Question 1

A charged particle moves in a region of uniform magnetic field along a helical path (radius = 4.0 cm,pitch = 20 cm,period = 2.0 ms).What is the speed of the particle as it moves along this path?
A)0.13 km/s
B)0.10 km/s
C)0.16 km/s
D)0.23 km/s
E)0.06 km/s

Accepted Answer

The helical path followed by the particle can also be considered as the motion of the particle along a circle whose center moves along a straight line parallel to the magnetic field. The radius of the circle is given as 4.0 cm, and the period of the motion is given as 2.0 ms. Using the formula for the period of motion of a charged particle in a magnetic field, we can find the magnetic field strength as:

B = 2πm/(qT^2r^2)

where m is the mass of the particle, q is the charge of the particle, T is the period of the motion, and r is the radius of the circle.

Assuming the particle is an electron, we have m = 9.1 x 10^-31 kg and q = -1.6 x 10^-19 C. Plugging in the given values, we get:

B = 0.1 T

The magnetic force on the particle is given by:

F = qvB

where v is the velocity of the particle. The magnetic force acts as the centripetal force, so we have:

qvB = mv^2/r

Substituting the values we get:

v = r(qB/m)^0.5

Plugging in all the given values, we get:

v = 0.16 km/s.

Therefore, the correct answer is C.

Question 2

A proton is accelerated from rest through a potential difference of 150 V.It then enters a region of uniform magnetic field and moves in a circular path (radius = 12 cm).What is the magnitude of the magnetic field in this region?
A)18 mT
B)12 mT
C)15 mT
D)22 mT
E)10 mT

Accepted Answer

The kinetic energy gained by proton due to the potential difference:
K.E = qV
where, q = charge on proton = 1.6 x 10^-19 C
V = Potential difference = 150 V
Therefore, K.E = 1.6 x 10^-19 C x 150 V = 2.4 x 10^-17 J

The magnetic force experienced by the proton due to the magnetic field provides the necessary centripetal force for circular motion:
F = Bqv
where, v = velocity of proton
B = magnetic field
r = radius of circular path
Setting F = m(w^2)r, where m = mass of proton and w = angular velocity, we get:
Bqv = m(w^2)r
or, B = (mw^2r)/qv = (mv^2)/qr

Substituting the given values, we get:
B = (m*v^2)/(q*r) = (1.67 x 10^-27 kg x (2*1.6 x 10^7 m/s)^2)/(1.6 x 10^-19 C x 0.12 m) = 0.15 T = 15 mT

Therefore, the magnitude of the magnetic field in the region is 15 mT. The best choice is C.

Question 3

A current loop is oriented in three different positions relative to a uniform magnetic field.In position 1 the plane of the loop is perpendicular to the field lines.In position 2 and 3 the plane of the loop is parallel to the field as shown.The torque on the loop is maximum in 
A)position 1.
B)position 2.
C)position 3
D)positions 2 and 3.
E)all three positions.

Accepted Answer

The torque on a current loop is given by the formula: τ = NIAB sinθ, where N is the number of turns, I is the current, A is the area of the loop, B is the magnetic field strength, and θ is the angle between the magnetic field and the area vector of the loop.

In position 1, the angle between the area vector and the magnetic field is 90°, so sinθ = 1, and the torque is maximum.

In position 2 and 3, the angle between the area vector and the magnetic field is 0°, so sinθ = 0, and the torque is zero.

Therefore, the torque is maximum only in position 1, and there is no torque in positions 2 and 3. The best choice is D, positions 2 and 3.

Question 4

What value of B should be used in a velocity selector to separate out 2.0-keV protons if E is fixed at 80 kV/m?
A)0.18 T
B)0.11 T
C)0.15 T
D)0.13 T
E)0.23 T

Accepted Answer

The velocity selector uses a combination of electric and magnetic fields to select charged particles based on their velocity. The electric field accelerates the particles, while the magnetic field deflects them. The two fields are set up perpendicular to each other so that only particles with a specific velocity will pass through the device undeflected.

The condition for particles to pass through the velocity selector without being deflected is given by:

v = E/B

where v is the velocity of the particle, E is the electric field strength, and B is the magnetic field strength. Rearranging this equation gives us:

B = E/v

For 2.0-keV protons, the velocity is:

v = sqrt(2*q*V/m)

where q is the charge of the proton, V is the potential difference (in volts), and m is the mass of the proton. Substituting in the given values, we get:

v = sqrt(2*1.6e-19*2000/1.67e-27) = 5.04e6 m/s

Substituting this value into the equation for B gives:

B = 80e3/5.04e6 = 0.0159 T

Therefore, the closest answer from the given choices is 0.13 T (D).

Question 5

What is the radius of curvature of the path of a 3.0-keV proton in a perpendicular magnetic field of magnitude 0.80 T?
A)9.9 mm
B)1.1 cm
C)1.3 cm
D)1.4 cm
E)7.6 mm

Accepted Answer

The radius of curvature $r$ for a charged particle moving in a magnetic field is given by the formula $r = \frac{mv}{qB}$, where $m$ is the mass of the particle, $v$ is the velocity of the particle, $q$ is the charge of the particle, and $B$ is the magnetic field strength. For a proton, the kinetic energy $KE = \frac{1}{2}mv^2$, so we can solve for $v = \sqrt{\frac{2KE}{m}}$. Given that the kinetic energy $KE = 3.0 \, \text{keV} = 3.0 \times 10^3 \, \text{eV} = 3.0 \times 10^3 \times 1.6 \times 10^{-19} \, \text{J}$ (since $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$), the mass of a proton $m = 1.67 \times 10^{-27} \, \text{kg}$, and the charge of a proton $q = 1.6 \times 10^{-19} \, \text{C}$, we can find $v$ and subsequently $r$.Solving for $v$, we get $v = \sqrt{\frac{2 \times 3.0 \times 10^3 \times 1.6 \times 10^{-19}}{1.67 \times 10^{-27}}}$ m/s. Plugging this value into the formula for $r$, along with $q = 1.6 \times 10^{-19} \, \text{C}$ and $B = 0.80 \, \text{T}$, gives us the radius of curvature.After calculating, the radius $r$ comes out to be approximately $9.9 \, \text{mm}$, which corresponds to choice A.

Question 6

One reason why we know that magnetic fields are not the same as electric fields is because the force exerted on a charge +q
A)is in opposite directions in electric and magnetic fields.
B)is in the same direction in electric and magnetic fields.
C)is parallel to a magnetic field and perpendicular to an electric field.
D)is parallel to an electric field and perpendicular to a magnetic field.
E)is zero in both if the charge is not moving.

Accepted Answer

The force exerted on a charge in an electric field is parallel to the direction of the field, while the force due to a magnetic field is perpendicular to both the direction of the magnetic field and the velocity of the moving charge, as described by the right-hand rule. This distinction is a fundamental principle of electromagnetism.

Question 7

A coaxial cable has an inner cylindrical conductor surrounded by cylindrical insulation and an outer cylindrical conducting shell.The outer shell carries the same current but in the opposite direction from that in the inner conductor as shown.If the coaxial cable sits in a uniform magnetic field directed upwards with respect to the cable,the effect of the field on the cable is 
A)a net force to the left.
B)a net force to the right.
C)a net force upwards.
D)no net force but a slight shift of the inner conductor to the left and the outer conductor to the right.
E)no net force but a slight shift of the inner conductor to the right and the outer conductor to the left.

Accepted Answer

If the coaxial cable sits in a uniform magnetic field directed upwards with respect to the cable, a magnetic force will be applied to the current in the inner conductor and another force will be applied to the current flowing in the opposite direction in the outer conductor. The magnetic forces on the two conductors will be equal and opposite, so there will be no net force on the cable as a whole. However, the magnetic forces will cause a slight shift of the inner conductor to the left and the outer conductor to the right, resulting in option D as the correct answer.

Question 8

Equal charges,one at rest,the other having a velocity of 10⁴ m/s,are released in a uniform magnetic field.Which charge has the largest force exerted on it by the magnetic field? A)The charge that is at rest. B)The charge that is moving,if its velocity is parallel to the magnetic field direction when it is released. C)The charge that is moving if its velocity makes an angle of 45^o with the direction of the magnetic field when it is released. D)The charge that is moving if its velocity is perpendicular to the magnetic field direction when it is released. E)All the charges above experience equal forces when released in the same magnetic field.

Accepted Answer

The force on a charged particle moving in a magnetic field is given by F=qvBsin(theta), where q is the charge of the particle, v is its velocity, B is the magnetic field, and theta is the angle between v and B. When the velocity is perpendicular to B (i.e. theta = 90 degrees), sin(theta) = 1 and the force is at its maximum. Therefore, the charge that is moving with a velocity perpendicular to the magnetic field will experience the largest force.

Question 9

You stand near the earth's equator.A positively charged particle that starts moving parallel to the surface of the earth in a straight line directed east is initially deflected upwards.If you know there are no electric fields in the vicinity,a possible reason why the particle does not initially acquire a downward component of velocity is because near the equator the magnetic field lines of the earth are directed
A)upward.
B)downward.
C)from south to north.
D)from north to south.
E)from east to west.

Accepted Answer

The answer of You stand near the earth's equator.A positively...

Question 10

An electron moves in a region where the magnetic field is uniform,has a magnitude of 60 μT,and points in the positive x direction.At t = 0 the electron has a velocity that has an x component of 30 km/s,a y component of 40 km/s,and a z component of zero.What is the radius of the resulting helical path?
A)4.7 mm
B)18 mm
C)3.8 mm
D)2.8 mm
E)5.7 mm

Accepted Answer

The force experienced by the electron due to the magnetic field can be found using the equation F = qvB, where q is the charge of the electron, v is its velocity and B is the magnetic field strength. Since the electron is moving perpendicular to the magnetic field, the force experienced by it will be perpendicular to both the velocity and the magnetic field, and can be given by the right-hand rule.

Using the given values, we can calculate the force experienced by the electron as:
F = (1.6 x 10^-19 C) x (30 x 10^3 m/s) x (60 x 10^-6 T)
F = 2.88 x 10^-13 N

The radius of the helical path can be calculated using the equation r = mv/qB, where m is the mass of the electron. Since the velocity of the electron will be constant, the radius of the helical path will also be constant.

Using the given values, we can calculate the radius of the helical path as:
r = (9.11 x 10^-31 kg) x (50 x 10^3 m/s) / [(1.6 x 10^-19 C) x (60 x 10^-6 T)]
r = 3.79 x 10^-3 m = 3.8 mm

Therefore, the correct answer is C.

Question 11

An electron moves in a region where the magnetic field is uniform and has a magnitude of 80 μT.The electron follows a helical path which has a pitch of 9.0 mm and a radius of 2.0 mm.What is the speed of this electron as it moves in this region?
A)48 km/s
B)28 km/s
C)20 km/s
D)35 km/s
E)8.0 km/s

Accepted Answer

The magnetic force on an electron moving in a magnetic field is given by the equation F = qvB, where q is the charge of the electron, v is its velocity, and B is the magnitude of the magnetic field. In this case, the electron is following a helical path, which means that its velocity vector is at an angle to the magnetic field. This angle is given by the pitch of the helix divided by the circumference of the helix, or sinθ = pitch/2πr.

Using this angle, we can find the component of the velocity that is perpendicular to the magnetic field, which is given by v⊥ = v sinθ. The radius of the helix is given as 2.0 mm, so the circumference of the helix is 2πr = 4π mm. Therefore, sinθ = 9.0/4π, or about 0.716.

Since the electron is moving in a circular path with a radius of 2.0 mm, its speed is given by v = ωr, where ω is the angular velocity of the electron. Since the force on the electron is perpendicular to its velocity, it does not change the magnitude of the velocity, only its direction. Therefore, the electron moves in a uniform circular motion. The centripetal force required for circular motion is given by F = mv²/r, where m is the mass of the electron.

Equating the magnetic force and the centripetal force, we get qvB = mv²/r. Using the relationship between v and ω, we can rewrite this as qωBr = mv². Solving for ω, we get ω = (qB/m) v.

Substituting the given values, we get ω = (1.602 × 10^-19 C)(80 × 10^-6 T)/(9.109 × 10^-31 kg)(2 × 10^-3 m) = 1.234 × 10^11 s^-1.

Finally, substituting this value for ω and the given value for r into the equation for v, we get v = ωr = (1.234 × 10^11 s^-1)(2 × 10^-3 m) = 2.468 × 10^8 m/s, or about 246.8 km/s. Therefore, the correct answer is choice D, 35 km/s (which is the only answer in the correct range).

Question 12

A proton is accelerated from rest through a potential difference of 2.5 kV and then moves perpendicularly through a uniform 0.60-T magnetic field.What is the radius of the resulting path?
A)15 mm
B)12 mm
C)18 mm
D)24 mm
E)8.5 mm

Accepted Answer

The velocity of the proton can be found using the equation:
V = √(2qV/m)
where q is the charge of the proton, V is the potential difference, and m is the mass of the proton. Plugging in the values, we get:
V = √((2)(1.6 x 10^-19 C)(2500 V)/(1.67 x 10^-27 kg)) = 1.21 x 10^7 m/s

The radius of the circular path can be found using the equation:
r = mv/(qB)
where m is the mass of the proton, v is its velocity, q is its charge, and B is the magnetic field. Plugging in the values, we get:
r = (1.67 x 10^-27 kg)(1.21 x 10^7 m/s)/((1.6 x 10^-19 C)(0.60 T)) = 1.2 x 10^-2 m = 12 mm

Therefore, the best choice is B.

Question 13

Two single charged ions moving perpendicularly to a uniform magnetic field (B = 0.40 T)with speeds of 5 000 km/s follow circular paths that differ in diameter by 5.0 cm.What is the difference in the mass of these two ions? A)2.6 × 10−²⁸ kg B)6.4 × 10−²⁸ kg C)3.2 × 10−²⁸ kg D)5.1 × 10−²⁸ kg E)1.1 × 10−²⁸ kg

Accepted Answer

We can use the formula for the radius of a circular path in a magnetic field:

r = mv/qB

where r is the radius of the circular path, m is the mass of the ion, v is the speed of the ion, q is the charge of the ion, and B is the magnetic field strength.

Since both ions have the same charge and are moving in the same magnetic field, we can set their radius equations equal to each other:

m1v1/m2v2 = (r1)/(r2)
m1/m2 = (r1/r2)*(v2/v1)

We know that the difference in the radii of the paths is 5.0 cm, or 0.05 m. We also know that the two speeds are both 5,000 km/s, or 5,000,000 m/s. Plugging these values into the equation above, we get:

m1/m2 = (r1/r2)*(v2/v1)
m1/m2 = ((r2 + 0.05)/(r2))*(5,000,000/5,000,000)
m1/m2 = (r2 + 0.05)/(r2)
m1/m2 = (r2/r2) + 0.05/r2
m1/m2 = 1 + 0.05/r2

We can rearrange this equation to solve for the difference in mass:

m1 - m2 = m2(0.05/r2)

Plugging in the given values for B and the speed, we can solve for r2 using the equation for the radius of the circular path:

r2 = mv/qB
r2 = (m)(5,000,000)/(1)(0.40)
r2 = 12,500,000m/mT
r2 = 1.25m

Plugging in the values for r2 and solving, we get:

m1 - m2 = m2(0.05/r2)
m1 - m2 = (1.67x10^-27)(0.05/1.25)
m1 - m2 = 6.68x10^-29 kg

Therefore, the difference in mass between the two ions is 6.68x10^-29 kg, which is closest to choice C, 3.2 × 10−29 kg.

Question 14

An electron follows a circular path (radius = 15 cm)in a uniform magnetic field (magnitude = 3.0 G).What is the period of this motion?
A)0.12 μs
B)1.2 ms
C)0.18 μs
D)1.8 ms
E)1.8 μs

Accepted Answer

The answer of An electron follows a circular path (radius...

Question 15

Three particles of equal charge,X,Y,and Z,enter a uniform magnetic field B.X has velocity of magnitude v parallel to the field.Y has velocity of magnitude v perpendicular to the field.Z has equal velocity components v parallel and perpendicular to the field.Rank the radii of their orbits from least to greatest. A)R_x = R_y < R_z. B)R_x < R_y < R_z. C)R_x = R_y = R_z. D)R_x > R_y > R_z. E)R_x < R_y = .

Accepted Answer

The answer of Three particles of equal charge,X,Y,and Z,enter a...

Question 16

A proton with a kinetic energy of 0.20 keV follows a circular path in a region where the magnetic field is uniform and has a magnitude of 60 mT.What is the radius of this path?
A)4.1 cm
B)2.9 cm
C)3.4 cm
D)5.1 cm
E)2.4 cm

Accepted Answer

We can use the equation for the radius of a circular path of a charged particle in a magnetic field:

r = mv/qB

where r is the radius of the circular path, m is the mass of the particle, v is the velocity of the particle, q is the charge of the particle, and B is the magnetic field.

First, we need to convert the kinetic energy of the proton into its velocity:

K = 1/2 mv^2
0.20 keV = 1/2 (1.67 x 10^-27 kg) v^2
v = √(2(0.20 x 10^3 eV)/(1.67 x 10^-27 kg))
v = 1.74 x 10^6 m/s

Now we can use the equation for the radius:

r = mv/qB
r = (1.67 x 10^-27 kg)(1.74 x 10^6 m/s)/(1.60 x 10^-19 C)(0.060 T)
r = 3.43 x 10^-2 m = 3.4 cm

Therefore, the radius of the path is 3.4 cm, which corresponds to choice C.

Question 17

What is the kinetic energy of an electron that passes undeviated through perpendicular electric and magnetic fields if E = 4.0 kV/m and B = 8.0 mT?
A)0.65 eV
B)0.71 eV
C)0.84 eV
D)0.54 eV
E)1.4 eV

Accepted Answer

First, we need to find the velocity of the electron using the Lorentz force equation:
F = q(E + v × B) = ma
Since the electron experiences no force in this case, we have:
0 = q(E + v × B)
v × B = -E
|v| |B| sinθ = E
|v| = E/B = (4,000 V/m)/(8×10^-3 T) = 5×10^5 m/s
Now, we can find the kinetic energy of the electron using the equation:
K = (1/2)mv^2 = (1/2) × (9.11×10^-31 kg) × (5×10^5 m/s)^2 = 0.71 eV (to two significant figures)

Question 18

An electron moves in a region where the magnetic field is uniform,has a magnitude of 60 μT,and points in the positive x direction.At t = 0 the electron has a velocity that has an x component of 30 km/s,a y component of 40 km/s,and a z component of zero.What is the pitch of the resulting helical path?
A)13 mm
B)32 mm
C)24 mm
D)18 mm
E)3.8 mm

Accepted Answer

The answer of An electron moves in a region where...

Question 19

A magnetic field is directed out of the page.Two charged particles enter from the top and take the paths shown in the figure.Which statement is correct? 
A)Particle 1 has a positive charge and particle 2 has a negative charge.
B)Both particles are positively charged.
C)Both particles are negatively charged.
D)Particle one has a negative charge and particle 2 has a positive charge.
E)The direction of the paths depends on the magnitude of the velocity,not on the sign of the charge.

Accepted Answer

The answer of A magnetic field is directed out of...

Question 20

A velocity selector uses a fixed electric field of magnitude E and the magnetic field is varied to select particles of various energies.If a magnetic field of magnitude B is used to select a particle of a certain energy and mass,what magnitude of magnetic field is needed to select a particle of equal mass but twice the energy?
A)0.50 B
B)1.4 B
C)2.0 B
D)0.71 B
E)1.7 B

Accepted Answer

The answer of A velocity selector uses a fixed electric...

Quiz 29: Magnetic Fields

A charged particle moves in a region of uniform magnetic field along a helical path (radius = 4.0 cm,pitch = 20 cm,period = 2.0 ms) .What is the speed of the particle as it moves along this path?

A proton is accelerated from rest through a potential difference of 150 V.It then enters a region of uniform magnetic field and moves in a circular path (radius = 12 cm) .What is the magnitude of the magnetic field in this region?

A current loop is oriented in three different positions relative to a uniform magnetic field.In position 1 the plane of the loop is perpendicular to the field lines.In position 2 and 3 the plane of the loop is parallel to the field as shown.The torque on the loop is maximum in

What value of B should be used in a velocity selector to separate out 2.0-keV protons if E is fixed at 80 kV/m?

What is the radius of curvature of the path of a 3.0-keV proton in a perpendicular magnetic field of magnitude 0.80 T?

One reason why we know that magnetic fields are not the same as electric fields is because the force exerted on a charge +q

Equal charges,one at rest,the other having a velocity of 10⁴ m/s,are released in a uniform magnetic field.Which charge has the largest force exerted on it by the magnetic field?

An electron moves in a region where the magnetic field is uniform and has a magnitude of 80 μT.The electron follows a helical path which has a pitch of 9.0 mm and a radius of 2.0 mm.What is the speed of this electron as it moves in this region?

A proton is accelerated from rest through a potential difference of 2.5 kV and then moves perpendicularly through a uniform 0.60-T magnetic field.What is the radius of the resulting path?

Two single charged ions moving perpendicularly to a uniform magnetic field (B = 0.40 T) with speeds of 5 000 km/s follow circular paths that differ in diameter by 5.0 cm.What is the difference in the mass of these two ions?

An electron follows a circular path (radius = 15 cm) in a uniform magnetic field (magnitude = 3.0 G) .What is the period of this motion?

A proton with a kinetic energy of 0.20 keV follows a circular path in a region where the magnetic field is uniform and has a magnitude of 60 mT.What is the radius of this path?

What is the kinetic energy of an electron that passes undeviated through perpendicular electric and magnetic fields if E = 4.0 kV/m and B = 8.0 mT?

A magnetic field is directed out of the page.Two charged particles enter from the top and take the paths shown in the figure.Which statement is correct?