Magnetic Force

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AP Physics 2 › Magnetic Force

Questions 1 - 10

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.

$1.76*10^{-21}N$

None of these

$1.95*10^{-21}N$

$4.05*10^{-21}N$

$1.22*10^{-21}N$

Explanation

Finding the magnetic field at the location of the proton.

$B=\frac{\mu_0*I}{2*\pi*r}$

Converting to and plugging in values

$B=\frac{4*\pi*10^{-7}*7.5}{2*\pi*.009}$

$B=1.67*10^{-4}T$

Using

$F_{magnetic}=qvB$

$F_{magnetic}=1.6*10^{-19}*66*1.67*10^{-4}$

$F_{magnetic}=1.76*10^{-21}N$

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.

None of these

$3*10^{-20}N$

$5*10^{-18}N$

$1.6*10^{-9}N$

$7*10^{-7}N$

Explanation

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.

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?

There is no net force on the charge

Explanation

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.

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

A proton traveling at $1*10^7\frac{m}{s}$ 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 $1.67*10^{-27}kg$ and its electric charge is $1.6*10^{-19}C$ .

Explanation

Mass_spec_fig

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:

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.

None of these

$5*10^{-19}N$

$1.7*10^{-19}N$

$9.8*10^{-9}N$

$1.8*10^{-25}N$

Explanation

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.

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?

There is no force on the particle

Explanation

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.

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

Zero

$Fsin\theta$

Explanation

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.

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?

There is no force on the charge

Explanation

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

$B=\frac{\mu_0}{4\pi}\frac{q*v}{r^2}$

$F_{mag}=q*v*B$

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

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?

$3*10^{8}\frac{m}{s}$

None of these

$9*10^{16}\frac{m}{s}$

$306\frac{m}{s}$

$1.73*10^{4}\frac{m}{s}$

Explanation

Using

$B=\frac{\mu_0}{4\pi}\frac{q_1*v_1}{r^2}$

$F_{mag}=q_2*v_2*B$

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

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:

$q_2*v_2*\frac{\mu_0}{4\pi}\frac{q_1*v_1}{r^2}=k\frac{q_1q_2}{r^2}$

$q_2*v_1*\frac{\mu_0}{4\pi}\frac{q_1*v_1}{r^2}=k\frac{q_1q_2}{r^2}$

Solving for

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

Plugging in values:

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

$v_1=3*10^{8}\frac{m}{s}$

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?

Explanation

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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