\text{Chapter 29}

\text{A current is set up in a wire loop that is formed as shown in FIGURE 10, where $R_I=$ $2.0 \mathrm{~cm}$ }

\text{and $R_2=4.0 \mathrm{~cm}$. The loop carries a current of $5.0 \mathrm{~A}$, as shown in the figure. }

\text{What is the magnetic field at the center of the loop (C)?}

\text{ A) $3.9 \times 10^{-5} \mathrm{~T}$ out of the page }

\text{ B) $3.9 \times 10^{-5} \mathrm{~T}$ into the page }

\text{ C) $1.2 \times 10^{-4} \mathrm{~T}$ out of the page }

\text{ D) $1.2 \times 10^{-4} \mathrm{~T}$ into the page }

\text{ E) $7.9 \times 10^{-5} \mathrm{~T}$ into of the page }

\text{A current is set up in a wire loop that is formed as shown in FIGURE 10, where $R_I=$ $2.0 \mathrm{~cm}$ }

\text{and $R_2=4.0 \mathrm{~cm}$. The loop carries a current of $5.0 \mathrm{~A}$, as shown in the figure. }

\text{What is the magnetic field at the center of the loop (C)?}

\begin{aligned} & \mathrm{B}_1=\frac{\mu_0 i_\phi}{4 \pi \mathrm{R}_1} \rightarrow \text { out of the page } \\ & \mathrm{B}_2=\frac{\mu_0 i_\phi}{4 \pi \mathrm{R}_2} \rightarrow \text { into the page } \\ & \mathrm{B}_1>\mathrm{B}_2 \text { because } \mathrm{R}_2>\mathrm{R}_1 \\ & \therefore \mathrm{B}_{\mathrm{c}}=\mathrm{B}_1-\mathrm{B}_2=\frac{\mu_0 i_\phi}{4 \pi}\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\ & =\frac{4 \pi \times 10^{-7} \times 5 \times \pi}{4 \pi}\left(\frac{1}{2}-\frac{1}{4}\right) \times 10^2=3.9 \times 10^5 \mathrm{~T} \rightarrow \text { out of the page } \end{aligned}

\text{In FIGURE 9, two infinitely long wires carry currents $\boldsymbol{i}$. Each follows a $90^{\circ}$ arc on the }

\text{circumference of the same circle of radius $\boldsymbol{R}$.}

\text{What is the magnitude of the net magnetic field at the center of the circle (point $\mathrm{C}$ )?}

\text{ A) $\frac{\mu_o i}{2 \pi R}$ }

\text{ B) $\frac{\mu_o i}{\pi R}$ }

\text{ C) $\frac{\mu_o i}{4 \pi R}$ }

\text{ D) $\frac{\mu_o i}{2 \pi R}+\frac{\mu_o i}{16 R}$ }

\text{ E) $\frac{\mu_o i}{\pi R}+\frac{\mu_o i}{16 R}$ }

\text{In FIGURE 9, two infinitely long wires carry currents $\boldsymbol{i}$. Each follows a $90^{\circ}$ arc on the }

\text{circumference of the same circle of radius $\boldsymbol{R}$.}

\text{What is the magnitude of the net magnetic field at the center of the circle (point $\mathrm{C}$ )?}

\text{ * The current in the straight inner wire does not produce any magnetic field }

\text{ * The magnetic fields due to the circular sections cancel. }

\text{ * We are left with the outer straight wires: }

\text{ $\mathrm{B}_{\mathrm{c}}=2 \times$ semi - infinite wires $=2 \times \frac{\mu_0 \mathrm{i}}{4 \pi R}=\frac{\mu_0 \mathrm{i}}{2 \pi R}$ }

\text{FIGURE 10 shows two long, thin wires, parallel to the $z$ axis, carrying currents in the positive $z$ }

\text{direction. The $50\mathrm{A}$ wire is in the $x-z$ plane and is $5 \mathrm{~m}$ from the $z$ axis. }

\text{The 40-A wire is in the $y-z$ plane and is $4 \mathrm{~m}$ from the $z$ axis. }

\text{What is the net magnetic field at the origin $\mathrm{O}$ due to the two wires?}

\text{ A) $(2 \hat{i}-2 \hat{j}) \mu \mathrm{T}$ }

\text{ B) $(2 \hat{i}+2 \hat{j}) \mu T$ }

\text{ C) $(2 \hat{i}+3 \hat{j}) \mu T$ }

\text{ D) $(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}) \mu \mathrm{T}$ }

\text{ E) $(3 \hat{i}+2 \hat{j}) \mu \mathrm{T}$ }

\text{FIGURE 10 shows two long, thin wires, parallel to the $z$ axis, carrying currents in the positive $z$ }

\text{direction. The $50\mathrm{A}$ wire is in the $x-z$ plane and is $5 \mathrm{~m}$ from the $z$ axis. }

\text{The 40-A wire is in the $y-z$ plane and is $4 \mathrm{~m}$ from the $z$ axis. }

\text{What is the net magnetic field at the origin $\mathrm{O}$ due to the two wires?}

\begin{aligned} & \overrightarrow{\mathrm{B}}_1=\frac{\mu_0 \mathrm{i}_1}{2 \pi \mathrm{d}_1}(-\hat{\mathrm{j}}) \\ & =\frac{4 \pi \times 10^{-7} \times 50}{2 \pi \times 5}(-\hat{\mathrm{j}})=-2 \times 10^{-6} \hat{\mathrm{j}}(\mathrm{T})=-2 \hat{\mathrm{j}}(\mu \mathrm{T}) \\ & \overrightarrow{\mathrm{B}}_2=+\frac{\mu_0 \mathrm{i}_2}{2 \pi \mathrm{d}_2} \hat{\mathrm{i}} \\ & =\frac{4 \pi \times 10^{-7} \times 40}{2 \pi \times 4}(\hat{\mathrm{i}})=+2 \hat{\mathrm{i}}(\mu \mathrm{T}) \\ & \therefore \overrightarrow{\mathrm{B}}_{\text {net }}=(2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}) \mu \mathrm{T} \end{aligned}

\text{Use the right hand rule for the direction of $\vec{B}$}

\text{FIGURE 11 shows a cross section of three parallel wires each carrying a current of $5.0 \mathrm{~A}$ out of }

\text{the paper. If the distance $\boldsymbol{d}=6.0 \mathrm{~mm}$, what is the magnitude of the net magnetic force on }

\text{a $2.0\mathrm{m}$ length of wire 1 ?}

\text{ A) $2.9 \mathrm{mN}$ }

\text{ B) $3.3 \mathrm{mN}$ }

\text{ C) $2.1 \mathrm{mN}$ }

\text{ D) $3.9 \mathrm{mN}$ }

\text{ E) $1.7 \mathrm{mN}$ }

\text{FIGURE 11 shows a cross section of three parallel wires each carrying a current of $5.0 \mathrm{~A}$ out of }

\text{the paper. If the distance $\boldsymbol{d}=6.0 \mathrm{~mm}$, what is the magnitude of the net magnetic force on }

\text{a $2.0\mathrm{m}$ length of wire 1 ?}

\begin{aligned} & \mathrm{F}=\frac{\mu_0 \mathrm{Li}^2}{2 \pi \mathrm{d}}=\frac{4 \pi \times 10^{-7} \times 2 \times 25}{2 \pi \times 6 \times 10^{-3}} \\ & =1.67 \times 10^{-3} \mathrm{~N} \\ & \mathrm{~F}_{\text {net }}=2 \text {. F } \cdot \sin 60=2.9 \times 10^{-3} \mathrm{~N} \\ & \end{aligned}

\text{A long, solid, cylindrical wire carries a uniformly distributed current. If the radius of the wire}

\text{is $3.5 \mathrm{~mm}$, and the magnitude of the current density is $1.5 \mathrm{~A} / \mathrm{cm}^2$, what is the magnitude of}

\text{the magnetic field at a distance of $2.5 \mathrm{~mm}$ from the axis of the wire?}

\text{ A) $2.4 \times 10^{-5} \mathrm{~T}$ }

\text{ B) $3.3 \times 10^{-5} \mathrm{~T}$ }

\text{ C) $1.3 \times 10^{-5} \mathrm{~T}$ }

\text{ D) $6.9 \times 10^{-5} \mathrm{~T}$ }

\text{ E) zero }

\text{A long, solid, cylindrical wire carries a uniformly distributed current. If the radius of the wire}

\text{is $3.5 \mathrm{~mm}$, and the magnitude of the current density is $1.5 \mathrm{~A} / \mathrm{cm}^2$, what is the magnitude of}

\text{the magnetic field at a distance of $2.5 \mathrm{~mm}$ from the axis of the wire?}

\begin{aligned} & \oint \vec{B} \cdot d \vec{s}==\mu_0 i_{\mathrm{enc}} \Rightarrow B(2 \pi r)=\mu_0 . J \cdot A=\mu_0 \mathrm{~J} . \pi r^2 \\ & B=\frac{\mu_0 \mathrm{Jr}}{2}=\frac{4 \pi \times 10^{-7} \times 1.5 \times 10^4 \times 2.5 \times 10^{-3}}{2}=2.4 \times 10^{-5} \mathrm{~T} \end{aligned}