\text{Chapter 30}

A) 16.74 \text{ }V

B) 13.26\text{ } V

C) 1.740\text{ } V

D) 15.00\text{ } V

E) 23.48 \text{ }V

\text { magnetic field, with half the area of the loop in the field. The loop contains an ideal }

\text{Chapter 30}

\begin{aligned} & \varepsilon_{\text {ind }}=+A \frac{d B}{d t} \\ & =\frac{2^2}{2}(0.87)=1.74 \mathrm{~V} \\ & \Rightarrow \varepsilon_{\text {net }}=15+1.74 \mathrm{~V}=16.74 \mathrm{~V} \end{aligned}

\begin{aligned} & \varepsilon_{\text {ind }}=B L v \\ & \mathrm{i}_{\text {ind }}=\frac{B L v}{\mathrm{R}}=\frac{0.3 \times 0.35 \times 2}{0.023 \mathrm{~A}} \\ & =0.023 \text { A counterclockwise } \end{aligned}

\begin{aligned} & 2 \pi r=20 \Rightarrow r=10 / \pi=3.182 \mathrm{~m} . \\ & \mathrm{A}=\pi \mathrm{r}^2=100 / \pi=31.83 \mathrm{~m}^2 \text {. } \\ & \varepsilon=A \frac{d B}{d t} \Rightarrow P=\frac{\varepsilon^2}{R} \\ & P=\frac{A^2\left(\frac{d B}{d t}\right)^2}{R} \\ & =\frac{(31.83)^2\left(5 \times 10^{-3}\right)^2}{10} \\ & =2532.7 \times 10^{-6}=2.53 \mathrm{~mW} \\ & \end{aligned}

\begin{aligned} & \varepsilon=-\frac{\Delta \phi}{\Delta \mathrm{t}}=-\mathrm{A} \frac{\Delta \mathrm{B}}{\Delta \mathrm{t}}=-\frac{\mathrm{A}\left(\mathrm{B}_{\mathrm{f}}-\mathrm{B}_{\mathrm{i}}\right)}{\Delta \mathrm{t}} \\ & 1.2=-\frac{0.065}{0.087}\left(0.3-\mathrm{B}_{\mathrm{i}}\right) \Rightarrow \mathrm{B}_{\mathrm{i}}=1.9 \mathrm{~T} \end{aligned}

\begin{aligned} \Phi & =\mathrm{B} \cdot \mathrm{A}=\mathrm{BA} \cos 45 \\ & =0.05 \times 100 \times 10^{-4} \times \cos 45=3.536 \times 10^{-4} \mathrm{~Wb} \end{aligned}

\begin{aligned} & B_{n e t}=B_2 \vec{\jmath}+B_1 \vec{k} \\ & B_1=\frac{\mu_{0 i}}{4 R_1}=\frac{4 \pi \times 10^{-7} \times 0.5}{4 \times 0.1}=15.71 \times 10^{-7} \mathrm{~T} \\ & B_2=\frac{\mu_{0 i}}{4 R_1}=\frac{4 \pi \times 10^{-7} \times 0.5}{4 \times 0.04}=39.27 \times 10^{-7} \mathrm{~T} \\ & \left|B_{n e t}\right|=\sqrt{B_1^2+B_2^2}=\sqrt{(15.71)^2+(39.27)^2} \times 10^{-7} \mathrm{~T}=4.23 \mu \mathrm{T} \end{aligned}

A) \text{ } 1.0 \times 10^{-5}T

B) \text{ }8.2 \times 10^{-5}T

C)\text{ } 1.7 \times 10^{-5}T

D)\text{ } 5.5 \times 10^{-5}T

E) \text{ }2.9 \times 10^{-5}T

\begin{aligned} & |B|=2\times \frac{ \mu_0 i}{2 \pi R} \cos 60^{\circ} \\ & =\frac{2 \times 4 \pi \times 10^{-7} \times 2}{2 \times \pi \times 4 \times 10^{-2}} \times \frac{1}{2} \\ & =1 \times 10^{-5} \mathrm{~T} \\ & \end{aligned}

A) \text{ } 1.0 \times 10^{-5}T

B) \text{ }8.2 \times 10^{-5}T

C)\text{ } 1.7 \times 10^{-5}T

D)\text{ } 5.5 \times 10^{-5}T

E) \text{ }2.9 \times 10^{-5}T

\theta

\theta

\theta

\theta=60^\circ