Chapter 31, Problem 63AP

(a)

To determine

The current in both the resistor.

Answer to Problem 63AP

The current in resistor $R_1$ is $3.50\text{A}$ in the upward direction and in resistor $R_2$ is $1.40\text{A}$ in the upward direction.

Explanation of Solution

Write the expression to calculate the emf induced at the end of the rods.

    $\epsilon =BLv$                                                                                                                      (I)

Here, $\epsilon$ is the induced emf, $v$ is the velocity of the rod, $B$ is the magnetic field and $L$ is the length of the rod.

Write the expression for the current.

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

Here, $R$ is the resistance and $I$ is the current.

Substitute $BLv$ for $\epsilon$ in equation (II).

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

Conclusion:

Substitute $2.50\text{T}$ for $B$, $35.0\text{cm}$ for $L$ and $8.00\text{m}/\text{s}$ for $v$ and $2.00\Omega$ for $R_1$ in equation (I) to solve for $I_1$.

    $\begin{array}{c}{I}_1=\frac{\left(2.50\text{T}\right)\left(35.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(8.00\text{m}/\text{s}\right)}{2.00\Omega }\\ =3.50\text{A}\end{array}$

Here, $I_1$ is the current through resistor $R_1$ and its direction is upward.

Substitute $2.50\text{T}$ for $B$, $35.0\text{cm}$ for $L$ and $8.00\text{m}/\text{s}$ for $v$ and $5.00\Omega$ for $R_2$ in equation (I) to solve for $I_2$.

    $\begin{array}{c}{I}_2=\frac{\left(2.50\text{T}\right)\left(35.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(8.00\text{m}/\text{s}\right)}{5.00\Omega }\\ =1.40\text{A}\end{array}$

Here, $I_2$ is the current through resistor $R_2$ and its direction is upward.

Therefore, the current in resistor $R_1$ is $3.50\text{A}$ in the upward direction and resistor $R_2$ is $1.40\text{A}$ in the upward direction.

(b)

To determine

The total power delivered to the resistance of the circuit.

Answer to Problem 63AP

The total power delivered to the resistance of the circuit is $34.3\text{W}$.

Explanation of Solution

Write the expression for equivalent resistance for the resistors connected in parallel.

    $R_{\text{eq}}=\frac{R_1{R}_2}{R_1+R_2}$                                                                                                         (IV)

Here, $R_{\text{eq}}$ is the equivalent resistance for the resistors connected in parallel.

Write the expression for power.

    $P=\frac{{\epsilon }^2}{R_{\text{eq}}}$                                                                                                                    (V)

Here, $P$ is the power.

Substitute $BLv$ for $\epsilon$ and $\frac{R_1{R}_2}{R_1+R_2}$ for $R_{\text{eq}}$ in equation (V).

    $\begin{array}{c}P=\frac{{\left(BLv\right)}^2}{\frac{R_1{R}_2}{R_1+R_2}}\\ P=\left(R_1+R_2\right)\frac{{\left(BLv\right)}^2}{R_1{R}_2}\end{array}$                                                                                             (VI)

Conclusion:

Substitute $2.50\text{T}$ for $B$, $35.0\text{cm}$ for $L$ and $8.00\text{m}/\text{s}$ for $v$, $2.00\Omega$ for $R_1$ and $5.00\Omega$ for $R_2$ in equation (VI) to solve for $P$.

    $\begin{array}{c}P=\left(2.00\Omega +5.00\Omega \right)\frac{{\left(\left(2.50\text{T}\right)\left(35.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(8.00\text{m}/\text{s}\right)\right)}^2}{\left(2.00\Omega \right)\left(5.00\Omega \right)}\\ =\left(\frac{\left(49\right)\left(7\right)}{10}\right)\text{W}\\ =34.3\text{W}\end{array}$

Therefore, the total power delivered to the resistance of the circuit is $34.3\text{W}$.

(c)

To determine

The force required to move the rod with the given velocity.

Answer to Problem 63AP

The force required to move the rod with the given velocity is $4.29\text{A}$.

Explanation of Solution

Write the expression to calculate the force required to move the rod.

    $F=BIL$                                                                                                                 (VII)

Here, $F$ is the force required to move the rod.

Substitute $\left(R_1+R_2\right)\frac{\left(BLv\right)}{R_1{R}_2}$ for $I$ in equation (VII) to solve for $F$.

    $F=B\left(\left(R_1+R_2\right)\frac{\left(BLv\right)}{R_1{R}_2}\right)L$                                                                                  (VIII)

Conclusion:

Substitute $2.50\text{T}$ for $B$, $35.0\text{cm}$ for $L$ and $8.00\text{m}/\text{s}$ for $v$, $2.00\Omega$ for $R_1$ and $5.00\Omega$ for $R_2$ in equation (VIII) to solve for $F$.

    $\begin{array}{c}F=\left(\begin{array}{l}\left(2.50\text{T}\right)\left(35.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(2.00\Omega +5.00\Omega \right)\\ \frac{\left(\left(2.50\text{T}\right)\left(35.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(8.00\text{m}/\text{s}\right)\right)}{\left(2.00\Omega \right)\left(5.00\Omega \right)}\end{array}\right)\\ =\left(\frac{\left(49\right)\left(7\right)}{10}\right)\text{W}\\ =4.29\text{N}\end{array}$

Therefore, the force required to move the rod with the given velocity is $4.29\text{A}$.