Chapter 21, Problem 9P

To determine

Part a:

The value of the EMF developed.

Answer to Problem 9P

Solution:

The EMF developed is equal to $1.728\times {10}^{-2}\text{ V}$.

Explanation of Solution

EMF is the abbreviated form of electromotive force, it is defined as the electrical intensity developed by a source of electrical energy.

Given:

The magnetic field is 0.800 T.

The length of the rod of the 12 cm.

The velocity with which the rod is moving is $18\text{ cm/s}$.

Formula used:

The expression for the EMF is as follows:

$\epsilon =Blv$

Here, B is the magnetic field, l is the length of rod, and v is the velocity.

Calculation:

Substitute 0.800 T for B, 12.0 cm for l, and $18\text{ cm/s}$ for v in the equation $\epsilon =Blv$.

$\begin{array}{c}\epsilon =\left(0.800\text{ T}\right)\left(12.0\text{ cm}\right)\left(18\text{ cm/s}\right)\\ =\left(0.800\text{ T}\right)\left(12.0\text{ cm}\right)\left(\frac{{10}^{-2}\text{ m}}{1.00\text{ cm}}\right)\left(18\text{ cm/s}\right)\left(\frac{{10}^{-2}\text{ m/s}}{1.00\text{ cm/s}}\right)\\ =1.728\times {10}^{-2}\text{ V}\end{array}$

To determine

Part b:

The electric field felt by the electrons in the rod.

Answer to Problem 9P

Solution:

The electric field felt by the electrons in the rod is equal to $0.144\text{ V/m}$.

Explanation of Solution

The region around the electrons within which the force would be exerted on the other charged particle is defined as electric field.

Given:

The length of the rod is 12.0 cm.

The magnetic field is 0.800 T.

Formula used:

The expression to calculate the electric field by the electrons on the rod is as follows:

$E=\frac{\epsilon }{l}$

Here, $\epsilon$ is the EMF developed or induced.

Calculation:

Substitute $1.728\times {10}^{-2}\text{ V}$ for $\epsilon$ and 12.0 cm for l in the equation $E=\frac{\epsilon }{l}$.

$\begin{array}{c}E=\frac{1.728\times {10}^{-2}\text{ V}}{12.0\text{ cm}}\\ =\frac{1.728\times {10}^{-2}\text{ V}}{12.0\text{ cm}\left(\frac{{10}^{-2}\text{ m}}{1.00\text{ cm}}\right)}\\ =0.144\text{ V/m}\end{array}$