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AP Physics 2 : Magnetic Force

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Example Question #11 : Magnetic Force

Mass of electron: $9.1*10^{-31}kg$

An electron enters a magnetic field at velocity $3*10^4\frac{m}{s}$ and experiences a force of $1*10^{-27}N$ . Determine the magnetic field.

Possible Answers:

$|\overrightarrow{B}|=4.65*10^{-13}T$

$|\overrightarrow{B}|=2.08*10^{-13}T$

$|\overrightarrow{B}|=3.96*10^{-13}T$

$|\overrightarrow{B}|=1.32*10^{-13}T$

$|\overrightarrow{B}|=7.65*10^{-13}T$

Correct answer:

$|\overrightarrow{B}|=2.08*10^{-13}T$

Explanation:

Use the magnetic force equation:

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

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

$|1*10^{-27}N|={1.6*10^{-19}}|3*10^4\times \overrightarrow{B}|$

$|\overrightarrow{B}|=2.08*10^{-13}T$

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Example Question #671 : Ap Physics 2

Simple circuit mag field

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

Possible Answers:

Down, towards the bottom of the screen

To the left

Out of the screen

Into the screen

Correct answer:

Down, towards the bottom of the screen

Explanation:

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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Example Question #27 : Electricity And Magnetism

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

Possible Answers:

The particle must be moving in a direction that is neither parallel or antiparallel to the direction of an external magnetic field

All of these conditions are needed for a magnetic force to act on a particle

The particle must have a charge

There must be a current

The particle must be of a certain size

Correct answer:

The particle must be of a certain size

Explanation:

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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Example Question #11 : Magnetic Force

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

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

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

Two electrons are traveling parallel to each other 1mm 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?

Possible Answers:

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

$306\frac{m}{s}$

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

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

None of these

Correct answer:

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

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

q_1 is the charge limited to traveling in a single dimension

q_2 is the free charge

is the distance between the charges

v_1 is the velocity of the first charge

v_2 is the velocity of the second charge

q_1 is the value of the first charge

q_2 is the value of the second charge

v_1 and v_2 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

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

Plugging in values:

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

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

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Example Question #29 : Electricity And Magnetism

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 7.5A . The proton and the wire are 9.0mm apart. Determine the magnetic force on the proton.

Possible Answers:

None of these

$4.05*10^{-21}N$

$1.95*10^{-21}N$

$1.76*10^{-21}N$

$1.22*10^{-21}N$

Correct answer:

$1.76*10^{-21}N$

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

$F_{magnetic}=qvB$

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

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

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Example Question #30 : Electricity And Magnetism

A circuit contains a 60V battery and a $20\Omega$ resistor in series. Determine the magnitude of the magnetic force outside of the loop 2mm away from the wire on an electron that is stationary.

Possible Answers:

$7*10^{-7}N$

$3*10^{-20}N$

None of these

$1.6*10^{-9}N$

$5*10^{-18}N$

Correct answer:

None of these

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.

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Example Question #11 : Magnetic Force

A circuit contains a 60V battery and a $20\Omega$ resistor in series. Determine the magnitude of the magnetic force outside of the loop 2mm away from the wire on an electron that is moving parallel to the magnetic field.

Possible Answers:

$1.8*10^{-25}N$

$9.8*10^{-9}N$

None of these

$5*10^{-19}N$

$1.7*10^{-19}N$

Correct answer:

None of these

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.

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Example Question #12 : Magnetic Force

A scientist wishes for an electron to move in a circle of radius .5m at $10\frac{rot}{s}$ . Determine the necessary magnetic field to make this happen.

Possible Answers:

$3.58*10^{-10}T$

None of these

$8.44*10^{-10}T$

$3.95*10^{-10}T$

$6.66*10^{-10}T$

Correct answer:

$3.58*10^{-10}T$

Explanation:

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:

$B=\frac{9.1*10^{-31}*31.4}{1.6*10^{-19}*.5}$

$B=3.58*10^{-10}T$

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AP Physics 2 : Magnetic Force

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #11 : Magnetic Force

Example Question #671 : Ap Physics 2

Example Question #27 : Electricity And Magnetism

Example Question #11 : Magnetic Force

Example Question #29 : Electricity And Magnetism

Example Question #30 : Electricity And Magnetism

Example Question #11 : Magnetic Force

Example Question #12 : Magnetic Force

All AP Physics 2 Resources

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