Question 1

If the charge carriers are positive instead of negative, what difference will result when measuring the Hall voltage?
A) There is no difference.
B) Hall voltages will be double for positive charge carriers.
C) Hall voltages will be double for negative charge carriers.
D) The polarity of the voltage will be reversed.
E) Hall voltages cannot be measured for positive charge carriers.

Accepted Answer

The Hall voltage is proportional to the product of the magnetic field, the current, and the carrier charge. If the charge carriers are positive instead of negative, the polarity of the Hall voltage will be reversed. This is because the positive charge carriers will deflect in the opposite direction compared to negative charge carriers due to the Lorentz force. Therefore, the Hall voltage will change sign.

Question 2

When measuring the Hall voltage, what happens to the reading when the current is increased 20%?
A) It increases 20%.
B) It increases 40%.
C) It increases 44%.
D) It decreases 20%.
E) It decreases 44%.

Accepted Answer

The Hall voltage (V_H) is directly proportional to the current (I) flowing through the conductor. Therefore, if the current is increased by 20%, the Hall voltage also increases by 20%.

Question 3

The magnetic forces exerted on each other by two parallel wires with unequal currents flowing in opposite directions are
A) repulsive and unequal in magnitude.
B) attractive and unequal in magnitude.
C) repulsive and equal in magnitude.
D) attractive and equal in magnitude.

Accepted Answer

According to Ampère's force law, two parallel wires carrying currents in opposite directions exert repulsive forces on each other. By Newton's third law, these forces are equal in magnitude and opposite in direction.

Question 4

In a velocity selector for ions, when the electric field is increased by 50%, what change in the magnetic field is necessary to select the same speed as before?
A) a 50% decrease
B) a 25% decrease
C) no change
D) a 25% increase
E) a 50% increase

Accepted Answer

To maintain the same speed in a velocity selector, the ratio of the electric field (E) to the magnetic field (B) must remain constant. If the electric field is increased by 50%, the magnetic field must also be increased by 50% to keep the ratio E/B unchanged.

Question 5

A velocity selector for ions uses perpendicular electric and magnetic fields. If the electric field used were 800 V/m, what magnetic field would allow the selection of a velocity of 4000 m/s?
A) 0.0400 T
B) 0.0800 T
C) 0.160 T
D) 0.200 T
E) 0.500 T

Accepted Answer

The velocity selector works by applying perpendicular electric and magnetic fields that are adjusted to select a specific velocity of ions passing through it. The magnetic field must be adjusted to balance the force on the ions due to the electric field, so that only ions with the desired velocity are allowed to pass through.

The formula for the velocity selector is:

v = E/B

where v is the velocity of ions, E is the electric field strength, and B is the magnetic field strength.

We are given that the electric field strength is 800 V/m and the desired velocity of ions is 4000 m/s. Therefore, we can rearrange the formula above to solve for the magnetic field strength:

B = E/v

B = (800 V/m)/(4000 m/s)

B = 0.2 T

Therefore, the magnetic field strength needed to select ions with a velocity of 4000 m/s is 0.2 T. The nearest value to this is 0.200 T, which is choice D.

Question 6

A long straight wire is carrying a current of 2.4 A in a region where a uniform magnetic field is perpendicular to the wire. If the magnetic field is 0.52 mT, what is the magnitude of the force on 3.0 m of the wire? A) 1.2 × 10⁻³ N B) 3.7 × 10⁻³ N C) 7.2 × 10⁻³ N D) 3.7 N E) 7.2 N

Accepted Answer

The magnitude of the force on a current-carrying wire in a magnetic field is given by F = BIL, where F is the force, B is the magnetic field strength, I is the current, and L is the length of the wire in the magnetic field. In this case, B = 0.52 mT, I = 2.4 A, and L = 3.0 m, so F = (0.52 mT)(2.4 A)(3.0 m) = 3.7 × 10⁻³ N.

Question 7

If a charged particle is moving through a magnetic field with its velocity at an angle of 45 degrees with respect to the field, what happens to the magnetic force on it if both the speed and magnetic field are doubled?
A) It stays the same.
B) It doubles.
C) It quadruples.
D) It halves.
E) It increases by a factor of 5.66.

Accepted Answer

The magnetic force on a charged particle moving through a magnetic field is given by $F = qvB\sin(\theta)$, where $q$ is the charge, $v$ is the velocity, $B$ is the magnetic field strength, and $\theta$ is the angle between the velocity and the magnetic field. Doubling both the speed ($v$) and the magnetic field ($B$) results in the force being quadrupled, since $F' = q(2v)(2B)\sin(\theta) = 4qvB\sin(\theta)$.

Question 8

In a velocity selector for ions, when the velocity of the ions is double the value for which the selector is set, how do the magnetic force F_M and the electric force F_E compare? A) F_M = 4 F_E B) F_M = 2 F_E C) F_M = F_E/2 D) F_M = F_E/4 E) F_M = 1.414 F_E

Accepted Answer

In a velocity selector for ions, the magnetic force F_M and electric force F_E are of equal magnitude and opposite direction, so they cancel each other out to allow only ions of a specific velocity to pass through. When the velocity of the ions doubles, the electric force remains the same because it depends only on the charge and electric field, which are constant. However, the magnetic force depends on the velocity of the ions and the strength of the magnetic field, so it becomes four times larger (since velocity is squared in the formula for magnetic force). Therefore, the magnetic force is twice as large as the electric force, and the answer is B.

Question 9

A long straight wire is carrying a current of 2.4 A in a region with a uniform magnetic field. If the magnetic field is 0.52 mT and makes an angle of 60° with the wire, what is the magnitude of the force on 3.0 m of the wire? A) 3.2 N B) 6.5 N C) 1.6 × 10⁻³ N D) 3.2 × 10⁻³ N E) 6.5 × 10⁻³ N

Accepted Answer

We can find the force on the wire using the formula:

$F = IlB\sin	heta$

where I is the current, l is the length of wire, B is the magnetic field strength, and θ is the angle between the wire and the magnetic field.

Substituting the given values, we get:

$F = (2.4~A)(3.0~m)(0.52~mT)\sin(60^\circ)$

$F = (2.4~A)(3.0~m)(0.52	imes10^{-3}~T)\sin(60^\circ)$  [converting mT to T]

$F = 3.2	imes10^{-3}~N$

Therefore, the answer is D: 3.2 × 10^-3 N.

Question 10

A flat conductor of thickness 0.500 mm is placed in a magnetic field of 0.200 T perpendicular to the flat surface. If there are 8.00 × 10²⁴ charge carriers (i.e., electrons) per m³, what is the Hall voltage generated when 3.40 A flows along the conductor? A) 0.266 mV B) 0.531 mV C) 0.797 mV D) 1.06 mV E) 2.12 mV

Accepted Answer

The answer of A flat conductor of thickness 0.500 mm...

Question 11

A 3.0 cm section of a horizontal wire, carrying a 6.7 A current from east to west, is placed in a horizontal north-to-south 0.42 T uniform magnetic field in the lab. What is the magnitude of the force on the 3.0 cm section of the wire?
A) 0.084 N
B) 0.17 N
C) 0.25 N
D) 0.0017 N
E) 0.0025 N

Accepted Answer

The magnitude of the force on a current-carrying wire in a magnetic field is given by $F = BIL\sin(\theta)$, where $B$ is the magnetic field strength, $I$ is the current, $L$ is the length of the wire, and $\theta$ is the angle between the direction of the current and the direction of the magnetic field. Here, $B = 0.42 \, \text{T}$, $I = 6.7 \, \text{A}$, $L = 0.03 \, \text{m}$ (since 3.0 cm = 0.03 m), and $\theta = 90^\circ$ (since the current flows east to west and the magnetic field is north to south, they are perpendicular). Thus, $\sin(90^\circ) = 1$, and the force is $F = 0.42 \times 6.7 \times 0.03 \times 1 = 0.0846 \, \text{N}$, which rounds to $0.084 \, \text{N}$.

Question 12

A 3.0 cm section of a horizontal wire, carrying a 6.7 A current from east to west, is placed in a horizontal north-to-south 0.42 T uniform magnetic field in the lab. In what direction is the magnetic force on the 3.0 cm section?
A) north
B) south
C) up
D) down
E) southwest

Accepted Answer

The direction of the magnetic force can be determined using the right-hand rule. If the right-hand thumb points in the direction of the current and the fingers point in the direction of the magnetic field, the palm points in the direction of the magnetic force. In this case, the magnetic force is upward, perpendicular to both the current direction (east-west) and the magnetic field direction (north-south).

Question 13

A 300-turn coil of area 25 cm² is placed in an 80 mT magnetic field. If the coil carries a current of 3.3 A, what is the magnitude of the maximum magnetic torque it could experience? A) 20 N·m B) 6.0 N·m C) 0.13 N·m D) 0.20 N·m E) 0.060 N·m

Accepted Answer

The maximum torque experienced by the coil is given by the equation τ = NIABsinθ, where N is the number of turns, I is the current, A is the area of the coil, B is the magnetic field strength, and θ is the angle between the normal to the coil and the magnetic field direction.
Thus, τ = (300)(3.3)(25×10^-4)(80×10^-3)sin90 = 0.20 N·m. 
Therefore, the correct answer is choice D.

Question 14

The magnetic field in a cyclotron is 0.50 T. What is the radius of the vacuum chamber for a maximum proton velocity of 1.0 × 10⁷ m/s? A) 0.11 m B) 1.2 m C) 0.21 m D) 0.013 m

Accepted Answer

The radius of the vacuum chamber can be found using the formula: 
r = mv / (qB), where m is the mass of the proton, v is the velocity, q is the charge of the proton, and B is the magnetic field.
Plugging in the given values, we get:
r = (1.67 x 10^-27 kg) x (1.0 x 10^7 m/s) / (1.6 x 10^-19 C) x (0.5 T)
r = 2.79 x 10^-2 m = 0.0279 m = 0.21 m (rounded to two significant figures)
Therefore, the answer is choice C.

Question 15

The magnetic field in a cyclotron is 0.50 T. Find the magnitude of the magnetic force on a proton with velocity of 1.0 × 10⁷ m/s in a plane perpendicular to the field. A) 6.0 × 10⁻¹⁴ N B) 6.0 × 10⁻¹³ N C) 8.0 × 10⁻¹⁴ N D) 8.0 × 10⁻¹³ N

Accepted Answer

The answer of The magnetic field in a cyclotron is...

Question 16

Cathode rays are
A) x-rays.
B) γ-rays.
C) electrons.
D) protons.
E) neutrons.

Accepted Answer

Cathode rays are streams of electrons that are emitted from the cathode in a vacuum tube.

Question 17

An electron is moving at 3.0 × 10⁶ m/s at an angle of 40° to a 0.80 T magnetic field. What is the magnitude of the acceleration of the electron? A) 2.7 × 10¹⁷ m/s² B) 1.4 × 10¹⁷ m/s² C) 5.2 × 10¹⁷ m/s² D) 7.8 × 10¹⁷ m/s² E) 9.8 × 10¹⁷ m/s²

Accepted Answer

The magnitude of the acceleration of an electron moving in a magnetic field is given by the formula $a = \frac{q|vB\sin(\theta)|}{m}$, where $q$ is the charge of the electron ($1.6 \times 10^{-19} C$), $v$ is the velocity of the electron ($3.0 \times 10^{6} m/s$), $B$ is the magnetic field strength ($0.80 T$), $\theta$ is the angle between the velocity and the magnetic field ($40^\circ$), and $m$ is the mass of the electron ($9.11 \times 10^{-31} kg$). Plugging in the values and solving for $a$ gives a value in the order of $10^{17} m/s^2$, which matches the magnitude of choice A.

Question 18

A 10 m long wire carrying a current of 22 A makes an angle of 37° with the Earth's magnetic field of 0.51 mT. What is the magnetic force on the wire?
A) 0.11 N
B) 0.068 N
C) 0.19 N
D) 0.090 N
E) 0.011 N

Accepted Answer

The answer of A 10 m long wire carrying a...

Question 19

A 300-turn coil of area 25 cm² is placed in an 80 mT magnetic field. If the coil carries a current of 3.3 A and if the normal to the area of the coil makes an angle of 20° with the direction of the magnetic field, what is the magnitude of the torque on the coil? A) 0.20 N·m B) 0.19 N·m C) 0.072 N·m D) 0.068 N·m E) 0.039 N·m

Accepted Answer

The torque on a current carrying coil in a magnetic field is given by the formula:

τ = NIABsinθ

where τ is torque, N is the number of turns in the coil, I is the current in the coil, A is the area of the coil, B is the strength of the magnetic field and θ is the angle between the normal to the area of the coil and the direction of the magnetic field.

Plugging in the given values,

N = 300
I = 3.3 A
A = 25 cm² = 0.0025 m²
B = 80 mT = 0.08 T
θ = 20°

τ = (300)(3.3)(0.0025)(0.08)sin(20°) = 0.068 N·m

Therefore, the answer is D.

Question 20

An electric field perpendicular to a magnetic field is used as a velocity selector for Li⁺ ions, selecting the velocity v₁. Next, Li⁺⁺ ions are sent through the same velocity selector, selecting velocity v₂. If neither the electric nor magnetic fields have changed between the two processes, how do v₁ and v₂ compare? A) v₂ = v₁ B) v₂ = 2 v₁ C) v₂ = v₁/2 D) v₂ = 4 v₁ E) v₂ = v₁/4

Accepted Answer

The velocity selector only depends on the ratio of the electric and magnetic field strengths, which is the same for both processes. Therefore, the selected velocities will be the same in both cases.

Quiz 19: Magnetic Forces and Fields

If the charge carriers are positive instead of negative, what difference will result when measuring the Hall voltage?

When measuring the Hall voltage, what happens to the reading when the current is increased 20%?

The magnetic forces exerted on each other by two parallel wires with unequal currents flowing in opposite directions are

In a velocity selector for ions, when the electric field is increased by 50%, what change in the magnetic field is necessary to select the same speed as before?

A velocity selector for ions uses perpendicular electric and magnetic fields. If the electric field used were 800 V/m, what magnetic field would allow the selection of a velocity of 4000 m/s?

A long straight wire is carrying a current of 2.4 A in a region where a uniform magnetic field is perpendicular to the wire. If the magnetic field is 0.52 mT, what is the magnitude of the force on 3.0 m of the wire?

If a charged particle is moving through a magnetic field with its velocity at an angle of 45 degrees with respect to the field, what happens to the magnetic force on it if both the speed and magnetic field are doubled?

In a velocity selector for ions, when the velocity of the ions is double the value for which the selector is set, how do the magnetic force FM and the electric force FE compare?

A long straight wire is carrying a current of 2.4 A in a region with a uniform magnetic field. If the magnetic field is 0.52 mT and makes an angle of 60° with the wire, what is the magnitude of the force on 3.0 m of the wire?

A flat conductor of thickness 0.500 mm is placed in a magnetic field of 0.200 T perpendicular to the flat surface. If there are 8.00 × 1024 charge carriers (i.e., electrons) per m3, what is the Hall voltage generated when 3.40 A flows along the conductor?

A 3.0 cm section of a horizontal wire, carrying a 6.7 A current from east to west, is placed in a horizontal north-to-south 0.42 T uniform magnetic field in the lab. What is the magnitude of the force on the 3.0 cm section of the wire?

A 3.0 cm section of a horizontal wire, carrying a 6.7 A current from east to west, is placed in a horizontal north-to-south 0.42 T uniform magnetic field in the lab. In what direction is the magnetic force on the 3.0 cm section?

A 300-turn coil of area 25 cm2 is placed in an 80 mT magnetic field. If the coil carries a current of 3.3 A, what is the magnitude of the maximum magnetic torque it could experience?

The magnetic field in a cyclotron is 0.50 T. What is the radius of the vacuum chamber for a maximum proton velocity of 1.0 × 107 m/s?

The magnetic field in a cyclotron is 0.50 T. Find the magnitude of the magnetic force on a proton with velocity of 1.0 × 107 m/s in a plane perpendicular to the field.

Cathode rays are

An electron is moving at 3.0 × 106 m/s at an angle of 40° to a 0.80 T magnetic field. What is the magnitude of the acceleration of the electron?

A 10 m long wire carrying a current of 22 A makes an angle of 37° with the Earth's magnetic field of 0.51 mT. What is the magnetic force on the wire?

A 300-turn coil of area 25 cm2 is placed in an 80 mT magnetic field. If the coil carries a current of 3.3 A and if the normal to the area of the coil makes an angle of 20° with the direction of the magnetic field, what is the magnitude of the torque on the coil?

In a velocity selector for ions, when the velocity of the ions is double the value for which the selector is set, how do the magnetic force F_M and the electric force F_E compare?

A flat conductor of thickness 0.500 mm is placed in a magnetic field of 0.200 T perpendicular to the flat surface. If there are 8.00 × 10²⁴ charge carriers (i.e., electrons) per m³, what is the Hall voltage generated when 3.40 A flows along the conductor?

A 300-turn coil of area 25 cm² is placed in an 80 mT magnetic field. If the coil carries a current of 3.3 A, what is the magnitude of the maximum magnetic torque it could experience?

The magnetic field in a cyclotron is 0.50 T. What is the radius of the vacuum chamber for a maximum proton velocity of 1.0 × 10⁷ m/s?

The magnetic field in a cyclotron is 0.50 T. Find the magnitude of the magnetic force on a proton with velocity of 1.0 × 10⁷ m/s in a plane perpendicular to the field.

An electron is moving at 3.0 × 10⁶ m/s at an angle of 40° to a 0.80 T magnetic field. What is the magnitude of the acceleration of the electron?

A 300-turn coil of area 25 cm² is placed in an 80 mT magnetic field. If the coil carries a current of 3.3 A and if the normal to the area of the coil makes an angle of 20° with the direction of the magnetic field, what is the magnitude of the torque on the coil?