The equation for finding the force on a moving charged particle in a magnetic field is as follows:

\(\displaystyle F=qv \times B\)

Here, \(\displaystyle F\) is the force in Newtons, \(\displaystyle q\) is the charge in Coulombs, \(\displaystyle v\) is the velocity in \(\displaystyle \frac{m}{s}\), and \(\displaystyle B\) is the magnetic field strength in Teslas.

Another way to write the equation without the cross-product is as follows:

\(\displaystyle F=qvBsin\theta\)

Here, \(\displaystyle \theta\) is the angle between the particles velocity and the magnetic field.

For our problem, because theta is \(\displaystyle 90^{\circ}\), \(\displaystyle sin\theta\) evaluates to 1, so we just need to perform multiplication.

\(\displaystyle F&=qvB\)

\(\displaystyle F=(2\cdot 10^{-6}C)(3\cdot 10^{6}\frac{m}{s})(0.05T)\)

\(\displaystyle F= 0.3N\)

Therefore, the force on the particle is 0.3N.

The equation for the force on a current carrying wire in a magnetic field is as follows:

\(\displaystyle F=IL\cdot B=ILBsin\theta\)

\(\displaystyle F\) is the force in Newtons, \(\displaystyle I\) is the current in amperes, \(\displaystyle B\) is the magnetic field strength in Teslas, and \(\displaystyle \theta\) is the angle from parallel to the magnetic field.

Because our wire is not fully perpendicular to the magnetic field, it does not experience the full possible force. Instead, it experiences \(\displaystyle sin(36^{\circ})\) times the maximum force value.

\(\displaystyle F=ILBsin\theta\)

\(\displaystyle F=(2A)(0.08m)(6T)sin(36^o)\)

\(\displaystyle F\approx 0.564N\)

A charged particle moving through a perpendicular magnetic field feels a Lorentz force equal to the formula:

\(\displaystyle F_{B} = qvB\)

\(\displaystyle q\) is the charge, \(\displaystyle v\) is the particle speed, and \(\displaystyle B\) is the magnetic field strength. This force is always directed perpendicular to the particle’s direction of travel at that moment, and thus acts as a centripetal force. This force is also given by the equation:

\(\displaystyle F_c=\frac{mv^2}{r}\)

We can set these two equations equal to one another, allowing us to solve for the radius of the arc.

\(\displaystyle qvB = \frac{mv^{2}}{r}\)

\(\displaystyle r = \frac{mv}{qB}\)

Once the particle travels through a semicircle, it is laterally one diameter in distance from where is started (i.e. twice the radius of the circle).

\(\displaystyle r = \frac{mv}{qB} = \frac{(1.67 \times 10^{-27} kg)(1.0 \times 10^{7} \frac{m}{s})}{(1.6 \times 10^{-19} C)(3 T)}= 34.8 mm\)

Twice this value is the lateral offset of its crash point from the entrance:

\(\displaystyle 2(34.8mm)=69.6mm\)

To find the force experienced by the charge, we use this equation:

\(\displaystyle F=qv\times B\)

Because the charge is moving at an angle from perpendicular, we need to take the cross product into account. 

\(\displaystyle F=qvB\sin\theta\)

Theta is the angle from perpendicular, which is \(\displaystyle 48^{\text o}\). Plug in known values and solve.

\(\displaystyle F= (15)(6.2\cdot10^7)(7)(\sin{48^{\text o})\)

\(\displaystyle F= 4837.87N\)

The equation for force on a charge moving through a magnetic field is:

\(\displaystyle F=qv\cdot B\)

Because the velocity is perpendicular to the field, the cross product doesn't matter, and we can do simple multiplication.

\(\displaystyle F&=qvB\)

\(\displaystyle F=(8\cdot 10^{-6})(2.5\cdot 10^7)(10)\)

\(\displaystyle F= 2000N\)

Therefore, the force on the charge is \(\displaystyle 2000 N\)

The equation for force on a charge moving through a magnetic field is:

\(\displaystyle F=qv\cdot B\).

The cross product is:

\(\displaystyle F=qvB\sin\theta\)

Above, \(\displaystyle \theta\) is the degree from parallel the charge is moving. The charge is moving at \(\displaystyle 43^{\circ}\) from parallel, so the equation, once we plug in our numbers, is:

\(\displaystyle F=qvB\sin\theta\)

\(\displaystyle F=(5\cdot 10^{-6})(3.1\cdot 10^7)(35)(\sin(43^{\circ}))\)

\(\displaystyle F= 3699.84N\)

The force on the charge is about \(\displaystyle 3700 N\).

The force experienced by a charged particle in a magnetic field is 

\(\displaystyle F=qv\cdot B\)

This means that when a charge particle is moving perpendicular to the field, due to the cross-product, it experiences the most amount of force (because \(\displaystyle v\cdot B\) is equal to \(\displaystyle vB\sin\theta\), and theta equals \(\displaystyle 90^{\circ}\) when it's perpendicular). This means that the charged particle will experience no force due to the magnetic field when it's parallel. We know there will be no force on the particle, and we also know that uncharged particles experience no force in magnetic fields, but we can't say for certain the particle has no net charge, only being told that it's moving parallel to the field.

The rod acts as a battery due to its motion in the magnetic field: \(\displaystyle V=Blv\). Because there is a closed circuit, this results in current flow:

\(\displaystyle I=\frac{V}{R}=\frac{Blv}{R}=\frac{0.035T*0.087m*15\frac{m}{s}}{0.0055\Omega }=8.3A\)

In this simple circuit, the current is the same everywhere, so the same current flows through the rod. Because of this current, the rod feels a force:

\(\displaystyle F=BIl=0.035T*8.3A*0.087m=0.025N\)

By the right-hand rule, this magnetic force is directed to the left, so the external force must be directed to the right. It is interesting to note that the power dissipated in the resistor:

\(\displaystyle P=I^{2}R=8.3^2* 0.0055=0.39W\) is the same as the power provided by the outside force:

\(\displaystyle P=Fv=0.025*15=0.39W\)

The universe conspires to conserve energy.

This question is presenting us with a scenario in which a charged particle is traveling with a certain velocity through a magnetic field. In this situation, we're being asked to determine what the force experienced by this particle is.

To solve this question, we'll need to determine what kind of force this particle is likely to experience. Since we're told that the particle is traveling in a magnetic field, it would make sense that this particle is going to be affected by a magnetic force. Thus, we'll need to use the equation for magnetic force.

\(\displaystyle F=qvBsin\Theta\)

Moreover, since we're told in the question stem that this particle is traveling perpendicularly to the magnetic field, we know that \(\displaystyle \Theta=90^{o}\) and thus \(\displaystyle sin\Theta=1\). This helps reduce the equation down.

\(\displaystyle F=qvB\)

Now, all we need to do is plug in the values given to us in order to calculate the resulting force.

\(\displaystyle F=(10 C)(2 \frac{m}{s})(5 T)\)

\(\displaystyle F=100 N\)

Charged particles only experience a magnetic force when some component of their velocity is perpendicular to the magnetic field. Here, the velocity is parallel to the magnetic field so the particle does not experience a force.