Chapter 21, Problem 12P

To determine

The induced EMF in the coil.

Answer to Problem 12P

Solution:

The induced EMF in the short coil is $\text{1}\text{.72}\times {\text{10}}^{-4}\text{ V}$.

Explanation of Solution

Given:

The length of solenoid id 25.0 cm.

The number of turns in the solenoid 600.

The number of turns in coil is 14.

The current changes from 0 A to 5.0 A.

The time taken is 0.60 s.

Formula used:

The expression for the induced EMF is as follows:

$\epsilon =NA\frac{\Delta B}{\Delta t}$

Here, $\Delta B$ is the change in the magnetic field, $\Delta t$ is the change in time, N is the number of loops in the coil, and A is the area.

The expression for the magnetic field of a Solenoid is as follows:

$B=\frac{{\mu }_{0}NI}{l}$

Here, ${\mu }_0$ is the magnetic permeability, N is the number of turns, I current, and l is the length of solenoid.

Substitute $\frac{{\mu }_{0}NI}{l}$ for B in the equation $\epsilon =NA\frac{\Delta B}{\Delta t}$.

$\begin{array}{c}\epsilon =NA\frac{\Delta \left(\frac{{\mu }_{0}NI}{l}\right)}{\Delta t}\\ =NA\frac{{\mu }_{0}N}{l}\frac{\Delta I}{\Delta t}\end{array}$

Therefore, the induced EMF is,

$\epsilon =NA\frac{{\mu }_{0}N}{l}\frac{\Delta I}{\Delta t}$

The expression for the area of coil is as follows:

$A=\pi r^2$

Here, r is the radius of coil.

The relation between the radius and the diameter is as follows:

$r=\frac{d}{2}$

Here, d is the diameter.

Therefore, the area is rewritten as follows:

$A=\pi {\left(\frac{d}{2}\right)}^2$

Now, substitute $\pi {\left(\frac{d}{2}\right)}^2$ for A in the equation $\epsilon =NA\frac{{\mu }_{0}N}{l}\frac{\Delta I}{\Delta t}$.

$\epsilon =N\pi {\left(\frac{d}{2}\right)}^2\frac{{\mu }_{0}N}{l}\frac{\Delta I}{\Delta t}$

Calculation:

Now, solve for $\epsilon$ by substituting 3.14 for $\pi$, $4\pi \times {10}^{-7}\text{ T}\cdot \text{m/A}$ for ${\mu }_0$, 14 for N, 25 cm for l, 2.5 cm for d, 0.60 s for $\Delta t$, and 5.0 A for $\Delta I$ in the equation $\epsilon =N\pi {\left(\frac{d}{2}\right)}^2\frac{{\mu }_{0}N}{l}\frac{\Delta I}{\Delta t}$.

$\begin{array}{c}\epsilon =\left(14\right)\left(3.14\right){\left(\frac{2.5\text{ cm}}{2}\right)}^2\frac{4\pi \times {10}^{-7}\text{ T}\cdot \text{m/A}\left(600\right)}{25.0\text{ cm}}\left(\frac{5.0\text{ A}}{0.60\text{ s}}\right)\\ =\left(14\right)\left(3.14\right){\left(\frac{2.5\text{ cm}\left(\frac{{10}^{-2}\text{ m}}{1.00\text{ cm}}\right)}{2}\right)}^2\frac{4\left(3.14\right)\times {10}^{-7}\text{ T}\cdot \text{m/A}\left(600\right)}{25.0\text{ cm}\left(\frac{{10}^{-2}\text{ m}}{1.00\text{ cm}}\right)}\left(\frac{5.0\text{ A}}{0.60\text{ s}}\right)\\ =172,543\times {10}^{-9}\text{ V}\\ =\text{1}\text{.72}\times {\text{10}}^{-4}\text{ V}\end{array}$