Chapter 30, Problem 47AP

A thin wire ℓ = 30.0 cm long is held parallel to and d = 80.0 cm above a long, thin wire carrying I = 200 A and fixed in position (Fig. P30.47). The 30.0-cm wire is released at the instant t = 0 and falls, remaining parallel to the current-carrying wire as it falls. Assume the falling wire accelerates at 9.80 m/s². (a) Derive an equation for the emf induced in it as a function of time. (b) What is the minimum value of the emf? (c) What is the maximum value? (d) What is the induced emf 0.300 s after the wire is released?

Figure P30.47

To determine

The equation for the emf induced in the wire as a function of time.

Answer to Problem 47AP

The equation for emf induced in the wire as function of time is $\frac{\left(1.18\times {10}^{-4}\right)t}{0.800-4.90t^2}$

Explanation of Solution

Given info: Length of wire is $30.0\text{cm}$, distance from parallel long wire is $80.0\text{cm}$ and current in the wire is $200\text{A}$.

The speed of the wire according to Newton’s law of motion can be given as,

$v=u+at$

Here,

$u$ is the initial speed of wire.

$a$ is the acceleration of wire.

$t$ is the time.

Substitute $0$ for $u$ and $9.80\text{m}/{\text{s}}^2$ for $a$ in the above equation,

$v=9.80t$

The distance covered by the wire can be given as,

$s=ut-\frac{1}{2}g{t}^2$

Here,

$s$ is the distance covered by the wire.

$g$ is the acceleration due to gravity.

Substitute $0$ for $u$ and $9.80\text{m}/{\text{s}}^2$ for $g$ in the above equation,

$\begin{array}{c}s=-\frac{1}{2}\left(9.80\text{m}/{\text{s}}^2\right)\\ =-4.90t^2\end{array}$

The total distance covered by wire can be given as,

$y=d+s$

Here,

$y$ is the total distance covered by the wire.

$d$ is the distance between wires.

Substitute $80.0\text{cm}$ for $d$ and $-4.90t^2$ for $s$ in the above equation,

$\begin{array}{c}y=\left(80.0\text{cm}\right)-4.90t^2\\ =0.800-4.90t^2\end{array}$

The magnetic field at a distance $y$ from the on a section of coil can be given as,

$B=\frac{{\mu }_{0}I}{2\pi y}$

Here,

$B$ is the magnetic field induced in the loop.

${\mu }_0$ is the permeability constant.

$I$ is the current induced in the loop.

$y$ is any arbitrary distance from the wire.

The emf induced in the wire can be given as,

$\epsilon =Blv$

Here,

$\epsilon$ is the emf induced in the wire.

$l$ is the length of the wire.

$v$ is the speed of the wire.

Substitute $\frac{{\mu }_{0}I}{2\pi y}$ for $B$ in the above equation,

$\epsilon =\left(\frac{{\mu }_{0}I}{2\pi y}\right)lv$

Substitute $4\pi \times {10}^{-7}\text{Tm}/\text{A}$ for ${\mu }_0$, $200\text{A}$ for $I$, $30.0\text{cm}$ for $l$, $9.80t$ for $v$ and $0.800-4.90t^2$ for $y$ in the above equation,

$\begin{array}{c}\epsilon =\frac{\left(4\pi \times {10}^{-7}\text{Tm}/\text{A}\right)\left(200\text{A}\right)\left(30.0\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\left(9.80t\right)}{2\pi \left(0.800-4.90t^2\right)}\\ =\frac{\left(1.18\times {10}^{-4}\right)t}{0.800-4.90t^2}\end{array}$ (1)

Conclusion:

Therefore, the equation for emf induced in the wire as function of time is $\frac{\left(1.18\times {10}^{-4}\right)t}{0.800-4.90t^2}$

To determine

The minimum value of the emf.

Answer to Problem 47AP

The minimum value of emf is $0.$

Explanation of Solution

Given info: Length of wire is $30.0\text{cm}$, distance from parallel long wire is $80.0\text{cm}$ and current in the wire is $200\text{A}$.

The expression for emf can be given as in equation (1),

$\epsilon =\frac{\left(1.18\times {10}^{-4}\right)t}{0.800-4.90t^2}$

At $t=0$,

$\begin{array}{c}\epsilon =\frac{\left(1.18\times {10}^{-4}\right)\times 0}{0.800-4.90\left(0^2\right)}\\ =0\end{array}$

Conclusion:

Therefore, the minimum value of emf is $0.$

To determine

The maximum value of emf.

Answer to Problem 47AP

The maximum value of emf is infinity.

Explanation of Solution

Given info: Length of wire is $30.0\text{cm}$, distance from parallel long wire is $80.0\text{cm}$ and current in the wire is $200\text{A}$.

The expression for emf can be given as in equation (1),

$\epsilon =\frac{\left(1.18\times {10}^{-4}\right)t}{0.800-4.90t^2}$

From the above equation, at $t=\infty$, $\epsilon =\infty$.

Conclusion:

Therefore, the minimum value of emf is $\infty$

To determine

The induced emf $0.300\text{s}$ after the wire is released.

Answer to Problem 47AP

The induced emf $0.300\text{s}$ after the wire is released is $98.3\text{μV}$.

Explanation of Solution

Given info: Length of wire is $30.0\text{cm}$, distance from parallel long wire is $80.0\text{cm}$ and current in the wire is $200\text{A}$, instant of time is $0.300\text{s}$.

The expression for emf can be given as in equation (1),

$\epsilon =\frac{\left(1.18\times {10}^{-4}\right)t}{0.800-4.90t^2}$

Substitute $0.300\text{s}$ for $t$ in the above equation,

$\begin{array}{c}\epsilon =\frac{\left(1.18\times {10}^{-4}\right)\left(0.300\text{s}\right)}{0.800-4.90{\left(0.300\text{s}\right)}^2}\\ =9.83\times {10}^{-5}\text{V}\\ =98.3\text{μV}\end{array}$

Conclusion:

Therefore, the induced emf $0.300\text{s}$ after the wire is released is $98.3\text{μV}$.