14.1: Magnetism
For static electric charges, the electromagnetic force is manifested by the Coulomb electric force alone. If charges are moving, an additional force emerges, called magnetism. The \begin{align*}19^{th}\end{align*} century realization that electricity and magnetism are dual aspects of the same force completely changed our understanding of the world we live in. Insights due to Ampere, Gauss, and Maxwell led to the understanding that moving charges – electric currents – create magnetic fields. Varying magnetic fields create electric fields. Thus a loop of wire in a changing magnetic field will have current induced in it. This is called electromagnetic induction. Magnetic fields are usually denoted by the letter \begin{align*}B\end{align*} and are measured in Teslas, in honor of the Serbian physicist Nikola Tesla.
Key Concepts
 Magnetic fields are generated by charged particles in motion.
 Magnetic fields exert magnetic forces on charged particles in motion.
 Permanent magnets (like refrigerator magnets) consist of atoms, such as iron, for which the magnetic moments (roughly spin) of the nuclei are “lined up” all across the atom. This means that their magnetic fields add up, rather than canceling each other out. The net effect is noticeable because so many atoms have lined up. This means that their magnetic fields add up, rather than canceling each other out. The net effect is noticeable because so many atoms have lined up. The magnetic field of such a magnet always points from the north pole to the south. The magnetic field of a bar magnet, for example, is illustrated below:
If we were to cut the magnet above in half, it would still have north and south poles; the resulting magnetic field would be qualitatively the same as the one above (but weaker).
 Changing magnetic fields passing through a loop of wire generate currents in that wire; this is how electric power generators work. Likewise, the changing amounts of current in a wire create a changing magnetic field; this is how speakers and electric motors work.
 Our world is in 3 dimensions including magnetic fiels and sometimes, it is necessary to represent such three dimensional fields on a two dimensional sheet of paper. The following example illustrates how this is done.
In the example above, a current is running along a wire towards the top of your page. The magnetic field is circling the wire in loops that are pierced through the center by the current. Where these loops intersect this piece of paper, we use the symbol \begin{align*}\bigodot\end{align*} to represent where the magnetic field is coming out of the page and the symbol \begin{align*}\bigotimes\end{align*} to represent where the magnetic field is going into the page. This convention can be used for all vector quantities: fields, forces, velocities, etc.
 Magnetic fields have a “3D” property, requiring you to use special vector rules (called right hand rules) to figure out the directions of forces, fields, and currents.
\begin{align*}\circ\end{align*} Right Hand Rule #1 A wire with electric current going through it produces a magnetic field going in circles around it. To find the direction of the magnetic field, point your thumb in the direction of the current. Then, curl your fingers around the wire. The direction your fingers curl tells you the direction that the magnetic field is pointing. Be sure to use your right hand!
\begin{align*}\circ\end{align*} Right Hand Rule #2
Point your index finger along the direction of the particle’s velocity or the direction of the current. Your middle finger goes in the direction of the magnetic field and your thumb tells you the direction of the force. NOTE: For negative charge reverse the direction of the force (or use your left hand)
Key Equations and Applications
\begin{align*} F_B = qvB\sin{\theta}\end{align*}
The force due to a magnetic field on a moving particle depends on the charge and speed of the particle. The angle \begin{align*} \theta \end{align*} is the angle between the velocity vector and the magnetic field vector. The direction of the force is given by RHR #2.
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.
Use this figure to remember the geometry of the angle \begin{align*}\theta\end{align*}.
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.
\begin{align*} F_{\text{wire}}= LIB\sin(\theta)\end{align*}
A currentcarrying wire is made up of a whole bunch of individually moving charges. Thus this wire will feel a force. In this equation, \begin{align*}L\end{align*} refers to the length of the wire, \begin{align*}B\end{align*} to the magnetic field strength and the angle \begin{align*}\theta\end{align*} is the angle between the direction of the current and the direction of the magnetic field. The direction of the force is given by RHR #2 as explained above.
\begin{align*} \mu_0=4 \pi \times 10^{7} \;\text{Tm/A}\end{align*}
Permeability of a Vacuum( approximately same for air also)
\begin{align*} B_{\text{wire}} = \frac{\mu_0 I}{2 \pi r} \end{align*}
The magnetic field generated by a currentcarrying wire depends on the strength of the current and the distance from the wire.
In the example to the left, a current is running along a wire towards the top of your page. The magnetic field is circling the wire in loops that are pierced through the center by the current. Where these loops intersect this piece of paper, we use the symbol \begin{align*}\bigodot\end{align*} to represent where the magnetic field is coming “out of the page”and the symbol \begin{align*}\bigotimes\end{align*} to represent where the magnetic field is going “into the page.” The distance r (for radius) is the distance from the wire.
Two currentcarrying wires next to each other each exert magnetic fields and thereby forces on each other.
\begin{align*} \Phi = N B\cdot A \end{align*}
If you have a closed, looped wire of area \begin{align*}A\end{align*} (measured in \begin{align*}m^2\end{align*}) and \begin{align*}N\end{align*} loops, and you pass a magnetic field \begin{align*}B\end{align*} through, the magnetic flux is \begin{align*}\Phi\end{align*}. The units of magnetic flux are \begin{align*}T \cdot m^2\end{align*}, also known as a Weber\begin{align*}(Wb)\end{align*}.
In the example to the left, there are four loops of wire \begin{align*}(N = 4)\end{align*} and each has area \begin{align*}\pi r^2\end{align*}. The magnetic field is pointing toward the top of the page, at a right angle to the loops. Think of the magnetic flux as the “bundle” of magnetic field lines “held” by the loop. Why does it matter? See the next equation
\begin{align*} emf = \frac{\Delta \Phi}{\Delta t} \end{align*}
The direction of the induced current is determined as follows: the current will flow so as to generate a magnetic field that opposes the change in flux. This is called Lenz’s Law.
If you change the amount of magnetic flux that is passing through a loop of wire, electrons in the wire will feel a force (called the electromotive force), and this will generate a current. The equivalent voltage (emf) that they feel is equal to the change in flux \begin{align*}\triangle \Phi\end{align*} divided by the amount of time \begin{align*}\triangle t\end{align*} it takes to change the flux by that amount. This is Faraday’s Law of Induction. The relative direction of the loops and the field matter; this relationship is preserved by creating an 'area vector': a vector whose magnitude is equal to the area of the loop and whose direction is perpendicular to the plane of the loop. The directions' influence can then be conveniently captured through a dot product:
\begin{align*} \Phi = N \vec{B} \cdot \vec{A} &&\text{ Electromagnetic Flux} \end{align*}
The units of magnetic flux are \begin{align*}\text{T} \times \mathrm{m^2}\end{align*}, also known as Webers\begin{align*}\text{(Wb)}\end{align*}.
In the example above, there are four loops of wire \begin{align*}(N = 4)\end{align*} and each has area \begin{align*}\pi r^2\end{align*} (horizontally hashed). The magnetic field is pointing at an angle \begin{align*} \theta \end{align*} to the area vector. If the magnetic field has magnitude \begin{align*} B \end{align*}, the flux through the loops will equal \begin{align*} 4 \cos \theta B \pi r^2\end{align*}. Think of the magnetic flux as the part of the “bundle” of magnetic field lines “held” by the loop that points along the area vector.
If the magnetic flux through a loop or loops changes, electrons in the wire will feel a force, and this will generate a current. The induced voltage (also called electromotive force, or emf) that they feel is equal to the change in flux \begin{align*}\triangle \Phi\end{align*} divided by the amount of time \begin{align*}\triangle t\end{align*} that change took. This relationship is called Faraday’s Law of Induction:
\begin{align*} emf = \frac{\Delta \Phi}{\Delta t}&& \text{Faraday's Law of Induction} \end{align*}
The direction of the induced current is determined as follows: the current will flow so as to generate a magnetic field that opposes the change in flux. This is called Lenz’s Law. Note that the electromotive force described above is not actually a force, since it is measured in Volts and acts like an induced potential difference. It was originally called that since it caused charged particles to move  hence electromotive  and the name stuck (it's somewhat analogous to calling an increase in a particle's gravitational potential energy difference a gravitomotive force).
Since only a changing flux can produce an induced potential difference, one or more of the variables in equation [5] must be changing if the ammeter in the picture above is to register any current. Specifically, the following can all induce a current in the loops of wire:
 Changing the direction or magnitude of the magnetic field.
 Changing the loops' orientation or area.
 Moving the loops out of the region with the magnetic field.
Example 1: Find the Magnetic Field
Question: 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?10^{12}\mathrm{N}\end{align*}. What is the magnitude and direction of the magnetic field this electron just passed through?
Answer: 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?10^{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\\ \intertext{Solving for B:}\\ 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 \intertext{Therefore,} 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, since \begin{align*} \theta \end{align*} cannot change due to such a magnetic force. (Why? Hint: what is the force perpendicular to? Read the paragraph above.)
Magnetism Problem Set
 Can you set a resting electron into motion with a stationary magnetic field? With an electric field? Explain.
 How is electrical energy produced in a dam using a hydroelectric generator? Explain in a short essay involving as many different ideas from physics as you need.
 A speaker consists of a diaphragm (a flat plate), which is attached to a magnet. A coil of wire surrounds the magnet. How can an electrical current be transformed into sound? Why is a coil better than a single loop? If you want to make music, what should you do to the current?
 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.
 Sketch the magnetic field lines for the horseshoe magnet shown here. Then, show the direction in which the two compasses (shown as circles) should point considering their positions. In other words, draw an arrow in the compass that represents North in relation to the compass magnet.
 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
 An electron is moving to the east at a speed of \begin{align*}1.8 \times 10^6 \;\mathrm{m/s}\end{align*}. It feels a force in the upward direction with a magnitude of \begin{align*}2.2 \times 10^{12} \;\mathrm{N}\end{align*}. What is the magnitude and direction of the magnetic field this electron just passed through?
 A vertical wire, with a current of \begin{align*}6.0 \;\mathrm{A}\end{align*} going towards the ground, is immersed in a magnetic field of \begin{align*}5.0 \;\mathrm{T}\end{align*} pointing to the right. What is the value and direction of the force on the wire? The length of the wire is \begin{align*}2.0 \;\mathrm{m}\end{align*}.
 A futuristic magnetocar uses the interaction between current flowing across the magneto car and magnetic fields to propel itself forward. The device consists of two fixed metal tracks and a freely moving metal car (see illustration above). A magnetic field is pointing downward with respect to the car, and has the strength of \begin{align*}5.00 \;\mathrm{T}\end{align*}. The car is \begin{align*}4.70 \;\mathrm{m}\end{align*} wide and has \begin{align*} 800 \;\mathrm{A}\end{align*} of current flowing through it. The arrows indicate the direction of the current flow.
 Find the direction and magnitude of the force on the car.
 If the car has a mass of \begin{align*}2050 \;\mathrm{kg}\end{align*}, what is its velocity after \begin{align*}10 \;\mathrm{s}\end{align*}, assuming it starts at rest?
 If you want double the force for the same magnetic field, how should the current change?
 A horizontal wire carries a current of \begin{align*}48\;\mathrm{A}\end{align*} towards the east. A second wire with mass \begin{align*}0.05 \;\mathrm{kg}\end{align*} runs parallel to the first, but lies \begin{align*}15 \;\mathrm{cm}\end{align*} below it. This second wire is held in suspension by the magnetic field of the first wire above it. If each wire has a length of half a meter, what is the magnitude and direction of the current in the lower wire?
 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.
 A bolt of lightening strikes the ground \begin{align*}200 \;\mathrm{m}\end{align*} away from a \begin{align*}100\end{align*}turn coil (see above). If the current in the lightening bolt falls from \begin{align*}6.0 \times 10^6 \;\mathrm{A}\end{align*} to \begin{align*}0.0 \;\mathrm{A}\end{align*} in \begin{align*}10 \;\mathrm{ms}\end{align*}, what is the average voltage, \begin{align*} \varepsilon\end{align*}, induced in the coil? What is the direction of the induced current in the coil? (Is it clockwise or counterclockwise?) Assume that the distance to the center of the coil determines the average magnetic induction at the coil’s position. Treat the lightning bolt as a vertical wire with the current flowing toward the ground.
 A coil of wire with \begin{align*}10\end{align*} loops and a radius of \begin{align*}0.2 \;\mathrm{m}\end{align*} is sitting on the lab bench with an electromagnet facing into the loop. For the purposes of your sketch, assume the magnetic field from the electromagnet is pointing out of the page. In \begin{align*}0.035 \;\mathrm{s}\end{align*}, the magnetic field drops from \begin{align*}0.42 \;\mathrm{T}\end{align*} to \begin{align*}0 \;\mathrm{T}\end{align*}.
 What is the voltage induced in the coil of wire?
 Sketch the direction of the current flowing in the loop as the magnetic field is turned off. (Answer as if you are looking down at the loop).
 A wire has \begin{align*}2 \;\mathrm{A}\end{align*} of current flowing in the upward direction.
 What is the value of the magnetic field \begin{align*}2 \;\mathrm{cm}\end{align*} away from the wire?
 Sketch the direction of the magnetic field lines in the picture to the right.
 If we turn on a magnetic field of \begin{align*}1.4 \;\mathrm{T}\end{align*}, pointing to the right, what is the value and direction of the force per meter acting on the wire of current?
 Instead of turning on a magnetic field, we decide to add a loop of wire (with radius \begin{align*}1 \;\mathrm{cm}\end{align*}) with its center \begin{align*}2 \;\mathrm{cm}\end{align*} from the original wire. If we then increase the current in the straight wire by \begin{align*}3 \;\mathrm{A}\end{align*} per second, what is the direction of the induced current flow in the loop of wire?
 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.
 Two long thin wires are on the same plane but perpendicular to each other. The wire on the \begin{align*}y\end{align*}axis carries a current of \begin{align*}6.0 \;\mathrm{A}\end{align*} in the \begin{align*}y\end{align*} direction. The wire on the \begin{align*}x\end{align*}axis carries a current of \begin{align*}2.0 \;\mathrm{A}\end{align*} in the \begin{align*}+ x\end{align*} direction. Point, \begin{align*}P\end{align*} has the coordinates of \begin{align*}(2.0, 2,0)\end{align*} in meters. A charged particle moves in a direction of \begin{align*}45^o\end{align*} away from the origin at point, \begin{align*}P\end{align*}, with a velocity of \begin{align*}1.0 \times 10^7 \;\mathrm{m/s}.
\end{align*}
 Find the magnitude and direction of the magnetic field at point, \begin{align*}P\end{align*}.
 If there is a magnetic force of \begin{align*}1.0 \times 10^{6} \;\mathrm{N}\end{align*} on the particle determine its charge.
 Determine the magnitude of an electric field that will cancel the magnetic force on the particle.
 A rectangular loop of wire \begin{align*} 8.0 \;\mathrm{m}\end{align*} long and \begin{align*}1.0 \;\mathrm{m}\end{align*} wide has a resistor of \begin{align*}5.0 \ \Omega\end{align*} on the \begin{align*}1.0\end{align*} side and moves out of a \begin{align*}0.40 \;\mathrm{T}\end{align*} magnetic field at a speed of \begin{align*}2.0 \;\mathrm{m/s}\end{align*} in the direction of the \begin{align*}8.0 \;\mathrm{m}\end{align*} side.
 Determine the induced voltage in the loop.
 Determine the direction of current.
 What would be the net force needed to keep the loop at a steady velocity?
 What is the electric field across the \begin{align*}.50 \;\mathrm{m}\end{align*} long resistor?
 What is the power dissipated in the resistor?
 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.
 A small rectangular loop of wire \begin{align*}2.00 \;\mathrm{m}\end{align*} by \begin{align*}3.00 \;\mathrm{m}\end{align*} moves with a velocity of \begin{align*}80.0 \;\mathrm{m/s}\end{align*} in a nonuniform field that diminishes in the direction of motion uniformly by \begin{align*}.0400 \;\mathrm{T/m}\end{align*}. Calculate the induced emf in the loop. What would be the direction of current?
 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.
 A long straight wire is on the \begin{align*}x\end{align*}axis and has a current of \begin{align*}12 \;\mathrm{A}\end{align*} in the \begin{align*}x\end{align*} direction. A point \begin{align*}P\end{align*}, is located \begin{align*}2.0 \;\mathrm{m}\end{align*} above the wire on the \begin{align*}y\end{align*}axis.
 What is the magnitude and direction of the magnetic field at \begin{align*}P\end{align*}.
 If an electron moves through \begin{align*}P\end{align*} in the\begin{align*}x\end{align*} direction at a speed of \begin{align*}8.0 \times 10^7 \;\mathrm{m/s}\end{align*} what is the magnitude and direction of the force on the electron?
 What would be the magnitude and direction of an electric field to be applied at \begin{align*}P\end{align*} that would counteract the magnetic force on the electron?
Answers to Selected Problems
 No: if \begin{align*}v = 0\end{align*} then \begin{align*}F = 0\end{align*}; yes: \begin{align*}F = qE\end{align*}
 .
 .
 Into the page
 Down the page
 Right
 Both pointing away from north
 .
 .
 \begin{align*}7.6 \;\mathrm{T}\end{align*}, south
 Down the page; \begin{align*}60 \;\mathrm{N}\end{align*}
 To the right, \begin{align*}1.88 \times 10^4 \;\mathrm{N}\end{align*}
 \begin{align*}91.7 \;\mathrm{m/s}\end{align*}
 It should be doubled
 East \begin{align*}1.5 \times 10^4 \;\mathrm{A}\end{align*}
 \begin{align*}0.00016 \;\mathrm{T}\end{align*}; if CCW motion, B is pointed into the ground.

\begin{align*}1.2 \times 105 \;\mathrm{V}\end{align*}, counterclockwise
 \begin{align*}15 \;\mathrm{V}\end{align*}
 Counterclockwise
 \begin{align*}2 \times 10^{5} \;\mathrm{T}\end{align*}
 Into the page
 \begin{align*}2.8 \;\mathrm{N/m}\end{align*}
 CW
 \begin{align*}2.42 \times 10^8 \;\mathrm{m/s}\end{align*}
 \begin{align*}9.69 \times 10^{12} \;\mathrm{N}\end{align*}
 \begin{align*}.0055 \;\mathrm{m}\end{align*}
 E/B
 \begin{align*}8 \times 10^{7} \;\mathrm{T}\end{align*}
 \begin{align*}1.3 \times 10^{6} \;\mathrm{C}\end{align*}
 \begin{align*}0 .8 \;\mathrm{V}\end{align*}
 CCW
 \begin{align*}.064 \;\mathrm{N}\end{align*}
 \begin{align*}.16 \;\mathrm{N/C}\end{align*}
 \begin{align*}.13 \;\mathrm{w}\end{align*}
 \begin{align*}1.11 \times 10^8 \;\mathrm{m/s}\end{align*}
 \begin{align*}9.1 \times 10^{30} \;\mathrm{N} < < 6.4 \times 10^{14} \;\mathrm{N}\end{align*}
 \begin{align*}.00364 \;\mathrm{T}\end{align*}
 \begin{align*}.173 \;\mathrm{m}\end{align*}
 \begin{align*}7.03 \times 1016 \;\mathrm{m/s}^2\end{align*}
 \begin{align*}3.27^\circ\end{align*}

\begin{align*}19.2 \;\mathrm{V}\end{align*}
 \begin{align*}8.39 \times 10^7 \;\mathrm{m/s}\end{align*}
 \begin{align*}2.68 \times 10^{13} \;\mathrm{N}, y\end{align*}
 \begin{align*}2.95 \times 10^17 \;\mathrm{m/s}^2\end{align*}
 \begin{align*}.00838 \;\mathrm{m}\end{align*}
 \begin{align*}1.68 \times 10^6 \;\mathrm{N/C}\end{align*}
 \begin{align*}16,800 \;\mathrm{V}\end{align*}
 \begin{align*}1.2 \times 10^{6} \;\mathrm{T}, +z\end{align*}
 \begin{align*}1.5 \times 10^{17} \;\mathrm{N}, y\end{align*}
 \begin{align*}96 \;\mathrm{N/C}, y\end{align*}
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