Worked Example 1
Worked Example 2
Worked Example 3
Worked Example 4
Worked Example 5

The Problem

Take a moment to work on the problem below. The following tab(s) will walk you through finding the solution to this problem, but you must complete the problem first to understand the example.

Consider two positive charges of 0.2 Coulomb each. What distance must they be separated for the force between them to be 10,000 N?

Example Question: Step 1

Since you are dealing with electric charges, use Coulomb’s law to solve this problem.

$LaTeX: F=k\frac{\vert q_1q_2\vert}{r^2}$

From this equation, solve for the distance between the charges.

$LaTeX: r^2=k\frac{\vert q_1q_2\vert}{F}=8.99\times10^9\frac{N\cdot m^2}{C^2}\frac{0.04C^2}{10,000 N}= 36,000 m^2$

Take the square root to find this distance:

r= 190 m

The Problem

Take a moment to work on the problem below. The following tab(s) will walk you through finding the solution to this problem, but you must complete the problem first to understand the example.

Consider a cylinder of radius 0.25 m and length 0.6 m. When a current of 0.5 A runs through this cylinder, you measure a voltage of 5 V. What is the resistivity of the cylinder?

Example Question: Step 1

You are given the measures of the cylinder. To find the resistivity, find the resistance. Use Ohm’s law:

$LaTeX: R=\frac{\varrho L}{A}\longrightarrow {\varrho}=\frac{AR}{L}=\frac{(0.2m^2)(10\Omega)}{0.6m}=3.33\Omega .m$

Next, find the cross-sectional area of the cylinder.

A = πr² = π(0.25 m)² = 0.2 m²

Now calculate the resistivity from the definition of resistance.

$LaTeX: R=\frac{\varrho L}{A}\longrightarrow \varrho = \frac{AR}{L}=\frac{(0.2m^2)(10\Omega)}{0.6m}=3.33\Omega .m$

The Problem

Take a moment to work on the problem below. The following tab(s) will walk you through finding the solution to this problem, but you must complete the problem first to understand the example.

A certain wire has a cross-section of 0.1 m² and resistivity of 1200 Ω.m. What should be the length of the wire to make it carry a current of 0.001 A when a voltage of 10 V is applied?

Example Question: Step 1

You are given the current and voltage going through the wire. Use Ohm’s law to find the required resistance of the wire.

$LaTeX: R=\frac{V}{I}=\frac{10V}{0.001 A}=10,000\Omega$

To find the length of the wire, solve for L from the resistivity equation.

$LaTeX: R=\frac{\varrho L}{A}\longrightarrow L = \frac{AR}{\varrho}=\frac{(0.1 m^2)(10,000\Omega)}{1200\Omega m} = 0.83 m$

You need a wire that is 0.83 meters long.

The Problem

Take a moment to work on the problem below. The following tab(s) will walk you through finding the solution to this problem, but you must complete the problem first to understand the example.

A long current-carrying wire generates a magnetic field of 0.0005 T at a distance of 0.02 meters away from the wire. What is the current going through the wire?

Example Question: Step 1

You know that the magnetic field generated by a long current carrying wire is:

$LaTeX: B=\frac{\mu_0 I}{2\pi r}$

You are given the values of the magnetic field B and the distance r. Solve for the current I from the above equation (remember that $LaTeX: \mu_0 =4\pi \times 10^{-7}T\cdot m/A$ is a known constant).

$LaTeX: I=2\pi r \frac{B}{\mu_0}2\pi(0.02m)\frac{0.0005T}{r\pi \times 10^{-7}T\cdot m/A}= 50 A$

The Problem

Take a moment to work on the problem below. The following tab(s) will walk you through finding the solution to this problem, but you must complete the problem first to understand the example.

A solenoid has a length of 1 meter. It has an internal magnetic field of 0.0001 T with a current of 0.005 A. How many loops of wire does this solenoid have?

Example Question: Step 1

The magnetic field inside of a solenoid is given by:

$LaTeX: B=\mu_0nI$

In this problem, you are not given the number of loops per unit length, only the total length. Use the definition n=N/L and place it into the equation for the magnetic field.

$LaTeX: B=\mu_0\frac{NI}{L}\longrightarrow N=\frac{LB}{/mu_0I}=\frac{(1m)(0.0001T)}{(4\pi\times10^{-7}T\cdot m/A)(0.005A)}=50,000$

The solenoid has a total of 50,000 loops.