Magnetism and Electromagnetism

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AP Physics 2 › Magnetism and Electromagnetism

Questions 1 - 10

A circuit contains a battery and a $20\Omega$ resistor in series. Determine the magnitude of the magnetic field outside of the loop away from the wire.

None of these

Explanation

Using

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

Converting to and plugging in values

$B=\frac{4\pi*10^{-7}*I}{2\pi*.002}$

Determining current:

$B=\frac{4\pi*10^{-7}*3}{2\pi*.002}$

A circuit contains a battery and a $20\Omega$ resistor in series. Determine the magnitude of the magnetic field outside of the loop away from the wire.

None of these

Explanation

Using

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

Converting to and plugging in values

$B=\frac{4\pi*10^{-7}*I}{2\pi*.002}$

Determining current:

$B=\frac{4\pi*10^{-7}*3}{2\pi*.002}$

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

$F_{magnetic}=qvB$

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

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

$1.76*10^{-21}N$

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

$F_{magnetic}=qvB$

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

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

Simple circuit mag field

Which direction will the force be on a proton moving left at location ?

Down, to the bottom of the screen

Up, towards the top of the screen

To the left

To the right

None of these

Explanation

Using the right hand rule for magnetic fields, it is seen that the magnetic field is point into the page at location . Using the right hand rule for force on a moving positively charged particle, it is seen that the force is acting down.

V blv

A conductive rod is moving through a region of magnetic field, as diagrammed above. As a result of its motion, mobile charge carriers in the conductor separate, creating an electric potential across the rod. When, if ever, do the charge carriers cease this motion?

The motion stops when the electric field created by the separated charges creates an equal and opposite force to the magnetic force created by the rod's motion.

The motion does not stop. Mobile charge carriers continue to separate as long as the rod remains in motion.

The motion stops when the electric potential equals the magnetic field strength.

The motion stops when the electric field strength created by the separated charge equals the magnetic field strength.

The motion stops when the magnetic field created by the separated charges equals the external magnetic field.

Explanation

The separated charge creates a potential $\varepsilon=Blv$ . This potential results in an electric field $E=\frac{\varepsilon}{l}=\frac{Blv}{l}=Bv$ When this induced electric field creates a force equal to the magnetic force on the mobile charge carriers, motion stops. Of course, if an electric circuit is created drawing current from the rod, motion will resume to rebuild the field.

V blv

The motion stops when the electric field created by the separated charges creates an equal and opposite force to the magnetic force created by the rod's motion.

The motion does not stop. Mobile charge carriers continue to separate as long as the rod remains in motion.

The motion stops when the electric potential equals the magnetic field strength.

The motion stops when the electric field strength created by the separated charge equals the magnetic field strength.

The motion stops when the magnetic field created by the separated charges equals the external magnetic field.

Explanation

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.

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.

Simple circuit mag field

Which direction will the force be on a proton moving left at location ?

Down, to the bottom of the screen

Up, towards the top of the screen

To the left

To the right

None of these

Explanation

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