Chapter 31, Problem 37P

To determine

The current in the resistor $R_3$.

Answer to Problem 37P

The net current in $R_3$ will be $145\mu \text{A}$ upward direction in the picture.

Explanation of Solution

Write the expression for the induced emf.

    $e=B_{\text{in}}vl$                                                                                                                     (I)

Here, $e$ is emf, $B_{\text{in}}$ is the magnetic field, $v$ is the velocity of the rod and $l$ is the length of the rod.

Conclusion:

Calculate the emf for the left rod.

Substitute $0.0100\text{T}$ for $B_{\text{in}}$, $10.0\text{cm}$ for $l$ and $4.00\text{m}/\text{s}$ for $v$ in equation (I) to find emf

    $\begin{array}{c}{e}_{\text{1}}=\left(0.0100\text{T}\right)\left(4.00\text{m}/\text{s}\right)\left(10.0\text{cm}\right)\\ =\left(0.0100\text{T}\right)\left(4.00\text{m}/\text{s}\right)\left(10.0\text{cm}\times \left(\frac{1\text{m}}{100\text{cm}}\right)\right)\\ =\left(0.0100\text{T}\right)\left(4.00\text{m}/\text{s}\right)\left(0.1\text{m}\right)\\ =0.004\text{V}\end{array}$

Here, $e_1$ is the emf in the leftward rod.

Calculate the emf for the sright rod.

Substitute $0.0100\text{T}$ for $B_{\text{in}}$, $10.0\text{cm}$ for $l$ and $2.00\text{m}/\text{s}$ for $v$ in equation (I) to find the emf.

    $\begin{array}{c}{e}_2=\left(0.0100\text{T}\right)\left(2.00\text{m}/\text{s}\right)\left(10.0\text{cm}\right)\\ =\left(0.0100\text{T}\right)\left(2.00\text{m}/\text{s}\right)\left(10.0\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\\ =\left(0.0100\text{T}\right)\left(2.00\text{m}/\text{s}\right)\left(0.1\text{m}\right)\\ =0.002\text{V}\end{array}$

Here, $e_2$ is the emf in the rightward rod.

Let the current in left loop be $I_1$ and in right loop be $I_2$.

It is given that $R_1$ is the resistance in the left rod, $R_2$ is the resistance in the right rod and $R_3$ is the resistance in between these two as shown in the picture.

Apply Kirchhoff voltage law in left loop.

    $-I_1{R}_1-e_1-I_1{R}_3+I_2{R}_3=0$                                                                                       (II)

Apply Kirchhoff voltage law in right loop.

    $-e_2-I_2{R}_2-I_2{R}_3+I_1{R}_3=0$                                                                                     (III)

Substitute $0.004\text{V}$ for $e_1$ $10.0\text{Ω}$ for $R_1$, $5.00\text{Ω}$ for $R_3$ in equation (II).

    $\begin{array}{c}-I_1\left(10.0\text{Ω}\right)-0.004\text{V}-I_1\left(5.00\text{Ω}\right)+I_2\left(5.00\text{Ω}\right)=0\\ -\left(15\Omega \right)I_1+\left(5\Omega \right)I_2=0.004\text{V}\\ I_2=\frac{0.004\text{V}+\left(15\Omega \right)I_1}{5\Omega }\end{array}$

Substitute $0.002\text{V}$ for $e_2$ $15.0\text{Ω}$ for $R_2$, $5.00\text{Ω}$ for $R_3$ in equation (III).

    $\begin{array}{c}-0.002\text{V}-I_2\left(15.0\Omega \right)-I_2\left(5.00\text{Ω}\right)+I_1\left(5.00\text{Ω}\right)=0\\ -\left(20\Omega \right)I_2+\left(5\Omega \right)I_1=0.002\text{V}\end{array}$

Further solve for $I_2$.

    $I_2=\frac{0.002\text{V}-\left(5\text{Ω}\right)I_1}{-20\text{Ω}}$                                                                                          (IV)

Substitute $\frac{0.004\text{V}+\left(15\Omega \right)I_1}{5\Omega }$ for $I_2$ in equation (IV).

    $\begin{array}{c}\frac{0.004\text{V}+\left(15\Omega \right)I_1}{5\Omega }=\frac{0.002\text{V}-\left(5\text{Ω}\right)I_1}{-20\text{Ω}}\\ -4\left(0.004\text{V}+\left(15\Omega \right)I_1\right)=0.002\text{V}-\left(5\text{Ω}\right)I_1\\ -0.016\text{V}-\left(55\text{Ω}\right)I_1=0.002\text{V}\\ \left(55\text{Ω}\right)I_1=-0.018\text{V}\end{array}$

Further solve for $I_1$.

    $\begin{array}{c}{I}_1=-3.27\times {10}^{-4}\text{A}\\ \text{=}-3.27\times {10}^{-4}\text{A}\times \left(\frac{1\mu \text{A}}{{10}^{-6}\text{A}}\right)\\ =-327\mu \text{A}\end{array}$

Therefore, current in the left loop is $-327\mu \text{A}$.

Substitute $-327\mu \text{A}$ for $I_1$ in equation (IV) to find $I_2$.

    $\begin{array}{c}{I}_2=\frac{0.002\text{V}-\left(5\text{Ω}\right){\left(-327\mu \text{A}\right)}_1}{-20\text{Ω}}\\ =-1.817\times {10}^{-4}\text{A}\\ \text{=}-1.817\times {10}^{-4}\text{A}\left(\frac{1\mu \text{A}}{{10}^{-6}\text{A}}\right)\\ =-182\mu \text{A}\end{array}$

Therefore current in the right loop is $-182\mu \text{A}$.

Net current $I_{\text{net}}$ in $R_3$ will be due to both the loops.

    $I_{\text{net}}=I_1-I_2$

Substitute $-327\mu \text{A}$ for $I_1$ and $-182\mu \text{A}$ for $I_2$ in above equation to find net current in $R_3$.

    $\begin{array}{c}{I}_{\text{net}}=\left(-327\mu \text{A}\right)-\left(-182\mu \text{A}\right)\\ =-327\mu \text{A}+182\mu \text{A}\\ =-145\mu \text{A}\end{array}$

Therefore, the net current in $R_3$ will be $145\mu \text{A}$ in the upward direction in picture.