Magnetism and Electromagnetism - AP Physics 2

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Question

There is a particle with a charge of $11\mu C$ moving $4.7 \cdot 10^7 \frac{m}{s}$ perpendicular through a magnetic field with a strength of . What is the force on the particle?

Answer

The equation for force on a moving charged particle in a magnetic field is

$F=qv\cdot B$ .

Because the charge is moving perpendicularly through the magnetic field, we don't have to worry about the cross product, and the equation becomes simple multiplication.

$F= (11\mu C)(4.7\cdot 10^7\frac{m}{s})(7T)$

Therefore, the force experienced by the particle is 3619N.

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Question

A conductive rod is moving through a region of magnetic field, directed out of the page as diagrammed above. As a result of its motion, the mobile charge in the rod separates, creating an electric potential across the length of the rod. The length of the rod is 0.12m and the magnitude of the magnetic field is 0.022T. If the rod is moving with velocity $v=17\frac{m}{s}$ , what is the magnitude and direction of the potential from one end of the rod to the other?

Answer

For a conductor moving in a magnetic field, cutting straight across the field lines, a potential is generated $\varepsilon=Blv$ where $\varepsilon$ is the potential or EMF, is the magnetic field strength, and is the conductor's velocity relative to the field.

$\varepsilon=0.022T* 0.12m*17\frac{m}{s} =0.045V$

As the rod moves, the positive charge feels an upward-directed force by the right-hand rule, and the negative a downward force resulting in the top being at a higher potential than the bottom of the rod.

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Question

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?

Answer

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.

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Question

A magnetic rod is inside a coil of wire which is connected to an ammeter. If the rod is stationary, which of the following statements are true?

Answer

Current is induced in wire when the magnetic flux changes. When the magnetic rod is in motion, the flux is changing, so current is induced. If the coil were expanded or contracted with the rod still there, the flux would change and current would be induced. In our case, the rod is stationary and the coil isn't changing shape. Therefore, the flux is not changing, so there is no current being induced. Additionally, there's no reason for the rod to lose its magnetic property.

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Question

Suppose the induced electromotive force of a double loop wire has a magnitude of $1.1\textup{V}$ when the change in magnetic flux is $0.683\textup{T} \textup{m}^2$ . How much time has elapsed for the flux change?

Answer

Write the formula for induced electromotive force.

$\left |\epsilon \right |=N\left | \frac{\Phi_{final}-\Phi_{initial} }{t_{final}-t_{initial} } \right |=N\left | \frac{\Delta \Phi }{\Delta t} \right |$

Since there are two loops, .

$\left |\epsilon \right |= 1.1\textup{V}$

$\Delta \Phi=0.683 \textup{T} \textup{m}^2$

Solve for $\Delta t$ .

$\Delta t=N\left | \frac{\Delta \Phi}{\epsilon } \right | = 2\left | \frac{0.683 \textup{T} \textup{m}^2}{1.1\textup{V}} \right |=1.24 \textup{s}$

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Question

A system consists of two rails, whose resistance is zero, a conductive bar, whose resistance is also zero, and a resistor whose resistance, 0.02 Ohms. The bar remains in contact with the rails as it slides to the right with a velocity $v=22\frac{m}{s}$ . The distance from one rail to the other is 0.07m. The magnetic field, which is directed out of the page, has a magnitude of 0.0015T. What is the magnitude and direction of current in the resistor?

Answer

When a conductor moves through a magnetic field in such a way that it cuts through magnetic field lines, the mobile charge carriers separate due to the magnetic force on them, creating a potential . Since there is no resistance anywhere else in the circuit, all of this potential is lost in the resistor, so we can apply Ohm's law:

$I=\frac{V}{R}$

$I=\frac{Blv}{R}$

$I=\frac{0.0015T0.07m22\frac{m}{s}}{0.02\Omega }=0.1155A$

Because the positive charge in the rod feels an upward force due to the right-hand rule, the top of the rod has a greater potential than the bottom, and current flows counterclockwise around the circuit, resulting in a downward direction in the resistor.

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Question

A $150\Omega$ resistor is connected to a coil with turns and a cross sectional area of $0.02m^{2}$ . A magnet is lowered as shown in the figure. The magnetic field increases from to in . Find the magnitude of the current going through the resistor.

Answer

There is a change in flux produced by the changing magnetic field which is given by

$\Delta \Phi=A \Delta B \cos\left ( \theta\right )$

where is the cross sectional area of the coil, $\Delta B$ is the change in magnetic field, and $\theta$ is the angle of the field lines relative to the normal of the cross section of the coil.

In this case the magnet is perpendicular to the cross section of the coil and so

$\theta=0 \rightarrow \cos\left ( 0\right )=1$

$\Delta \Phi=A \Delta B$

The change in magnetic field is just the final given value minus the intial value. Faraday's Law says that an emf will be generated by a change in flux,

$\varepsilon=N\left \left | \frac{\Delta \Phi}{\Delta t}\right |$

where is the number of turns in the coil. Plugging in the change in flux gives

$\varepsilon=N \left | A\frac{\Delta B}{\Delta t}\right |$

The change in time is just since we can start our clock at zero. The current can be found using Ohm's Law where

$I=\frac{\varepsilon}{R}=\frac{N A \Delta B}{R \Delta t} =0.01A$

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Question

Suppose that a proton moves perpendicularly through a magnetic field at a speed of $5* 10^{3} \frac{m}{s}$ . If this proton experiences a magnetic force of $1.2* 10^{-18}N$ , what is the strength of the magnetic field?

$e=+1.6* 10^{-19}C$ .

Answer

To solve this question, we need to relate the speed and charge of the particle with the magnetic force it experiences in order to solve for the magnetic field strength. Thus, we'll need to use the following equation:

$F_{B}=qvBsin(\theta )$

Also, we are told that the particle is moving perpendicularly to the magnetic field.

$\theta=90^{o}$

$\sin \left ( 90^{o}\right )=1$

$F_{B}=qvB$

Rearrange to solve for the magnetic field, then plug in known values and solve.

$B=\frac{F_{B}}{qv}$

$B=\frac{1.210^{-18} N}{\left ( 1.6 10^{-19}C\right )(5* 10^{3}\frac{m}{s})}$

$B=1.5* 10^{-3}\frac{N\cdot s}{C\cdot m}$

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Question

Suppose that a positively charged particle with charge moves in a circular path of radius in a constant magnetic field of strength . If the magnetic field strength is doubled to , what effect does this have on the radius of the circular path that this charge takes?

Answer

To answer this question, we need to realize that the particle is moving in a circular path because of some sort of centripetal force. Since the charge is moving while within a constant magnetic field, we can conclude that it is the magnetic force that is responsible for the centripetal force that keeps this charge moving in a circle. Thus, we need to relate the centripetal force to the magnetic force.

$F_{c}=F_{B}$

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

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

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

The above equation shows us that the radius of the circular path is directy proportional to the mass and velocity of the particle, and inversely proportional to the charge of the particle and the magnetic field strength. Thus, if the value of the magnetic field is doubled, the above equation predicts that the value of the radius would be cut in half.

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Question

loops of current carrying wire form a solenoid of length $.600\textup{ m}$ that carries $3\textup{ A}$ and have radius $45\textup{ cm}$ . Determine the magnetic field at the center of the solenoid.

Answer

Using:

$B=\mu_0\frac{NI}{L}$

Where:

is the magnetic field

is the number of coils

is the current in the solenoid

is the length of the solenoid

$\mu_0$ is $1.25710^{-6}\frac{N}{A^2}$

Plugging in values:

$B=1.25710^{-6}\frac{6003}{.600}$

$B=3.77*10^{-3}T$

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Question

There is a loop with a radius of $0.03\textup{ m}$ and a current of $0.250\textup{ A}$ . Determine the magnitude of the magnetic field at the center of the loop.

Answer

Using the Biot-Savart law:

$B=\frac{\mu_0}{4\pi}\frac{2\pi R^2 I}{(z^2+R^2)^{(3/2)}}$

Where is the radius of the loop

is the current

is the distance from the center of the loop

$\mu_0=4\pi10^{-7}\frac{T\cdot m}{A}$

Plugging in values:

$B=\frac{4\pi10^{-7}}{4\pi}\frac{2\pi.03^2.250}{(0^2+.03^2)^{(3/2)}}$

$B=5.24*10^{-6}\textup{ T}$

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Question

A circular circuit is powered by a $9\textup{ V}$ battery. How will the magnetic field change if the battery is removed and placed in the opposite direction?

Answer

Reversing the battery will reverse the direction of the current. Using the right hand rule, it can be seen that this will also reverse the direction of the magnetic field. Since the magnitude of the current stays the same, the magnitude of the magnetic field will as well.

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Question

A circular circuit is powered by a $9\textup{ V}$ battery. How will the magnetic field change if a second $9\textup{ V}$ battery is added in the same direction as the first?

Answer

Based on the Biot-Savart law:

$B=\frac{\mu_0}{4\pi}*\frac{2\pi R^2 I}{(z^2+R^2)^{(3/2)}}$

Doubling the voltage will double the current, which will double the magnetic field. The direction will stay the same.

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Question

If the north end of a magnetic points towards the geographic north pole, that means that the geographic north pole is a magnetic __________ pole.

Answer

Magnets will align themselves with the surrounding magnetic field. Thus, if the north pole of a magnet is pointing north, the direction of the magnetic field must be pointing north. Magnetic fields point towards magnetic south poles, so the geographic north pole is actually a magnetic south pole.

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Question

An infinitely long wire carries a current of determine the magnitude of the magnetic field away.

Answer

Magnetic field of an infinitely long wire:

$B=\frac{\mu_0I}{2\pi r}$

Where $\mu_0=4\pi10^{-7}\frac{Tm}{A}$

Plugging in values:

$B=\frac{410^{-7}3}{2\pi .015}$

$B=1.27*10^{-5}T$

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Question

Radius of moon:

A loop of current carrying wire runs along the equator of the moon. Determine the magnetic field at the center of the loop if are traveling through it

Answer

Using

$B=\frac{\mu_0 I}{2R}$

Plugging in values

$B=\frac{4\pi10^{-7} 110^6}{21.7410^6}$

$B=3.6110^{-7}T$

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Question

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.

Answer

Using

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

Converting to and plugging in values

$B=\frac{4\pi10^{-7}I}{2\pi.002}$

Determining current:

$B=\frac{4\pi10^{-7}3}{2\pi.002}$

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Question

How strong would a magnetic field need to be in order to make a particle with a mass of $4\cdot10^{-11}:kg$ and a charge of move in a circular path with a speed of $400:\frac{m}{s}$ and a radius of ?

Answer

For this question, we are being asked to determine the magnetic field necessary to make a particle of a given mass and charge to move in a circular path with a given speed and radius.

To begin with, we can realize that the particle will be moving in a circular path. Thus, there is going to be a centripetal force associated with this circular motion. Moreover, because we know the particle will be present in a magnetic field, we can infer that the magnetic force will be the source of the centripetal force. Thus, we can start by writing out the expression for each of these forces, and then setting them equal to one another.

$F_{B}=qvB$

$F_{c}=\frac{mv^{2}}{r}$

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

Rearranging the above expression to isolate the term for magnetic field, we arrive at the following expression.

$B=\frac{mv^{2}}{qvr}=\frac{mv}{qr}$

Now, we can plug in the values given to us in the question stem to solve for the magnetic field strength.

$B=\frac{(4\cdot10^{-11}:kg)(400:\frac{m}{s})}{(8\cdot 10^{-9}:C)(0.5:m)}$

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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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