Magnetic Force - AP Physics 2

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Question

A proton traveling at $110^7\frac{m}{s}$ in a horizontal plane passes through an opening into a mass spectrometer with a uniform magnetic field directed upward. The particle then moves in a circular path through $180^{\circ}}$ and crashes into the wall of the spectrometer adjacent to the entrance opening. How far down from the entrance is the proton when it crashes into the wall?

The proton’s mass is $1.6710^{-27}kg$ and its electric charge is $1.6*10^{-19}C$ .

Answer

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

$F_{B} = qvB$

is the charge, is the particle speed, and 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:

$F_c=\frac{mv^2}{r}$

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

$qvB = \frac{mv^{2}}{r}$

$r = \frac{mv}{qB}$

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).

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

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

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Question

A particle with a charge of $2\mu C$ is moving at $3 \cdot 10^{6} \frac{m}{s}$ perpendicularly through a magnetic field with a strength of . What is the magnitude of the force on the particle?

Answer

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

$F=qv \times B$

Here, is the force in Newtons, is the charge in Coulombs, is the velocity in $\frac{m}{s}$ , and is the magnetic field strength in Teslas.

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

$F=qvBsin\theta$

Here, $\theta$ is the angle between the particles velocity and the magnetic field.

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

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

Therefore, the force on the particle is 0.3N.

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Question

An wire with a current of is oriented $36^{\circ}$ from parallel to a magnetic field with a strength of . What is the force on the wire?

Answer

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

$F=IL\cdot B=ILBsin\theta$

is the force in Newtons, is the current in amperes, is the magnetic field strength in Teslas, and $\theta$ 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 $sin(36^{\circ})$ times the maximum force value.

$F=ILBsin\theta$

$F\approx 0.564N$

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Question

An $8\mu C$ charge is moving through a magnetic field at a speed of $2.5\cdot 10^7 \frac{m}{s}$ perpendicular to the direction of the field. What is the force on the charge?

Answer

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

$F=qv\cdot B$

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

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

Therefore, the force on the charge is

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Question

A $5\mu C$ charge is moving through a magnetic field at a speed of $3.1\cdot 10^7\frac{m}{s}$ $43^{\text o}$ from parallel to the magnetic field. What is the force on the charge?

Answer

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

$F=qv\cdot B$ .

The cross product is:

$F=qvB\sin\theta$

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

$F=qvB\sin\theta$

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

The force on the charge is about .

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Question

A particle is moving parallel to a uniform magnetic field. Which of the following statements are true?

Answer

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

$F=qv\cdot B$

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 $v\cdot B$ is equal to $vB\sin\theta$ , and theta equals $90^{\circ}$ 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.

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Question

What is the force experienced by a $15\mu C$ charge moving at $6.2\cdot 10^7\frac{m}{s}$ through a magnetic field with strength at $48^{\text o}$ from perpendicular to the field?

Answer

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

$F=qv\times B$

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

$F=qvB\sin\theta$

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

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

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Question

A rail system is formed in a magnetic field directed out of the page as diagrammed above. The rod remains in contact with the rails with zero friction as it moves to the right at a constant velocity of $v=15\frac{m}{s}$ due to an external force. The distance from one rail to the other is 0.087m. The rails and the rod have no resistance, but the resistor has a resistance of 0.0055 Ohms. The magnetic field has a magnitude of 0.035T. What is the magnitude and direction of the external force required to keep the rod moving at a constant velocity?

Answer

The rod acts as a battery due to its motion in the magnetic field: . Because there is a closed circuit, this results in current flow:

$I=\frac{V}{R}=\frac{Blv}{R}=\frac{0.035T0.087m15\frac{m}{s}}{0.0055\Omega }=8.3A$

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:

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:

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

The universe conspires to conserve energy.

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Question

A charged particle, Q, is traveling along a magnetic field, B, with speed v. What is the magnitude of the force the particle experiences?

Answer

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.

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Question

If a 10C charged particle is traveling perpendicularly to a magnetic field of 5T at a speed of $2: \frac{m}{s}$ , what is the force experienced by this charged particle?

Answer

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.

$F=qvBsin\Theta$

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

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

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

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Question

Mass of electron: $9.110^{-31}kg$

An electron enters a magnetic field at velocity $310^4\frac{m}{s}$ and experiences a force of $1*10^{-27}N$ . Determine the magnetic field.

Answer

Use the magnetic force equation:

$|\overrightarrow{F_M}|=q|\overrightarrow{v}\times \overrightarrow{B}|$

Plug in known values and solve for $\overrightarrow{B}$

$|110^{-27}N|={1.610^{-19}}|310^4\times \overrightarrow{B}|$

$|\overrightarrow{B}|=2.0810^{-13}T$

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Question

What direction of force would a negative charge at location moving left experience due to the magnetic field?

Answer

Using the right hand rule for a current carrying wire shows that the magnetic field is pointing out of the screen. Using the right hand rule for magnetic force on a negatively charged particle shows the force acting downward.

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Question

Which of the following conditions is not needed in order for a particle to experience a magnetic force?

Answer

In this question, we're asked to determine which answer choice falsely represents a necessary condition for a magnetic force to act on a particle.

To answer this, it is useful to look at the equation for magnetic force.

$F=qvBsin(\Theta )$

What this equation shows is that the magnetic force on a particle is dependent upon that particle's charge, its velocity, and on the strength of the magnetic field. Already we can rule out a few of the answer choices.

Additionally, this equation also shows that $\Theta$ cannot be equal to zero. What this means is that the particle's direction of motion cannot be along the magnetic field lines. In other words, the particle cannot be travelling in a direction that is parallel or antiparallel with the magnetic field.

Lastly, we can see that no where in the above equation is there a variable for the size of the particle. Thus, size is not a requirement for a particle to experience a magnetic force.

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Question

$B=\frac{\mu_0}{4\pi}\frac{qv}{r^2}$

$F_{mag}=qv*B$

$F_{elec}=k\frac{q_1q_2}{r^2}$

Two electrons are traveling parallel to each other apart. The distance between them is perpendicular to their motion. One of them is on a track that prevents it from moving side to side. The other one is able to movie in all directions. At what velocity would the magnetic attractive force equal the repulsive electric force?

Answer

Using

$B=\frac{\mu_0}{4\pi}\frac{q_1v_1}{r^2}$

$F_{mag}=q_2v_2B$

$F_{elec}=k\frac{q_1q_2}{r^2}$

Where

is the charge limited to traveling in a single dimension

is the free charge

is the distance between the charges

is the velocity of the first charge

is the velocity of the second charge

is the value of the first charge

is the value of the second charge

and are equivalent as the charges are running parallel to each other

$F_{elec}=F_{mag}$

Combining equations:

$q_2v_2\frac{\mu_0}{4\pi}\frac{q_1v_1}{r^2}=k\frac{q_1q_2}{r^2}$

$q_2v_1\frac{\mu_0}{4\pi}\frac{q_1v_1}{r^2}=k\frac{q_1q_2}{r^2}$

Solving for

$v_1=\sqrt{k\frac{4\pi}{\mu_0}}$

Plugging in values:

$v_1=\sqrt{910^9\frac{4\pi}{4\pi 10^{-7}}}$

$v_1=310^{8}\frac{m}{s}$

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Question

A proton is traveling parallel to a wire in the same direction as the conventional current. The proton is traveling at $66\frac{m}{s}$ . The current in the wire is . The proton and the wire are apart. Determine the magnetic force on the proton.

Answer

Finding the magnetic field at the location of the proton.

$B=\frac{\mu_0I}{2\pir}$

Converting to and plugging in values

$B=\frac{4\pi10^{-7}7.5}{2\pi.009}$

$B=1.6710^{-4}T$

Using

$F_{magnetic}=qvB$

$F_{magnetic}=1.610^{-19}661.6710^{-4}$

$F_{magnetic}=1.7610^{-21}N$

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Question

A scientist wishes for an electron to move in a circle of radius at $10\frac{rot}{s}$ . Determine the necessary magnetic field to make this happen.

Answer

The centripetal force will need to be equal to the magnetic force.

$qvB=m\frac{v^2}{r}$

Solving for

$B=\frac{mv}{qr}$

Converting:

$\frac{10 rot}{s}\frac{2\pi.5}{1 rot}=31.4\frac{m}{s}$

Plugging in values:

$B=\frac{9.110^{-31}31.4}{1.610^{-19}.5}$

$B=3.5810^{-10}T$

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Question

A circuit contains a battery and a $20\Omega$ resistor in series. Determine the magnitude of the magnetic force outside of the loop away from the wire on an electron that is stationary.

Answer

Since the electron is stationary, there will be no magnetic force, as magnetic force requires the particle to be both charged and to be moving.

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Question

A circuit contains a battery and a $20\Omega$ resistor in series. Determine the magnitude of the magnetic force outside of the loop away from the wire on an electron that is moving parallel to the magnetic field.

Answer

Magnetic fields do not affect charges that are moving parallel to them.

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

Thus, the magnetic force will be at a maximum when moving perpendicular to the field, and at zero when moving parallel to the field.

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