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$$
\begin{array}{l}{\text { An air-filled toroidal solenoid has a mean radius of } 15.0 \mathrm{cm}} \\ {\text { and a cross-sectional area of } 5.00 \mathrm{cm}^{2} . \text { When the current is } 12.0 \mathrm{A} \text { , }} \\ {\text { the energy stored is } 0.390 \mathrm{J} \text { . How many turns does the winding }} \\ {\text { have? }}\end{array}
$$

$$
U = \frac{1}{2} L I^{2}
$$

$$
\longrightarrow L=\frac{24}{I^{2}}
$$

$$
\longrightarrow L=\frac{M_{0} N^{2} A}{2 \pi r}
$$$$
\longrightarrow N=\sqrt{\frac{2 \pi r L}{M_{0} A}}
$$

$$
U = \frac{1}{2} L I^{2} \rightarrow L=
$$$$
\frac{24}{I^{2}}=\frac{2(0.39)}{(12){2}}=5.417*10^{-3} \mathrm{H}
$$

$$
L=\frac{M_{0} N^{2} A}{2 \pi r} \longrightarrow
$$$$
N=\sqrt{\frac{2 \pi i L}{M_{0} A}}
$$$$
=\sqrt{\frac{2 \pi(0 \cdot 15)\left(5.417*10^{-3}\right)}{4 \pi * 10^{-7}\left(5 * 10^{-4}\right)}}
$$

\(∴ N=2850 \text { turns. } \)

$$
\begin{array}{l}{\text { An air-filled toroidal solenoid has } 300 \text { turns of wire, a }} \\ {\text { mean radius of } 12.0 \mathrm{cm}, \text { and a cross-sectional area of } 4.00 \mathrm{cm}^{2} . \text { If }} \\ {\text { the current is } 5.00 \text { A, calculate: (a) the magnetic field in the sole- - }} \\ {\text { noid; (b) the self-inductance of the solenoid; (c) the energy stored }}\end{array}
$$

$$
\begin{array}{l}{\text { in the magnetic field; (d) the energy density in the magnetic field. }} \\ {\text { (e) Check your answer for part (d) by dividing your answer to part }} \\ {\text { (c) by the volume of the solenoid. }}\end{array}
$$

(a) $$
B=\frac{M_{0} N I}{2 \pi r}=\frac{4 \pi*10^{-7}*300*5}{2 \pi(0.12)}
$$$$
=2.5 * 10^{-3} T=2.5 \mathrm{mT}
$$

(b) $$
L=\frac{M_{0} N^{2} A}{2 \pi r}=
$$$$
\frac{4 \pi*10^{-7}*300^{2}*4 * 10^{-4}}{2 \pi(0.12)}=6 * 10^{-5} \mathrm{H}
$$

(c) $$
U_{L}=\frac{1}{2} L I^{2}=
$$$$
=\frac{1}{2}\left(6 * 10^{-5}\right)(5)^{2}=7.5 * 10^{-4} \mathrm{J}
$$

(d) $$
u=\frac{B^{2}}{2 M_{0}}=
$$$$
\frac{\left(2.5 * 10^{-3}\right)^{2}}{2\left(4 \pi * 10^{-7}\right)}=2.49 \mathrm{J} / \mathrm{m}^{3}
$$

(e) $$
u=\frac{energy}{\text { volume }}=
$$$$
\frac{\text { energy }}{2 \pi r \mathrm{A}}=
$$$$
\frac{7.5 * 10^{-4}}{2 \pi(0.12)\left(4 * 10^{-4}\right)}=2.49 \mathrm{J} / \mathrm{m}^{3}
$$

$$
\begin{array}{l}{\text { In a proton accelerator used in elementary particle }} \\ {\text { physics experiments, the trajectories of protons are controlled by }} \\ {\text { bending magnets that produce a magnetic field of } 4.80 \text { T. What is }} \\ {\text { the magnetic-field energy in a } 10.0-\mathrm{cm}^{3} \text { volume of space where }} \\ {B=4.80 \mathrm{T} ?}\end{array}
$$

$$
u=\frac{B^{2}}{2 M_{0}}
$$

$$
U=u V \leftarrow
$$

$$
\rightarrow u=\frac{B^{2}}{2 M_{0}}
$$$$
=\frac{(4.8)^{2}}{2 * 4 \pi*10^{-7}}=9.167 * 10^{6} J / m^ 3
$$

$$
\rightarrow U=9.167 * 10^{6} * 10 * 10^{-6}=91.7 \mathrm{J}
$$

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