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• Notes

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

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

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

$\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$

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