Chapter 20, Problem 89P

(a)

To determine

The magnitude and direction of induced emf, induced current, and force on the loop.

Answer to Problem 89P

The magnitude of induced emf is $\underset{\_}{\epsilon =0}$, induced current is $\underset{\_}{I=0}$, and magnetic force is $\underset{\_}{\text{zero}}$.

Explanation of Solution

The loop is at rest and magnetic field is constant, and average velocity of electron is also zero

Write the expression for magnetic force

  $F_B=q\left(\text{v}\times \text{B}\right)$                                                                                                           (I)

Here, $F_B$ is the magnetic force, $q$ is the charge, $\text{v}$ is the velocity, and $B$ is the magnetic field strength

Since velocity of electrons is zero, according to equation (I) the magnetic force will be zero.

Write the expression for induced emf

  $\epsilon ={\left(\overrightarrow{\text{v}}\times \overrightarrow{\text{B}}\right)}_{\parallel }L$                                                                                                          (II)

Here, $L$ is the length of loop, $\epsilon$ is the induced emf

From equation (II) when velocity of electrons become zero, the emf is also become zero.

The expression for induced current

  $I=\frac{\epsilon }{R}$                                                                                                                     (III)

Here, $R$ is the resistance of loop

Since emf is zero, induced current is also zero according to equation (I)

Conclusion:

Therefore, the magnitude of induced emf is $\underset{\_}{\epsilon =0}$, induced current is $\underset{\_}{I=0}$, and magnetic force is $\underset{\_}{\text{zero}}$.

(b)

To determine

Repeat part (a) for a loop moving to right with constant speed $45\text{cm/s}$.

Answer to Problem 89P

The magnitude of induced emf is $\underset{\_}{{\epsilon }_{\text{net}}=32\text{nV}}$, induced current is $\underset{\_}{I_{\text{loop}}=400\text{pA}}$, and magnetic force is $\underset{\_}{\text{2}\text{.9}\times {\text{10}}^{-17}\text{N}}$.

Explanation of Solution

In this case electrons are moving through the loop and there is a magnetic field associated with the motion of electrons. According to right hand rule, when a magnetic field is in the direction shown by curled fingers, the direction of force is along the right thumb. Since negatively charged electrons moves to right, the direction of field is downward in the plane of paper.

Write the expression for magnetic field in a loop

  $\overrightarrow{\text{B}}=\frac{{\mu }_{0}I}{2\pi r}$                                                                                                                (IV)

Here, $I$ is the current in loop, $B$ is the magnetic field, $r$ is the radius of the loop

The expression for magnetic force is

  $\overrightarrow{\text{F}}=-e\overrightarrow{\text{v}}\times \overrightarrow{\text{B}}$                                                                                                             (V)

The force will develop an emf in loop in the left and right side, since left side is close to the wire, the magnetic field will be greater at right side, hence electron circulate in counterclockwise direction, and current flows in clockwise direction.

Write the expression for induced emf

  ${\epsilon }_{\text{net}}=vL\left(B_L-B_R\right)$                                                                                                 (VI)

Here, $B_L$ is the magnetic field at left, and $B_R$ is the magnetic field at right

Substitute, $\frac{{\mu }_{0}I}{2\pi R_L}$ for $B_L$, and $\frac{{\mu }_{0}I}{2\pi R_R}$ for $B_R$ in equation (VI)

  ${\epsilon }_{\text{net}}=\frac{{\mu }_{0}I}{2\pi }vL\left(\frac{1}{R_L}-\frac{1}{R_R}\right)$                                                                                  (VII)

The current in the loop is

  $I_{\text{loop}}=\frac{{\epsilon }_{\text{net}}}{R}$                                                                                                           (VIII)

The magnetic force acting on each segment of loop be $\overrightarrow{\text{F}}=I_{\text{loop}}\overrightarrow{\text{L}}\times \overrightarrow{\text{B}}$, the top and bottom of loop experiences equal and opposite forces, and force on left acting to left and right to right, with force on left is greater.

The expression for magnitude of force is

  ${\displaystyle \sum F}=I_{\text{loop}}L\left(B_L-B_R\right)$                                                                                          (IX)

Substitute, $\frac{{\mu }_{0}I}{2\pi R_L}$ for $B_L$, and $\frac{{\mu }_{0}I}{2\pi R_R}$ for $B_R$ in equation (IX)

  ${\displaystyle \sum F}=\frac{{\mu }_{0}IL{I}_{\text{loop}}}{2\pi }\left(\frac{1}{R_L}-\frac{1}{R_R}\right)$                                                                                   (X)

Conclusion:

Substitute, $4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}$ for ${\mu }_0$, $6.8\text{A}$ for $I$, $0.45\text{m/s}$ for $v$, $0.023\text{m}$ for $L$, $0.090\text{m}$ for $R_L$, and $0.113\text{m}$ for $R_R$ in equation (VII)

  $\begin{array}{c}{\epsilon }_{\text{net}}=\frac{4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\times 6.8\text{A}}{2\times 3.14}\left(0.45\text{m/s}\right)\left(0.023\text{m}\right)\left(\frac{1}{0.090\text{m}}-\frac{1}{0.113\text{m}}\right)\\ =32\text{nV}\end{array}$

Substitute, $3.183\times {10}^{-8}\text{V}$ for ${\epsilon }_{\text{net}}$, and $79\Omega$ for $R$ in equation (VIII)

  $\begin{array}{c}{I}_{\text{loop}}=\frac{3.183\times {10}^{-8}\text{V}}{79\Omega }\\ =400\text{pA}\end{array}$

Substitute, $4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}$ for ${\mu }_0$, $6.8\text{A}$ for $I$, $0.45\text{m/s}$ for $v$, $0.023\text{m}$ for $L$, $0.090\text{m}$ for $R_L$, $4.03\times {10}^{-10}\text{A}$ for $I_{\text{loop}}$, and $0.113\text{m}$ for $R_R$ in equation (X)

  $\begin{array}{c}{\displaystyle \sum F}=\frac{4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\times 6.8\text{A}}{2\times 3.14}\left(4.03\times {10}^{-10}\text{A}\right)\left(0.45\text{m/s}\right)\left(0.023\text{m}\right)\left(\frac{1}{0.090\text{m}}-\frac{1}{0.113\text{m}}\right)\\ =2.9\times {10}^{-17}\text{N}\end{array}$

The force is acting to left.

Therefore, the magnitude of induced emf is $\underset{\_}{{\epsilon }_{\text{net}}=32\text{nV}}$, induced current is $\underset{\_}{I_{\text{loop}}=400\text{pA}}$, and magnetic force is $\underset{\_}{\text{2}\text{.9}\times {\text{10}}^{-17}\text{N}}$.

(c)

To determine

The electric power dissipated in the loop, and show that it is equal to the rate of external force.

Answer to Problem 89P

The electric power dissipated in the loop is $\underset{\_}{P_E=1.3\times {10}^{-17}\text{W}}$, and it is equal to power dissipated by external force.

Explanation of Solution

Write the expression for power dissipated in the loop

  $P_E=I\epsilon$                                                                                                                 (XI)

Here, $I$ is the induced current, $\epsilon$ is the induced emf

Write the expression for power dissipated by external force

  $P_F=Fv$                                                                                                   (XII)

Here, $F$ is the magnetic force, $v$ is the velocity of electrons

Conclusion:

Substitute, $4.03\times {10}^{-10}\text{A}$ for $I$, and $3.183\times {10}^{-8}\text{V}$ for $\epsilon$ in equation (XI)

  $\begin{array}{c}{P}_E=\left(4.03\times {10}^{-10}\text{A}\right)\left(3.183\times {10}^{-8}\text{V}\right)\\ =1.3\times {10}^{-17}\text{W}\end{array}$

Substitute, $2.85\times {10}^{-17}\text{N}$ for $F$, and $0.45\text{m/s}$ for $v$ in equation (XII)

  $\begin{array}{c}{P}_F=\left(2.85\times {10}^{-17}\text{N}\right)\left(0.45\text{m/s}\right)\\ =1.3\times {10}^{-17}\text{W}\end{array}$

Thus, $P_E=P_F$.

Therefore, the electric power dissipated in the loop is $\underset{\_}{P_E=1.3\times {10}^{-17}\text{W}}$, and it is equal to power dissipated by external force.