To explain this problem and how is force affected by the velocity vector if the charged object is in a magnetic field, we remember the term for the magnitude of magnetic force for this case:

$$
F=|q| v B sin theta
$$

Where $theta$ is exactly the angle between the velocity and magnetic field vector. This can also be represented by the cross product:

$$
vec{F}=q vec{v} times vec{B}
$$

From this, we can conclude, as if the velocity vector and magnetic field vector are perpendiculars, the force will have maximum value, as $sin 90^{circ}=1$. But if they are parallel it will give $sin 0^{circ}=0$, then the force will be zero.

We look here to which two vectors is the magnetic force perpendicular to. Since the magnetic force can be expressed as:

$$
F=|q| v B sin theta
$$

where $theta$ is the angle between velocity and the magnetic field vector, we can see that $textbf{velocity}$ and $textbf{magnetic field}$ are perpendicular to the magnetic force vector.

The current-carrying wire is going to experience a magnetic force when it is in an external magnetic field. It is important $textbf{to emphasize}$ that a current-carrying wire inside because of moving charges is producing its own magnetic field, which direction can be calculated by the right-hand rule and the following relation:

$$
F=|q| v B sin theta
$$

But it is $textbf{not this field}$ that is meant to produce a magnetic force. So magnetic field produced by a current-carrying wire is not producing a total magnetic force on the wire itself (the total force due to the wire produced magnetic field is zero).

What is meant here that if the wire is inside the external magnetic field it is going to experience the magnetic force. Because the external magnetic field acts on each charge that is moving in a wire (due to the current), the total force which is the sum of all the magnetic forces experienced by the individual charges moving within it and can be calculated with the relation:

$$
F=I L B sin theta
$$

The direction of this magnetic force can be determined by the $textbf{right – hand}$ rule, where we simply point our fingers in the direction of the current flow and bend our fingers in the direction of the magnetic field. Then, the thumb is going to show the direction of the magnetic force.

The connection between moving charges and the magnetic field is that if we have a charge that is moving we collectively watch it as a current. This current is going to produce a magnetic field around itself, which can be determined by the $textbf{right-hand rule}$. For example, this produces a magnetic field for a straight wire that looks like:

$$
B=frac{mu_{0} I}{2 pi r}
$$

We can see that the magnetic field will be stronger as the current is higher in magnitude, while it is decreasing radially.

When it comes to the connection between the magnetic field and the moving charges we know that in the presence of the external magnetic field a moving charge experiences a force, that will tend to make it move in the circular motion. Force can be calculated with the relation:

$$
F=|q| v B sin theta
$$

Even if the current-carrying wire is in the presence of the external magnetic field, the wire is going to experience a force that can be calculated with:

$$
F=I L B sin theta
$$

And the direction can be calculated with the right-hand rule, as well. Since the definition is coming from the cross product.

For this problem, we know that electron has a constant velocity and it is moving in the region of the $textbf{zero electric fields}$. Regarding the magnetic field, we know that since the electron is charged particle it would make it go in the circle, while inside the magnetic field.

Also, the force on the electron could be calculated as:

$$
F=|q| v B sin theta
$$

we see that, if electron velocity would be parallel or antiparallel to the vector of the magnetic field, then it would be no force, even the field is present because $theta$ is the angle between velocity and magnetic field vector. By this argument, we can not surely determine the answer.

If we read the problem statement again, we know that the velocity is $textbf{zero}$. Since velocity is a vector it has magnitude and direction. When an electron would be inside the magnetic field it would stay inside the field with a constant magnitude of velocity. But, since the magnetic field would make an electron go in a circular motion, we know that there is $textbf{no present magnetic field}$, as there a $textbf{constant}$ velocity and there is no circular motion, so the magnetic field is indeed zero.

The force on the electron will be towards west. From the right thumb rule we can find out that the $vec{V}times vec{B}$ will be towards the east. But since the electron is negatively charged, the direction will be towards the west.

The magnitude of the force exerted on an electron that is moving perpendicular to a magnetic field of $B=0.12 :T$ with speed $v = 2.8 times 10^4:m/s$ is given by

$F = |e| v B = 1.6 times 10^{-19} times 2.8 times 10^4 times 0.12 = 0.54 times 10^{-15}:N$

The magnitude of the force exerted on a wire of length $L$ carrying current $I$ lying at an angle $theta$ relative to a magnetic field of magnitude $B$ is given by

$F = IBL sin theta$

Given data are

$L=1.5:m$, $I =5.3:A$, $B=0.68:T$ and $theta = 45^o$

So, $F = 5.3 times 0.68 times 1.5 times sin 45^o = dfrac{5.3 times 0.68 times 1.5}{sqrt{2}} =3.8:N$

Kinetic energy of the proton is given as $4.9 times 10^{-16}:J$. That is,

$dfrac{1}{2} m_pv^2 = 4.9 times 10^{-16}:J$

So the speed of the proton is

$v = sqrt{dfrac{2 times 4.9 times 10^{-16}}{m_p}}$

where $m_p = 1.673 times 10^{-27}:kg$ is the mass of the proton.

So, $v = sqrt{dfrac{2 times 4.9 times 10^{-16}}{1.673 times 10^{-27}}} = 7.65 times 10^5:m/s$

If this proton moves perpendicular to a magnetic field of $0.26:T$, then the radius of its circular orbit is

$r = dfrac{m_p v}{e B} = dfrac{1.673 times 10^{-27} times 7.65 times 10^5}{1.6 times 10^{-19} times 0.26} = 30.7 times 10^{-3}:m = 30.7:mm$

The magnitude of the force exerted, on a rod of length $L$ carrying current $I$, by a magnetic field of magnitude $B$ is given by

$F = IBL sin theta$

where $theta$ is relative angle between orientation of the rod and the direction of the field.

From Newton’s second law of motion, we have

$mg = IBL sin theta$

where $mg$ is the weight of the rod. So the magnetic field required to levitate the rod is given by

$B = dfrac{mg}{IL sin theta}$

This field is minimum only when $sin theta = 1$ or $theta = 90^o$.

So field is perpendicular to the rod.

$B = dfrac{mg}{IL} = dfrac{0.17 times 9.8}{11 times 0.45} = 0.34:T$

So the magnitude of the required magnetic field is $0.34:T$.

Current flows in the positive $x$ direction. To lift the rod, the magnetic force has to act in the positive $z$ direction, that is, opposite to the weight of the rod. So magnetic field points in the positive $y$ direction.