Chapter 31, Problem 69AP

(a)

To determine

The magnitude of the washer as given in $69$ AP.

Answer to Problem 69AP

Induced emf is $97.4\text{nV}$.

Explanation of Solution

Write the expression for the magnetic field.

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

Here, $r$ is the distance from the point to the current flow, ${\mu }_0$ is the permeability of free space, $B$ is the magnetic field and $I$ is the current flow.

Write the expression for distance with relation to time, gravity and speed.

    $h=ut+\frac{1}{2}g{t}^2$                                                                                                           (II)

Here, $h$ is the distance, $g$ is acceleration due to gravity, $u$ is the initial speed and $t$ is the time.

Magnetic field, when washer is held near the wire is.

Substitute $r_1$ for $r$ in equation (I) to calculate $B_1$.

    $B_1=\frac{{\mu }_{0}I}{2\pi r_1}$

Magnetic field, when washer is above the table is.

Substitute $r_2$ for $r$ in equation (I) to calculate $B_2$.

    $B_2=\frac{{\mu }_{0}I}{2\pi r_2}$

Write the expression for circular area.

    $A=\pi r^2$                                                                                                                  (III)

Here, $r$ is the radius and $A$ is the area.

Write the expression for rate of change of magnetic field.

    $\frac{\Delta B}{\Delta t}=\frac{B_2-B_1}{t}$                                                                                                          (IV)

Here, $\frac{\Delta B}{\Delta t}$ is the rate of change of magnetic field.

Substitute $\frac{{\mu }_{0}I}{2\pi r_1}$ for $B_1$ and $\frac{{\mu }_{0}I}{2\pi r_2}$ for $B_2$ in Equation (IV) to calculate $\frac{\Delta B}{\Delta t}$.

    $\begin{array}{c}\frac{\Delta B}{\Delta t}=\frac{\left(\frac{{\mu }_{0}I}{2\pi r_1}\right)-\left(\frac{{\mu }_{0}I}{2\pi r_2}\right)}{t}\\ =\left(\frac{{\mu }_{0}I}{2\pi }\right)\left(\frac{r_2-r_1}{r_1{r}_{2}t}\right)\end{array}$                                                                                            (V)

Write the expression for emf induced.

    $\epsilon =\frac{-A\Delta B}{\Delta t}$                                                                                                              (VI)

Here, $\epsilon$ is the emf induced.

Conclusion:

Substitute $0$ for $u$, $9.8\text{m}/{\text{s}}^2$ for $g$ and $0.500\text{m}$ for $h$ in Equation (II) to calculate $t$.

    $\begin{array}{c}0.500\text{m}=\left(0\right)t+\frac{1}{2}\left(9.8\text{m}/{\text{s}}^2\right)t^2\\ 0.500\text{m}=\left(4.9\text{m}/{\text{s}}^2\right)t^2\\ t^2=\frac{0.500\text{m}}{\left(4.9\text{m}/{\text{s}}^2\right)}\\ t^2=0.102{\text{s}}^2\end{array}$

Further solve the above equation.

    $\begin{array}{c}t=\sqrt{0.102{\text{s}}^2}\\ t=0.319\text{s}\end{array}$

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ for ${\mu }_0$, $0.319\text{s}$ for $t$, $0.500\text{cm}$ for $r_1$, $0.500\text{m}$ for $r_2$ and $10.0\text{A}$ for $I$ in Equation (V) to calculate $\frac{\Delta B}{\Delta t}$.

    $\begin{array}{c}\frac{\Delta B}{\Delta t}=\left(\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(10.0\text{A}\right)}{2\pi }\right)\left(\frac{\left(0.500\text{m}\right)-\left[\left(0.500\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}{\left(0.500\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\left(0.500\text{m}\right)\left(0.319\text{s}\right)}\right)\\ =\left(20\text{T}\cdot \text{m}\times {10}^{-7}\right)\left(\frac{\left(0.500\text{m}\right)-\left(0.005\text{m}\right)}{\left(0.005\text{m}\right)\left(0.500\text{m}\right)\left(0.319\text{s}\right)}\right)\\ =\left(20\times {10}^{-7}\right)\left(\frac{\left(0.495\right)}{7.975\times {10}^{-4}}\right)\text{T}/\text{s}\\ =\left(20\times {10}^{-7}\right)\left(620.69\right)\text{T}/\text{s}\end{array}$

Further solve the above equation.

    $\frac{\Delta B}{\Delta t}=12413.79\times {10}^{-7}\text{T}/\text{s}$

Substitute $0.500\text{cm}$ for r in Equation (III) to calculate $A$.

    $\begin{array}{c}A=\pi {\left(0.500\text{cm}\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\\ =\pi {\left(0.005\text{m}\right)}^2\\ =7.85\times {10}^{-5}{\text{m}}^2\end{array}$

Substitute $12413.79\times {10}^{-7}\text{T}/\text{s}$ for $\frac{\Delta B}{\Delta t}$ and $7.85\times {10}^{-5}{\text{m}}^2$ for $A$ in Equation (VI) to calculate $\epsilon$.

    $\begin{array}{c}\epsilon =\left(12413.79\times {10}^{-7}\text{T}/\text{s}\right)\left(7.85\times {10}^{-5}{\text{m}}^2\right)\\ =97.4\times {10}^{-9}\text{V}\left(\frac{1\text{nV}}{1\times {10}^{-9}\text{V}}\right)\\ =97.4\text{nV}\end{array}$

Therefore, emf induced is $97.4\text{nV}$.

(b)

To determine

The direction of the induced current in the washer.

Answer to Problem 69AP

Current flow in washer is in clockwise direction.

Explanation of Solution

From ampere’s right hand rule, the magnetic field on washer due to the wire is going into the page. According to Fleming’s right hand rule, thumb figure shows the direction of motion, index finger shows the direction of magnetic field and the middle finger shows the direction of current.

Therefore, from Fleming’s right hand rule, the current flow in the washer is in clock wise.