section of the two solenoids. (b) If } N_{1}=500} \\ {\text { turns, } N_{2}=300 \text { turns, } r=10.0 \mathrm{cm}, \text { and } A=0.800 \mathrm{cm}^{2}, \text { what is }} \\ {\text { the value of the mutual inductance? }}\end{array}
$$

$$
B_{1}=\frac{M_{0} N_{1} i_{1}}{2 \pi r}
$$

$$
M=\frac{N_{2}\left|\Phi_{B 2}\right|}{i_{1}}
$$

$$
\rightarrow \Phi_{B_{2}}=B_{1} A
$$

(a) $$
\Phi_{B 2}=\frac{M_{0} N_{1} i_{1}}{2 \pi r}A
$$

$$
M=\frac{N _2 M_{0} N_1i_1 A}{2 \pi r i_1}\rightarrow
$$$$
M=\frac{N_{2} M_{0} N_{1} A}{2 \pi r}
$$$$
\rightarrow M=\frac{M_{0} N_{1} N_{2} A}{2 \pi  r}
$$

(b) $$
M=\frac{2*10^{-7} * 500 * 300 * 0.8*10^{-4}}{0.1}
$$$$
=2.4 * 10^{-5} \mathrm{H}=24 \mathrm{MH}
$$

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