Question 1

Which is NOT one of the differences between a cyclotron and a synchrotron?

Accepted Answer

The statement "In general, synchrotrons are much smaller than cyclotrons" is incorrect, as synchrotrons are typically much larger than cyclotrons due to their need for multiple magnets and beam lines.

Question 2

A loop of wire carrying a current of 2.0 A is in the shape of a right triangle with two equal sides, each 15 cm long.A 0.7 T uniform magnetic field is in the plane of the triangle and is perpendicular to the hypotenuse.The resultant magnetic force on the two equal sides has a magnitude of:

Accepted Answer

The answer of A loop of wire carrying a current...

Question 3

Electrons (mass m, charge -e)are accelerated from rest through a potential difference V and are then deflected by a magnetic field  that is perpendicular to their velocity.The radius of the resulting electron trajectory is:

Accepted Answer

The radius of an electron trajectory in a magnetic field is given by the formula:

r = mv/eB

where 
m = mass of electron 
v = velocity of electron 
e = charge on electron 
B = magnetic field

The velocity of the electron after being accelerated through a potential difference V is given by the formula:

v = sqrt(2eV/m)

Substituting this value of v in the above formula for r, we get:

r = m(sqrt(2eV/m))/eB

Simplifying:

r = sqrt(2mV)/Be

Since the magnetic field is perpendicular to the velocity, we can use the right-hand rule to determine that the force on the electron is directed towards the center of the circle, and is given by:

F = evB

Setting this equal to the centripetal force:

F = mv^2/r

We get:

evB = mv^2/r

Substituting v from earlier:

eB(sqrt(2eV/m)) = m(2eV/m)/r

Simplifying:

r = sqrt(2mV)/(eB)

Thus the correct option is C.

Question 4

An electron is launched with velocity in a uniform magnetic field .The angle $\theta$0between and is between 0 and 90^o.As a result, the electron follows a helix, its velocity vector returning to its initial value in a time interval of:

Accepted Answer

The answer of An electron is launched with velocity ...

Question 5

The figure shows the motion of electrons in a wire which is near the N pole of a magnet.The wire will be pushed:

Accepted Answer

The answer of The figure shows the motion of electrons...

Question 6

A conducting strip of width 1.5 mm is in a magnetic field.As a result, there is a potential difference of 4.3 mV across the width of the strip.What is the magnitude of the electric field in the strip?

Accepted Answer

We can use the formula for the Hall voltage: $V_H=Bvd$, where $B$ is the magnetic field, $v$ is the velocity of the charge carriers, and $d$ is the width of the strip. Rearranging this formula to solve for the electric field, we get $E=\frac{V_H}{d}\cdot\frac{1}{B}$.

Plugging in the values given, we get $E=\frac{4.3	ext{ mV}}{1.5	ext{ mm}}\cdot\frac{1}{B}$. We don't know the value of $B$, but we can find it using the fact that the Hall voltage is caused by the Lorentz force $F_B=qvB$, where $q$ is the charge of the carriers. If we assume that the charge carriers are electrons, then $q=-1.6	imes10^{-19}	ext{ C}$.

We also know that the current density $J$ is related to the carrier density $n$, the electron charge $q$, and the carrier velocity $v$ by the formula $J=nqv$. Rearranging this formula, we can solve for $v=\frac{J}{nq}$, and then use this expression in the formula for the Lorentz force: $F_B=q\frac{J}{nq}B=\frac{JB}{n}$.

The Hall voltage is equal to the electric field times the width of the strip times the charge carrier density $n$: $V_H=E\cdot d\cdot nq$. Rearranging this formula, we can solve for $n=\frac{V_H}{Eqd}$.

Now we can combine all of these formulas to solve for $B$ and then $E$. We have:

$\frac{JB}{n}=-F_B=-\frac{V_H}{d}=-\frac{4.3	ext{ mV}}{1.5	ext{ mm}}$
$n=\frac{J}{qv}=\frac{J}{qn^{-1}\frac{J}{nq}}=n$
$V_H=E\cdot d\cdot nq=\frac{E\cdot d}{B}\cdot\frac{JB}{q}\cdot q$

Plugging in the given values of $d$, $V_H$, and $q$, and using the known value of the current density $J=I/A$, where $I$ is the current and $A$ is the cross-sectional area of the strip, we get:

$n=\frac{I}{qA}\cdot\frac{1}{E}$

$\frac{JB}{n}=-\frac{V_H}{d}$ becomes:

$\frac{B}{n}=-\frac{V_H}{dJ}\cdot\frac{1}{q}=-\frac{4.3	ext{ mV}}{1.5	ext{ mm}}\cdot\frac{1}{1.5	ext{ mm}\cdot I/A}\cdot\frac{1}{1.6	imes10^{-19}	ext{ C}}=-\frac{1.07}{IA}$

$\frac{B}{n}=-\frac{1.07}{IA}$ can be substituted into $n=\frac{I}{qA}\cdot\frac{1}{E}$ to get:

$n=-\frac{qA}{1.07I}\cdot E$

Now we can solve for $E$:

$E=\frac{V_H}{d}\cdot\frac{1}{B}=\frac{4.3	ext{ mV}}{1.5	ext{ mm}}\cdot\frac{1}{-\frac{1.07}{nqIA}}=2.9	ext{ V/m}$

Therefore, the magnitude of the electric field in the strip is 2.9 V/m. The correct answer is D).

Question 7

An electron and a proton both each travel with equal speeds around circular orbits in the same uniform magnetic field, as shown in the diagram (not to scale).The field is into the page on the diagram.Because the electron is less massive than the proton and because the electron is negatively charged and the proton is positively charged:

Accepted Answer

The answer of An electron and a proton both each...

Question 8

A cyclotron operates with a given magnetic field and at a given frequency.If R denotes the radius of the final orbit, the final particle energy is proportional to:

Accepted Answer

The final particle energy in a cyclotron is proportional to the square of the radius of the final orbit, as the kinetic energy is given by $E = \frac{1}{2}mv^2$ and the velocity $v$ is proportional to the radius $R$ in a magnetic field.

Question 9

The figure shows a uniform magnetic field  directed to the left and a wire carrying a current into the page.The magnetic force acting on the wire is:

Accepted Answer

The answer of The figure shows a uniform magnetic field...

Question 10

The diagram shows a straight wire carrying a flow of electrons into the page.The wire is between the poles of a permanent magnet.The direction of the magnetic force exerted on the wire is:

Accepted Answer

The answer of The diagram shows a straight wire carrying...

Question 11

A loop of wire carrying a current of 2.0 A is in the shape of a right triangle with two equal sides, each 15 cm long.A 0.7 T uniform magnetic field is parallel to the hypotenuse.The total magnetic force on the two equal sides has a magnitude of:

Accepted Answer

The magnetic force on a current-carrying wire is given by $F = ILB \sin(\theta)$, where $I$ is the current, $L$ is the length of the wire, $B$ is the magnetic field strength, and $\theta$ is the angle between the wire and the magnetic field. In this case, the magnetic field is parallel to the hypotenuse, meaning it is perpendicular to the two equal sides of the triangle. However, since the forces on the two equal sides are equal in magnitude but opposite in direction (due to the symmetry of the situation and the direction of the current), they cancel each other out, resulting in a net force of 0 N.

Question 12

The diagram shows a straight wire carrying current i in a uniform magnetic field.The magnetic force on the wire is indicated by an arrow but the magnetic field is not shown.Of the following possibilities, the direction of the magnetic field is:

Accepted Answer

The answer of The diagram shows a straight wire carrying...

Question 13

A uniform magnetic field is directed into the page.A charged particle, moving in the plane of the page, follows a clockwise spiral of decreasing radius as shown.A reasonable explanation is:

Accepted Answer

The charged particle is moving in a spiral of decreasing radius, which means that its speed is decreasing. In a uniform magnetic field, the force on a charged particle is perpendicular to both its velocity and the magnetic field. Since the particle is moving clockwise and the magnetic field is directed into the page, the force on the particle is directed outward (i.e. away from the center of the spiral). This means that the charge must be negative, since the force on a positive charge would be directed inward. Furthermore, since the particle's speed is decreasing, its kinetic energy is decreasing, and it must be losing energy. Therefore, the particle is slowing down.

Question 14

The Hall effect can be used to calculate the charge-carrier number density in a conductor.If a conductor carrying a current of 2.0 A is 0.5 mm thick, and the Hall effect voltage is 4.5 µV when it is in a uniform magnetic field of 1.2 T, what is the density of charge carriers in the conductor?

Accepted Answer

The equation for calculating the charge-carrier density using the Hall effect is: 
$n=\frac{IB}{qV_Hd}$, where 
$n$ = charge-carrier number density in the conductor 
$I$ = current through the conductor (in amperes) 
$B$ = magnetic field strength (in teslas) 
$q$ = charge of the carrier (in coulombs) 
$V_H$ = Hall voltage (in volts) 
$d$ = thickness of the conductor (in meters)

Substituting the given values:

$n=\frac{(2.0A)(1.2T)}{(1.6	imes10^{-19} C)(4.5	imes10^{-6} V)(0.5	imes10^{-3}m)}\approx6.7	imes10^{27}/m^3$

Therefore, the answer is B.

Question 15

In a certain mass spectrograph, an ion beam passes through a velocity filter consisting of mutually perpendicular fields  and  .The beam then enters a region of another magnetic field  perpendicular to the beam.The radius of curvature of the resulting ion beam is proportional to:

Accepted Answer

The answer of In a certain mass spectrograph, an ion...

Question 16

The resonance condition in a cyclotron states that:

Accepted Answer

The resonance condition in a cyclotron states that the frequency of the proton orbits equals the frequency of the electrical oscillator. This is because the oscillator provides the electric field that accelerates the protons, and if the frequency of the oscillator matches the frequency of the proton orbits, then the protons will continue to gain kinetic energy with each orbit.

Question 17

The current is from left to right in the conductor shown.The magnetic field is into the page and point S is at a higher potential than point T.The charge carriers are:

Accepted Answer

The answer of The current is from left to right...

Question 18

At one instant an electron is moving in the positive x direction along the x axis in a region where there is a uniform magnetic field in the positive z direction.When viewed from a point on the positive z axis, it subsequent motion is:

Accepted Answer

Since the magnetic field is in the positive z direction and the electron is moving in the positive x direction, the magnetic force acting on the electron will be in the positive y direction due to the cross product of the velocity and magnetic field vectors. This will cause the electron to move in a circular path in the xy plane, with the direction of rotation determined by the right-hand rule. Since the magnetic force is perpendicular to the velocity, the speed of the electron will remain constant as it moves in a circle.

Question 19

An electron is launched with velocity in a uniform magnetic field .The angle $\theta$between and is between 0 and 90^o.As a result, the electron follows a helical path.The pitch of the helix is:

Accepted Answer

The pitch of the helix is defined as the distance between adjacent turns of the helix.

Question 20

A strip 1.2 mm wide is moving at a speed of 25 cm/s through a uniform magnetic field of 5.6 T.What is the maximum Hall voltage across the strip?

Accepted Answer

The maximum Hall voltage (V_H) can be calculated using the formula V_H = Bvl, where B is the magnetic field strength (in Tesla), v is the velocity of the strip (in meters per second), and l is the width of the strip (in meters). Here, B = 5.6 T, v = 25 cm/s = 0.25 m/s, and l = 1.2 mm = 0.0012 m. Thus, V_H = 5.6 * 0.25 * 0.0012 = 0.00168 V = 1.68 mV, which rounds to 1.7 mV.

Quiz 28: Magnetic Fields

Which is NOT one of the differences between a cyclotron and a synchrotron?

A loop of wire carrying a current of 2.0 A is in the shape of a right triangle with two equal sides, each 15 cm long.A 0.7 T uniform magnetic field is in the plane of the triangle and is perpendicular to the hypotenuse.The resultant magnetic force on the two equal sides has a magnitude of:

Electrons (mass m, charge -e) are accelerated from rest through a potential difference V and are then deflected by a magnetic field that is perpendicular to their velocity.The radius of the resulting electron trajectory is:

The figure shows the motion of electrons in a wire which is near the N pole of a magnet.The wire will be pushed:

A conducting strip of width 1.5 mm is in a magnetic field.As a result, there is a potential difference of 4.3 mV across the width of the strip.What is the magnitude of the electric field in the strip?

A cyclotron operates with a given magnetic field and at a given frequency.If R denotes the radius of the final orbit, the final particle energy is proportional to:

The figure shows a uniform magnetic field directed to the left and a wire carrying a current into the page.The magnetic force acting on the wire is:

A loop of wire carrying a current of 2.0 A is in the shape of a right triangle with two equal sides, each 15 cm long.A 0.7 T uniform magnetic field is parallel to the hypotenuse.The total magnetic force on the two equal sides has a magnitude of:

The diagram shows a straight wire carrying current i in a uniform magnetic field.The magnetic force on the wire is indicated by an arrow but the magnetic field is not shown.Of the following possibilities, the direction of the magnetic field is:

A uniform magnetic field is directed into the page.A charged particle, moving in the plane of the page, follows a clockwise spiral of decreasing radius as shown.A reasonable explanation is:

The Hall effect can be used to calculate the charge-carrier number density in a conductor.If a conductor carrying a current of 2.0 A is 0.5 mm thick, and the Hall effect voltage is 4.5 µV when it is in a uniform magnetic field of 1.2 T, what is the density of charge carriers in the conductor?

In a certain mass spectrograph, an ion beam passes through a velocity filter consisting of mutually perpendicular fields and .The beam then enters a region of another magnetic field perpendicular to the beam.The radius of curvature of the resulting ion beam is proportional to:

The resonance condition in a cyclotron states that:

The current is from left to right in the conductor shown.The magnetic field is into the page and point S is at a higher potential than point T.The charge carriers are:

At one instant an electron is moving in the positive x direction along the x axis in a region where there is a uniform magnetic field in the positive z direction.When viewed from a point on the positive z axis, it subsequent motion is:

A strip 1.2 mm wide is moving at a speed of 25 cm/s through a uniform magnetic field of 5.6 T.What is the maximum Hall voltage across the strip?

Quiz 28: Magnetic Fields

Which is NOT one of the differences between a cyclotron and a synchrotron?

A loop of wire carrying a current of 2.0 A is in the shape of a right triangle with two equal sides, each 15 cm long.A 0.7 T uniform magnetic field is in the plane of the triangle and is perpendicular to the hypotenuse.The resultant magnetic force on the two equal sides has a magnitude of:

Electrons (mass m, charge -e) are accelerated from rest through a potential difference V and are then deflected by a magnetic field that is perpendicular to their velocity.The radius of the resulting electron trajectory is:

An electron is launched with velocity in a uniform magnetic field .The angle θ\thetaθ 0between and is between 0 and 90o.As a result, the electron follows a helix, its velocity vector returning to its initial value in a time interval of:

The figure shows the motion of electrons in a wire which is near the N pole of a magnet.The wire will be pushed:

A conducting strip of width 1.5 mm is in a magnetic field.As a result, there is a potential difference of 4.3 mV across the width of the strip.What is the magnitude of the electric field in the strip?

A cyclotron operates with a given magnetic field and at a given frequency.If R denotes the radius of the final orbit, the final particle energy is proportional to:

The figure shows a uniform magnetic field directed to the left and a wire carrying a current into the page.The magnetic force acting on the wire is:

The diagram shows a straight wire carrying a flow of electrons into the page.The wire is between the poles of a permanent magnet.The direction of the magnetic force exerted on the wire is:

A loop of wire carrying a current of 2.0 A is in the shape of a right triangle with two equal sides, each 15 cm long.A 0.7 T uniform magnetic field is parallel to the hypotenuse.The total magnetic force on the two equal sides has a magnitude of:

The diagram shows a straight wire carrying current i in a uniform magnetic field.The magnetic force on the wire is indicated by an arrow but the magnetic field is not shown.Of the following possibilities, the direction of the magnetic field is:

A uniform magnetic field is directed into the page.A charged particle, moving in the plane of the page, follows a clockwise spiral of decreasing radius as shown.A reasonable explanation is:

The Hall effect can be used to calculate the charge-carrier number density in a conductor.If a conductor carrying a current of 2.0 A is 0.5 mm thick, and the Hall effect voltage is 4.5 µV when it is in a uniform magnetic field of 1.2 T, what is the density of charge carriers in the conductor?

In a certain mass spectrograph, an ion beam passes through a velocity filter consisting of mutually perpendicular fields and .The beam then enters a region of another magnetic field perpendicular to the beam.The radius of curvature of the resulting ion beam is proportional to:

The resonance condition in a cyclotron states that:

The current is from left to right in the conductor shown.The magnetic field is into the page and point S is at a higher potential than point T.The charge carriers are:

At one instant an electron is moving in the positive x direction along the x axis in a region where there is a uniform magnetic field in the positive z direction.When viewed from a point on the positive z axis, it subsequent motion is:

An electron is launched with velocity in a uniform magnetic field .The angle θ\thetaθ between and is between 0 and 90o.As a result, the electron follows a helical path.The pitch of the helix is:

A strip 1.2 mm wide is moving at a speed of 25 cm/s through a uniform magnetic field of 5.6 T.What is the maximum Hall voltage across the strip?