As moving charges create magnetic fields, so they experience forces from magnetic fields generated by other materials. The magnitude of the force experienced by a particle traveling in a magnetic field depends on the charge of the particle \begin{align*} (q) \end{align*}, the velocity of the particle \begin{align*} (v) \end{align*}, the strength of the field \begin{align*}(B) \end{align*}, and, importantly, the angle between their relative directions \begin{align*} (\theta)\end{align*}:

There is a **second right hand rule** that will show the direction of the force on a positive charge in a magnetic field: point your index finger along the direction of the particle’s velocity . If your middle finger points along the magnetic field, your thumb will point in the direction of the force. NOTE: For negative charge reverse the direction of the force (or use your left hand)

For instance, if a positively charged particle is moving to the right, and it enters a magnetic field pointing towards the top of your page, it feels a force going out of the page, while if a positively charged particle is moving to the left, and it enters a magnetic field pointing towards the top of your page, it feels a force going into the page:

where q is the charge of the particle, v is the velocity of the particle, *B* is the magnetic field value and \begin{align*} \theta \end{align*} is the angle between the velocity vector and the magnetic field vector. Note that for problems where the direction of the particle and the direction of the magnetic field are perpendicular then

\begin{align*} F_B = qvB \end{align*}

also, recall that

\begin{align*} F = qE \end{align*}

\begin{align*} E = V \Delta x && \text{ for a constant electric field}\end{align*}

#### Example 1: Find the Magnetic Field

An electron is moving to the east at a speed of \begin{align*}1.8\times10^6\;\mathrm{m/s}\end{align*}. It feels a force in the upward direction with a magnitude of \begin{align*}2.2\times10^{-12}\;\mathrm{N}\end{align*}. What is the magnitude and direction of the magnetic field this electron just passed through?

There are two parts to this question, the magnitude of the electric field and the direction. We will first focus on the magnitude.

To find the magnitude we will use the equation \begin{align*} F_B=qvBsin\theta \end{align*} We were given the force of the magnetic field \begin{align*}(2.2\times10^{-12}\;\mathrm{N})\end{align*} and the velocity that the electron is traveling \begin{align*}(1.8\times10^6\;\mathrm{m/s})\end{align*}. We also know the charge of the electron \begin{align*}(1.6\times10^{-19}\;\mathrm{C})\end{align*}. Also, because the electron's velocity is perpendicular to the field, we do not have to deal with \begin{align*}\sin\theta\end{align*} because \begin{align*}\sin\theta\end{align*} of \begin{align*}90\end{align*} degrees is \begin{align*}1\end{align*}. Therefore all we have to do is solve for B and plug in the known values to get the answer.

\begin{align*} F_B=qvB\sin\theta \end{align*}

Solving for B:

\begin{align*}B = \frac{F_B}{qv\sin\theta}\end{align*}

Now, plugging the known values we have:

\begin{align*} B=\frac{F_B}{qv\sin\theta}=\frac{2.2?10^{-12}\:\mathrm{N}}{1.6\times10^{-19}\:\mathrm{C} \times 1.8\times10^6\:\mathrm{m/s}\times1}=7.6\:\mathrm{T} \end{align*}

Now we will find the direction of the field. We know the direction of the velocity (east) and the direction of the force due to the magnetic field (up, out of the page). Therefore we can use the second right hand rule (we will use the left hand, since an electron's charge is negative). Point the pointer finger to the right to represent the velocity and the thumb up to represent the force. This forces the middle finger, which represents the direction of the magnetic field, to point south. Alternatively, we could recognize that this situation is illustrated for a *positive* particle in the right half of the drawing above; for a negative particle to experience the same force, the field has to point in the opposite direction: south.

#### Example 2: Circular Motion in Magnetic Fields

Consider the following problem: a positively charged particle with an initial velocity of \begin{align*} \vec{v}_1 \end{align*}, charge \begin{align*} q \end{align*} and mass \begin{align*} m \end{align*} traveling in the plane of this page enters a region with a constant magnetic field \begin{align*} \vec{B} \end{align*} pointing into the page. We are interested in finding the trajectory of this particle.

Since the force on a charged particle in a magnetic field is always perpendicular to both its velocity vector and the field vector (check this using the second right hand rule above), a *constant* magnetic field will provide a centripetal force --- that is, a constant force that is always directed perpendicular to the direction of motion. Two such force/velocity combinations are illustrated above. According to our study of rotational motion, this implies that as long as the particle does not leave the region of the magnetic field, it will travel in a circle. To find the radius of the circle, we set the magnitude of the centripetal force equal to the magnitude of the magnetic force and solve for \begin{align*} r \end{align*}:

\begin{align*}F_c = \frac{mv^2}{r} = F_B = qvB\sin \theta = qvB \end{align*}

Therefore,

\begin{align*}r = \frac{mv^2}{qvB}\end{align*} In the examples above, \begin{align*} \theta \end{align*} was conveniently 90 degrees, which made \begin{align*} \sin \theta = 1\end{align*}. But that does not really matter; in a constant magnetic fields a different \begin{align*} \theta \end{align*} will simply decrease the force by a constant factor and will not change the qualitative behavior of the particle.

### Interactive Simulation

### Review

- For each of the arrangements of velocity \begin{align*}v\end{align*} and magnetic field \begin{align*}B\end{align*} below, determine the direction of the force. Assume the moving particle has a positive charge.
- As an electron that is traveling in the positive \begin{align*}x-\end{align*}direction encounters a magnetic field, it begins to turn in the upward direction (positive \begin{align*}y-\end{align*}direction). What is the direction of the magnetic field?
- -"\begin{align*}x\end{align*}"-direction
- +"\begin{align*}y\end{align*}"-direction (towards the top of the page)
- -"\begin{align*}z\end{align*}"-direction (i.e. into the page)
- +"\begin{align*}z\end{align*}"-direction (i.e. out of the page)
- none of the above

- A positively charged hydrogen ion turns upward as it enters a magnetic field that points into the page. What direction was the ion going before it entered the field?
- -"\begin{align*}x\end{align*}"-direction
- +"\begin{align*}x\end{align*}"-direction
- -"\begin{align*}y\end{align*}"-direction (towards the bottom of the page)
- +"\begin{align*}z\end{align*}"-direction (i.e. out of the page)
- none of the above

- Protons with momentum \begin{align*}5.1 \times 10^{-20} \;\mathrm{kg} \cdot \;\mathrm{m/s}\end{align*} are magnetically steered clockwise in a circular path. The path is \begin{align*}2.0 \;\mathrm{km}\end{align*} in diameter. (This takes place at the Dann International Accelerator Laboratory, to be built in 2057 in San Francisco.) Find the magnitude and direction of the magnetic field acting on the protons.
- An electron is accelerated from rest through a potential difference of \begin{align*}1.67 \times 10^5\end{align*} volts. It then enters a region traveling perpendicular to a magnetic field of \begin{align*}0.25 \;\mathrm{T}\end{align*}.
- Calculate the velocity of the electron.
- Calculate the magnitude of the magnetic force on the electron.
- Calculate the radius of the circle of the electron’s path in the region of the magnetic field

- A beam of charged particles travel in a straight line through mutually perpendicular electric and magnetic fields. One of the particles has a charge, \begin{align*}q\end{align*}; the magnetic field is \begin{align*}B\end{align*} and the electric field is \begin{align*}E\end{align*}. Find the velocity of the particle.
- A positron (same mass, opposite charge as an electron) is accelerated through \begin{align*}35,000\end{align*} volts and enters the center of a \begin{align*}1.00 \;\mathrm{cm}\end{align*} long and \begin{align*}1.00 \;\mathrm{mm}\end{align*} wide capacitor, which is charged to \begin{align*}400\end{align*} volts. A magnetic filed is applied to keep the positron in a straight line in the capacitor. The same field is applied to the region (region II) the positron enters after the capacitor.
- What is the speed of the positron as it enters the capacitor?
- Show all forces on the positron.
- Prove that the force of gravity can be safely ignored in this problem.
- Calculate the magnitude and direction of the magnetic field necessary.
- Show the path and calculate the radius of the positron in region II.
- Now the magnetic field is removed; calculate the acceleration of the positron away from the center.
- Calculate the angle away from the center with which it would enter region II if the magnetic field were to be removed.

- An electron is accelerated through \begin{align*}20,000 \;\mathrm{V}\end{align*} and moves along the positive \begin{align*}x-\end{align*}axis through a plate \begin{align*}1.00\;\mathrm{cm}\end{align*} wide and \begin{align*}2.00 \;\mathrm{cm}\end{align*} long. A magnetic field of \begin{align*}0.020 \;\mathrm{T}\end{align*} is applied in the \begin{align*}-z\end{align*} direction.
- Calculate the velocity with which the electron enters the plate.
- Calculate the magnitude and direction of the magnetic force on the electron.
- Calculate the acceleration of the electron.
- Calculate the deviation in the \begin{align*}y\end{align*} direction of the electron form the center.
- Calculate the electric field necessary to keep the electron on a straight path.
- Calculate the necessary voltage that must be applied to the plate.

### Review (Answers)

- a. Into the page b. Down the page c. Right
- d
- b
- \begin{align*}0.00016 \;\mathrm{T}\end{align*}; if CCW motion, B is pointed into the ground.
- a. \begin{align*}2.42 \times 10^8 \;\mathrm{m/s}\end{align*} b. \begin{align*}9.69 \times 10^{-12} \;\mathrm{N}\end{align*} c. \begin{align*}.0055 \;\mathrm{m}\end{align*}
- E/B
- a. \begin{align*}1.11 \times 10^8 \;\mathrm{m/s}\end{align*} b. \begin{align*}9.1 \times 10^{-30} \;\mathrm{N} < < 6.4 \times 10^{-14} \;\mathrm{N}\end{align*} d. \begin{align*}.00364 \;\mathrm{T}\end{align*} e. \begin{align*}.173 \;\mathrm{m}\end{align*} f. \begin{align*}7.03 \times 1016 \;\mathrm{m/s}^2\end{align*} g. \begin{align*}3.27^\circ\end{align*}
- a.\begin{align*}8.39 \times 10^7 \;\mathrm{m/s}\end{align*} b. \begin{align*}2.68 \times 10^{-13} \;\mathrm{N}, -y\end{align*} c. \begin{align*}2.95 \times 10^17 \;\mathrm{m/s}^2\end{align*} d. \begin{align*}.00838 \;\mathrm{m}\end{align*} e. \begin{align*}1.68 \times 10^6 \;\mathrm{N/C}\end{align*} f. \begin{align*}16,800 \;\mathrm{V}\end{align*}