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PAU Physics (Aragón) study materials
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PAU Physics (Aragón) practice questions

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Question 1Gravity

Two point masses of m1 = 2000 kg and m2 = 5000 kg are separated by a distance of r = 5.0 m in a vacuum. Using Newton's Law of Universal Gravitation with G = 6.67 x 10^-11 N m^2/kg^2, what is the magnitude of the gravitational force between them?

  1. 1.33 x 10^-4 N
  2. 5.34 x 10^-6 N
  3. 2.67 x 10^-7 N
  4. 2.67 x 10^-5 N Correct answer.

Explanation

Applying Newton's law of universal gravitation: F = G * m1 * m2 / r^2 = (6.67 x 10^-11) * (2000) * (5000) / (5.0)^2 = (6.67 x 10^-11) * (1.0 x 10^7) / 25 = 2.668 x 10^-5 N.

Question 2Gravity

If the gravitational field strength at Earth's surface is g0 = 9.80 N/kg, what is the field strength g at an altitude h = 2 R_E above Earth's surface, where R_E is Earth's radius?

  1. 3.27 N/kg
  2. 2.45 N/kg
  3. 4.90 N/kg
  4. 1.09 N/kg Correct answer.

Explanation

The distance from Earth's center is r = R_E + h = R_E + 2 R_E = 3 R_E. Since g = G M_E / r^2, g = G M_E / (3 R_E)^2 = g0 / 9 = 9.80 / 9 = 1.089 N/kg.

Question 3Gravity

A satellite moves in a circular orbit around Earth at an altitude h = R_E equal to Earth's radius (R_E = 6.37 x 10^6 m). Taking g0 = 9.80 m/s^2 at the surface, what is the satellite's orbital speed?

  1. 5.59 km/s Correct answer.
  2. 11.2 km/s
  3. 3.95 km/s
  4. 7.91 km/s

Explanation

At radial distance r = 2 R_E, centripetal force equals gravitational force: m v^2 / r = G M_E m / r^2 => v = sqrt(G M_E / r) = sqrt(g0 R_E^2 / (2 R_E)) = sqrt(g0 R_E / 2) = sqrt(9.80 * 6.37 x 10^6 / 2) = sqrt(3.1213 x 10^7) = 5587 m/s = 5.59 km/s.

Question 4Gravity

Given the Moon's mass M_M = 7.35 x 10^22 kg and radius R_M = 1.74 x 10^6 m, what is the escape velocity from the lunar surface? (G = 6.67 x 10^-11 N m^2/kg^2)

  1. 2.37 km/s Correct answer.
  2. 3.35 km/s
  3. 4.74 km/s
  4. 1.68 km/s

Explanation

Escape velocity is v_esc = sqrt(2 G M / R) = sqrt(2 * 6.67 x 10^-11 * 7.35 x 10^22 / 1.74 x 10^6) = sqrt(5.6356 x 10^6) = 2374 m/s = 2.37 km/s.

Question 5Gravity

What is the gravitational potential energy of a 500 kg satellite in a circular orbit of radius r = 1.28 x 10^7 m around Earth? (M_E = 5.97 x 10^24 kg, G = 6.67 x 10^-11 N m^2/kg^2)

  1. -3.11 x 10^10 J
  2. -7.78 x 10^9 J
  3. +1.56 x 10^10 J
  4. -1.56 x 10^10 J Correct answer.

Explanation

Gravitational potential energy is E_p = -G M_E m / r = -(6.67 x 10^-11 * 5.97 x 10^24 * 500) / (1.28 x 10^7) = -(1.991 x 10^17) / (1.28 x 10^7) = -1.555 x 10^10 J.

Question 6Gravity

Planet A orbits a star with radius r_A and orbital period T_A = 1.0 year. Planet B orbits the same star at a radius r_B = 4.0 r_A. According to Kepler's Third Law, what is planet B's orbital period?

  1. 4.0 years
  2. 16.0 years
  3. 8.0 years Correct answer.
  4. 2.0 years

Explanation

Kepler's Third Law states T^2 / r^3 = constant. Thus (T_B / T_A)^2 = (r_B / r_A)^3 = 4.0^3 = 64. Taking square roots gives T_B / T_A = 8.0, so T_B = 8.0 years.

Question 7Gravity

What is the total mechanical energy of a satellite of mass m = 1000 kg orbiting Earth in a circular path of radius r = 2.0 x 10^7 m? (M_E = 5.97 x 10^24 kg, G = 6.67 x 10^-11 N m^2/kg^2)

  1. -4.98 x 10^9 J
  2. -9.95 x 10^9 J Correct answer.
  3. -1.99 x 10^10 J
  4. +9.95 x 10^9 J

Explanation

Total mechanical energy of a circular orbit is E = K + E_p = -G M_E m / (2 r) = -(6.67 x 10^-11 * 5.97 x 10^24 * 1000) / (2 * 2.0 x 10^7) = -3.982 x 10^17 / 4.0 x 10^7 = -9.955 x 10^9 J.

Question 8Gravity

At what distance from Earth's center along the Earth-Moon line is the net gravitational field intensity equal to zero? (Earth mass M_E = 81 M_M, distance D between centers = 3.84 x 10^8 m)

  1. 3.46 x 10^8 m Correct answer.
  2. 1.92 x 10^8 m
  3. 3.07 x 10^8 m
  4. 3.70 x 10^8 m

Explanation

Setting field magnitudes equal: G M_E / x^2 = G M_M / (D - x)^2 => 81 / x^2 = 1 / (D - x)^2. Taking square roots: 9 / x = 1 / (D - x) => 9 D - 9 x = x => 10 x = 9 D => x = 0.9 D = 0.9 * 3.84 x 10^8 = 3.456 x 10^8 m.

Question 9Gravity

What is the orbital radius of a geostationary satellite around Earth with period T = 24 hours (86,400 s)? (M_E = 5.97 x 10^24 kg, G = 6.67 x 10^-11 N m^2/kg^2)

  1. 4.22 x 10^7 m Correct answer.
  2. 3.58 x 10^7 m
  3. 6.37 x 10^6 m
  4. 1.28 x 10^8 m

Explanation

Using Kepler's Third Law r^3 = G M_E T^2 / (4 pi^2) = (6.67 x 10^-11 * 5.97 x 10^24 * 86400^2) / (39.478) = 2.973 x 10^24 / 39.478 = 7.53 x 10^22 m^3. Taking cube root: r = 4.22 x 10^7 m (42,200 km).

Question 10Gravity

What altitude above Earth's surface corresponds to a geostationary orbit? (Earth radius R_E = 6.37 x 10^6 m, geostationary radius r = 4.22 x 10^7 m)

  1. 4.22 x 10^7 m
  2. 3.58 x 10^7 m Correct answer.
  3. 2.58 x 10^7 m
  4. 1.79 x 10^7 m

Explanation

Altitude h is the distance from Earth's surface: h = r - R_E = 4.22 x 10^7 m - 0.637 x 10^7 m = 3.583 x 10^7 m (35,830 km).

Question 11Gravity

What is the ratio of kinetic energy K to the magnitude of potential energy |E_p| for a satellite in a circular gravitational orbit?

  1. 2.0
  2. 0.5 Correct answer.
  3. 0.25
  4. 1.0

Explanation

For a circular orbit, K = G M m / (2r) and |E_p| = G M m / r. Thus K / |E_p| = (1/2) / 1 = 0.5.

Question 12Gravity

How much energy is required to launch a satellite of mass m = 1000 kg from Earth's surface into a circular orbit at radius r = 2 R_E? (g0 = 9.80 m/s^2, R_E = 6.37 x 10^6 m)

  1. 4.68 x 10^10 J Correct answer.
  2. 6.24 x 10^10 J
  3. 3.12 x 10^10 J
  4. 1.56 x 10^10 J

Explanation

Initial energy at surface: E_i = -G M_E m / R_E = -m g0 R_E. Final orbital energy: E_f = -G M_E m / (2 r) = -G M_E m / (4 R_E) = -1/4 m g0 R_E. Energy required delta E = E_f - E_i = 3/4 m g0 R_E = 0.75 * 1000 * 9.80 * 6.37 x 10^6 = 4.682 x 10^10 J.

Question 13Gravity

Two point masses m = 1000 kg each are located at coordinates (3.0 m, 0) and (-3.0 m, 0). What is the gravitational field vector g at point P (0, 4.0 m)? (G = 6.67 x 10^-11 N m^2/kg^2)

  1. -5.34 x 10^-9 j N/kg
  2. -4.27 x 10^-9 j N/kg Correct answer.
  3. 0 N/kg
  4. -2.67 x 10^-9 j N/kg

Explanation

Distance from each mass to P: r = sqrt(3^2 + 4^2) = 5.0 m. Field magnitude from each mass: g1 = G m / r^2 = 6.67 x 10^-11 * 1000 / 25 = 2.668 x 10^-9 N/kg. Horizontal components cancel by symmetry. Vertical components add along -j: g_y = -2 * g1 * cos(theta) where cos(theta) = 4/5 = 0.8. Thus g_y = -2 * 2.668 x 10^-9 * 0.8 = -4.269 x 10^-9 j N/kg.

Question 14Gravity

A comet moves in an elliptical orbit around the Sun. At perihelion it is at distance r_p = 1.0 x 10^11 m with speed v_p = 60 km/s. What is its speed v_a at aphelion where r_a = 3.0 x 10^11 m?

  1. 35 km/s
  2. 10 km/s
  3. 20 km/s Correct answer.
  4. 180 km/s

Explanation

By conservation of angular momentum (Kepler's Second Law), r_p * v_p = r_a * v_a. Thus v_a = v_p * (r_p / r_a) = 60 * (1.0 x 10^11 / 3.0 x 10^11) = 20 km/s.

Question 15Gravity

Which physical law states that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time?

  1. Kepler's First Law
  2. Kepler's Second Law Correct answer.
  3. Kepler's Third Law
  4. Newton's Law of Inertia

Explanation

Kepler's Second Law (Law of Equal Areas) states that a planet sweeps out equal area per unit time, reflecting conservation of orbital angular momentum.

Question 16Gravity

A hypothetical planet has double the Earth's mass (M_p = 2 M_E) and double the Earth's radius (R_p = 2 R_E). What is the surface weight of an astronaut who weighs 700 N on Earth?

  1. 350 N Correct answer.
  2. 700 N
  3. 1400 N
  4. 175 N

Explanation

Surface gravity g_p = G (2 M_E) / (2 R_E)^2 = 2 / 4 * (G M_E / R_E^2) = 0.5 g0. Therefore, weight on the planet is W_p = 0.5 * W_E = 0.5 * 700 N = 350 N.

Question 17Gravity

What is the gravitational potential difference delta V = V(2 R_E) - V(R_E) between an altitude h = R_E and Earth's surface? (g0 = 9.80 N/kg, R_E = 6.37 x 10^6 m)

  1. +3.12 x 10^7 J/kg Correct answer.
  2. -3.12 x 10^7 J/kg
  3. +6.24 x 10^7 J/kg
  4. +1.56 x 10^7 J/kg

Explanation

Gravitational potential V(r) = -G M_E / r. delta V = V(2 R_E) - V(R_E) = -G M_E / (2 R_E) - (-G M_E / R_E) = G M_E / (2 R_E) = 0.5 g0 R_E = 0.5 * 9.80 * 6.37 x 10^6 = +3.121 x 10^7 J/kg.

Question 18Gravity

Why is the gravitational field described as a conservative vector field?

  1. Because gravitational forces are always repulsive
  2. Because work done along any closed path is zero Correct answer.
  3. Because energy is converted into thermal radiation during orbital motion
  4. Because field strength is constant everywhere

Explanation

A vector field is conservative if line integral work around any closed path is zero (oint f * dr = 0), or equivalently, work depends only on initial and final positions.

Question 19Gravity

A 200 kg mass is moved by an external agent at constant speed from Earth's surface to infinity. How much work must be done by the external agent? (g0 = 9.80 m/s^2, R_E = 6.37 x 10^6 m)

  1. 6.24 x 10^9 J
  2. 1.25 x 10^10 J Correct answer.
  3. 0 J
  4. 2.50 x 10^10 J

Explanation

Work required by external agent W_ext = E_p(infinity) - E_p(R_E) = 0 - (-G M_E m / R_E) = m g0 R_E = 200 * 9.80 * 6.37 x 10^6 = 1.2485 x 10^10 J.

Question 20Gravity

If a satellite in circular orbit decreases its orbital radius r, how do its orbital speed v and period T change?

  1. Speed v increases, period T increases
  2. Speed v increases, period T decreases Correct answer.
  3. Speed v decreases, period T decreases
  4. Speed v decreases, period T increases

Explanation

Orbital speed v = sqrt(G M / r) increases as r decreases. Orbital period T = 2 pi r^(3/2) / sqrt(G M) decreases as r decreases.

Question 21Electromagnetic Fields

Two point charges q1 = +4.0 uC and q2 = -2.0 uC are separated by r = 0.30 m in a vacuum. What is the electrostatic force between them? (k_e = 8.99 x 10^9 N m^2/C^2)

  1. 0.24 N (attractive)
  2. 0.80 N (repulsive)
  3. 2.40 N (attractive)
  4. 0.80 N (attractive) Correct answer.

Explanation

Coulomb's Law: F = k_e * |q1 * q2| / r^2 = (8.99 x 10^9) * (4.0 x 10^-6) * (2.0 x 10^-6) / (0.30)^2 = (8.99 x 10^9 * 8.0 x 10^-12) / 0.09 = 0.07192 / 0.09 = 0.799 N ≈ 0.80 N. Opposite charges attract.

Question 22Electromagnetic Fields

What is the electric field magnitude at a distance r = 0.20 m from a point charge q = +5.0 nC in air? (k_e = 8.99 x 10^9 N m^2/C^2)

  1. 562 N/C
  2. 1124 N/C Correct answer.
  3. 4500 N/C
  4. 225 N/C

Explanation

Electric field E = k_e * q / r^2 = (8.99 x 10^9) * (5.0 x 10^-9) / (0.20)^2 = 44.95 / 0.04 = 1123.75 N/C ≈ 1124 N/C.

Question 23Electromagnetic Fields

What is the electrostatic potential V at point P located r = 0.50 m from a point charge q = +6.0 uC in a vacuum? (k_e = 8.99 x 10^9 N m^2/C^2)

  1. 1.08 x 10^5 V Correct answer.
  2. 2.16 x 10^5 V
  3. 5.39 x 10^4 V
  4. 1.08 x 10^6 V

Explanation

Electrostatic potential V = k_e * q / r = (8.99 x 10^9) * (6.0 x 10^-6) / 0.50 = 53,940 / 0.50 = 1.0788 x 10^5 V ≈ 1.08 x 10^5 V.

Question 24Electromagnetic Fields

An electron (q = -1.60 x 10^-19 C, m = 9.11 x 10^-31 kg) is accelerated from rest through a potential difference delta V = 500 V. What is its final kinetic energy?

  1. 8.00 x 10^-17 J Correct answer.
  2. 1.60 x 10^-17 J
  3. 4.00 x 10^-17 J
  4. 3.20 x 10^-17 J

Explanation

Kinetic energy gained K = |q| * delta V = (1.60 x 10^-19 C) * (500 V) = 8.00 x 10^-17 J.

Question 25Electromagnetic Fields

What is the speed of an electron accelerated from rest through a potential difference of 500 V? (m_e = 9.11 x 10^-31 kg, e = 1.60 x 10^-19 C)

  1. 1.33 x 10^7 m/s Correct answer.
  2. 1.76 x 10^14 m/s
  3. 9.37 x 10^6 m/s
  4. 2.65 x 10^7 m/s

Explanation

Using 1/2 m_e v^2 = e * delta V => v = sqrt(2 e delta V / m_e) = sqrt(2 * 1.60 x 10^-19 * 500 / 9.11 x 10^-31) = sqrt(1.7563 x 10^14) = 1.325 x 10^7 m/s.

Question 26Electromagnetic Fields

Two parallel metal plates separated by d = 0.05 m are connected to a 200 V battery creating a uniform electric field. What is the electric field strength E between the plates?

  1. 4000 V/m Correct answer.
  2. 10 V/m
  3. 200 V/m
  4. 1000 V/m

Explanation

In a uniform electric field, E = V / d = 200 V / 0.05 m = 4000 V/m.

Question 27Electromagnetic Fields

According to Gauss's Law, what is the net electric flux through a closed spherical surface enclosing a net charge Q = +8.85 nC in a vacuum? (epsilon_0 = 8.85 x 10^-12 C^2/(N m^2))

  1. 1000 N m^2/C Correct answer.
  2. 78.3 N m^2/C
  3. 1.00 N m^2/C
  4. 8.85 x 10^3 N m^2/C

Explanation

Gauss's Law states net electric flux Phi_E = Q_enclosed / epsilon_0 = (8.85 x 10^-9 C) / (8.85 x 10^-12 C^2/(N m^2)) = 1000 N m^2/C.

Question 28Electromagnetic Fields

Two charges q1 = +3.0 uC and q2 = +3.0 uC are located at (0, 0.30 m) and (0, -0.30 m). What is the net electric potential at P (0.40 m, 0)? (k_e = 8.99 x 10^9 N m^2/C^2)

  1. 0 V
  2. 2.16 x 10^5 V
  3. 1.08 x 10^5 V Correct answer.
  4. 5.39 x 10^4 V

Explanation

Distance from each charge to P: r = sqrt(0.30^2 + 0.40^2) = 0.50 m. Since potential is a scalar: V_net = V1 + V2 = 2 * (k_e * q / r) = 2 * (8.99 x 10^9 * 3.0 x 10^-6 / 0.50) = 2 * (53,940) = 1.0788 x 10^5 V ≈ 1.08 x 10^5 V.

Question 29Electromagnetic Fields

In the scenario of Question 028, what is the net electric field vector E at P (0.40 m, 0)?

  1. 0 N/C
  2. 2.16 x 10^5 i N/C
  3. 1.73 x 10^5 i N/C Correct answer.
  4. 1.30 x 10^5 i N/C

Explanation

Each charge produces field E1 = k_e q / r^2 = 8.99 x 10^9 * 3.0 x 10^-6 / 0.25 = 1.0788 x 10^5 N/C. Vertical components cancel by symmetry. Horizontal components along +i add: E_x = 2 * E1 * cos(theta) where cos(theta) = 0.40 / 0.50 = 0.8. Thus E_x = 2 * 1.0788 x 10^5 * 0.8 = 1.726 x 10^5 i N/C.

Question 30Electromagnetic Fields

How much work is required to move a charge q = +2.0 uC from point A (where V_A = 100 V) to point B (where V_B = 400 V) in an electrostatic field?

  1. 2.0 x 10^-4 J
  2. 8.0 x 10^-4 J
  3. -6.0 x 10^-4 J
  4. 6.0 x 10^-4 J Correct answer.

Explanation

Work done by external agent is W_ext = q * (V_B - V_A) = (2.0 x 10^-6 C) * (400 V - 100 V) = 2.0 x 10^-6 * 300 = 6.0 x 10^-4 J.

Question 31Electromagnetic Fields

What is the electrostatic potential energy of a system of two point charges q1 = +5.0 uC and q2 = +4.0 uC separated by r = 0.20 m in a vacuum? (k_e = 8.99 x 10^9 N m^2/C^2)

  1. 0.90 J Correct answer.
  2. 4.50 J
  3. 0.18 J
  4. -0.90 J

Explanation

U = k_e * q1 * q2 / r = (8.99 x 10^9) * (5.0 x 10^-6) * (4.0 x 10^-6) / 0.20 = 0.1798 / 0.20 = 0.899 J ≈ 0.90 J.

Question 32Electromagnetic Fields

What is the electric flux through one face of a cube that encloses a single point charge Q at its center?

  1. 6 Q / epsilon_0
  2. Q / epsilon_0
  3. Q / (4 pi epsilon_0)
  4. Q / (6 epsilon_0) Correct answer.

Explanation

Total flux through all 6 identical faces of the cube is Phi_total = Q / epsilon_0. By symmetry, flux through one face is Phi = Q / (6 epsilon_0).

Question 33Electromagnetic Fields

A proton (q = +1.60 x 10^-19 C) enters a uniform electric field E = 2000 N/C directed along +i. What is the acceleration of the proton? (m_p = 1.67 x 10^-27 kg)

  1. 1.92 x 10^11 i m/s^2 Correct answer.
  2. 3.20 x 10^-16 i m/s^2
  3. 1.15 x 10^8 i m/s^2
  4. 3.84 x 10^11 i m/s^2

Explanation

Force F = q E = (1.60 x 10^-19 C) * (2000 N/C) = 3.20 x 10^-16 N. Acceleration a = F / m_p = (3.20 x 10^-16) / (1.67 x 10^-27) = 1.916 x 10^11 i m/s^2.

Question 34Electromagnetic Fields

What is the mathematical relationship between electric field vector E and electrostatic potential V?

  1. E = V / r^2
  2. E = grad V
  3. V = -grad E
  4. E = -grad V Correct answer.

Explanation

Electric field is the negative gradient of electric potential: E = -grad V = -(dV/dx i + dV/dy j + dV/dz k).

Question 35Electromagnetic Fields

An electric dipole consists of charges +q and -q separated by distance d. What is the electrostatic potential at any point on the perpendicular bisector of the dipole axis?

  1. -k_e q / d
  2. k_e q / d
  3. 0 V Correct answer.
  4. 2 k_e q / d

Explanation

Every point on the perpendicular bisector is equidistant (r1 = r2 = r) from +q and -q. V = k_e (+q)/r + k_e (-q)/r = 0 V.

Question 36Electromagnetic Fields

What work is done by the electric field when a charge q = -4.0 uC moves through an equipotential surface from V = 50 V to another point on the same surface?

  1. 200 J
  2. 0 J Correct answer.
  3. -2.0 x 10^-4 J
  4. +2.0 x 10^-4 J

Explanation

On an equipotential surface, delta V = 0. Therefore work W = -q * delta V = 0 J.

Question 37Electromagnetic Fields

An infinite flat plane of uniform charge density sigma = +1.77 x 10^-8 C/m^2 is in a vacuum. What is the electric field magnitude E near the plane? (epsilon_0 = 8.85 x 10^-12 C^2/(N m^2))

  1. 1000 N/C Correct answer.
  2. 2000 N/C
  3. 500 N/C
  4. 150 N/C

Explanation

Electric field near an infinite charged plane is E = sigma / (2 epsilon_0) = (1.77 x 10^-8) / (2 * 8.85 x 10^-12) = 1.77 x 10^-8 / 1.77 x 10^-11 = 1000 N/C.

Question 38Electromagnetic Fields

Three point charges q1 = +2.0 uC, q2 = -3.0 uC, and q3 = +5.0 uC are located inside a closed surface. What is the total electric flux passing through the surface? (epsilon_0 = 8.85 x 10^-12 C^2/(N m^2))

  1. 1.13 x 10^6 N m^2/C
  2. 9.04 x 10^5 N m^2/C
  3. 4.52 x 10^5 N m^2/C Correct answer.
  4. 2.26 x 10^5 N m^2/C

Explanation

Net enclosed charge Q_enc = q1 + q2 + q3 = (+2.0 - 3.0 + 5.0) uC = +4.0 uC = 4.0 x 10^-6 C. Net flux Phi = Q_enc / epsilon_0 = (4.0 x 10^-6) / (8.85 x 10^-12) = 4.5198 x 10^5 N m^2/C.

Question 39Electromagnetic Fields

A charged particle moves perpendicular to a uniform electric field E. What is the shape of its trajectory?

  1. Helical
  2. Parabolic Correct answer.
  3. Circular
  4. Straight line

Explanation

Under constant electrostatic force (F = q E) perpendicular to initial velocity, motion has constant velocity along initial direction and constant acceleration along field, resulting in a parabolic trajectory.

Question 40Electromagnetic Fields

A proton (q = +1.60 x 10^-19 C) moves with velocity v = 3.0 x 10^6 i m/s into a uniform magnetic field B = 0.50 k T. What is the Lorentz magnetic force F on the proton?

  1. +2.40 x 10^-13 k N
  2. 0 N
  3. -2.40 x 10^-13 j N Correct answer.
  4. +2.40 x 10^-13 j N

Explanation

Lorentz force F = q (v x B) = (1.60 x 10^-19) * (3.0 x 10^6 i x 0.50 k) = 2.40 x 10^-13 (i x k) = -2.40 x 10^-13 j N, using right-hand rule (i x k = -j).

Question 41Electromagnetic Fields

An electron (m_e = 9.11 x 10^-31 kg, e = 1.60 x 10^-19 C) enters perpendicularly into a uniform magnetic field B = 0.020 T with speed v = 4.0 x 10^6 m/s. What is the radius r of its circular orbit?

  1. 2.28 mm
  2. 0.57 mm
  3. 4.56 mm
  4. 1.14 mm Correct answer.

Explanation

Centripetal magnetic force m v^2 / r = q v B => r = m v / (q B) = (9.11 x 10^-31 * 4.0 x 10^6) / (1.60 x 10^-19 * 0.020) = 3.644 x 10^-24 / 3.20 x 10^-21 = 1.13875 x 10^-3 m = 1.14 mm.

Question 42Electromagnetic Fields

What is the cyclotron frequency f = v / (2 pi r) of the electron in Question 042?

  1. 1.12 x 10^9 Hz
  2. 5.60 x 10^8 Hz Correct answer.
  3. 2.80 x 10^8 Hz
  4. 7.00 x 10^7 Hz

Explanation

Cyclotron frequency f = q B / (2 pi m) = (1.60 x 10^-19 * 0.020) / (2 * pi * 9.11 x 10^-31) = 3.20 x 10^-21 / 5.724 x 10^-30 = 5.590 x 10^8 Hz (559 MHz).

Question 43Electromagnetic Fields

A long straight wire carries a current I = 10 A in a vacuum. What is the magnetic field magnitude B at a distance r = 0.05 m from the wire? (mu_0 = 4 pi x 10^-7 T m/A)

  1. 1.26 x 10^-4 T
  2. 4.0 x 10^-5 T Correct answer.
  3. 8.0 x 10^-5 T
  4. 2.0 x 10^-5 T

Explanation

Ampère's Law / Biot-Savart Law for long wire: B = mu_0 I / (2 pi r) = (4 pi x 10^-7 * 10) / (2 pi * 0.05) = (2 x 10^-6) / 0.05 = 4.0 x 10^-5 T (40 uT).

Question 44Electromagnetic Fields

Two long parallel wires separated by d = 0.10 m carry currents I1 = 5.0 A and I2 = 10.0 A in the same direction. What is the magnetic force per unit length F/L between them?

  1. 1.0 x 10^-4 N/m (attractive) Correct answer.
  2. 1.0 x 10^-4 N/m (repulsive)
  3. 2.0 x 10^-4 N/m (attractive)
  4. 5.0 x 10^-5 N/m (attractive)

Explanation

Force per unit length: F/L = mu_0 I1 I2 / (2 pi d) = (4 pi x 10^-7 * 5.0 * 10.0) / (2 pi * 0.10) = (2 x 10^-6 * 50) / 0.10 = 1.0 x 10^-4 N/m. Parallel currents in the same direction attract.

Question 45Electromagnetic Fields

A circular wire loop of N = 50 turns and area A = 0.040 m^2 is placed perpendicular to a uniform magnetic field B = 0.30 T. What is the magnetic flux Phi through the coil?

  1. 0.012 Wb
  2. 1.20 Wb
  3. 0 Wb
  4. 0.60 Wb Correct answer.

Explanation

Total magnetic flux for N turns: Phi = N B A cos(theta). Perpendicular field means theta = 0° (field parallel to normal vector), so cos 0° = 1. Phi = 50 * 0.30 * 0.040 = 0.60 Wb.

Question 46Electromagnetic Fields

In Question 046, if the magnetic field decreases uniformly from 0.30 T to 0 T in delta t = 0.15 s, what is the magnitude of the induced electromotive force (EMF)?

  1. 4.0 V Correct answer.
  2. 0.08 V
  3. 2.0 V
  4. 8.0 V

Explanation

Faraday's Law of Induction: |EMF| = |delta Phi / delta t| = |(0 - 0.60 Wb) / 0.15 s| = 4.0 V.

Question 47Electromagnetic Fields

What principle governs the direction of induced current specified by Lenz's Law?

  1. Conservation of electric charge
  2. Conservation of energy Correct answer.
  3. Conservation of linear momentum
  4. Gauss's law for magnetism

Explanation

Lenz's Law states that induced current opposes the change in magnetic flux creating it. This opposition ensures that work must be done to induce current, fulfilling conservation of energy.

Question 48Electromagnetic Fields

A straight conducting rod of length L = 0.40 m moves at speed v = 5.0 m/s perpendicular to a uniform magnetic field B = 0.25 T. What motional EMF is induced across the rod?

  1. 0.25 V
  2. 0.50 V Correct answer.
  3. 2.00 V
  4. 1.00 V

Explanation

Motional EMF: EMF = B L v = 0.25 T * 0.40 m * 5.0 m/s = 0.50 V.

Question 49Electromagnetic Fields

A solenoid of length L = 0.50 m with N = 1000 turns carries current I = 4.0 A. What is the magnetic field magnitude B near the center inside the solenoid? (mu_0 = 4 pi x 10^-7 T m/A)

  1. 5.03 x 10^-3 T
  2. 2.01 x 10^-2 T
  3. 1.01 x 10^-2 T Correct answer.
  4. 1.01 x 10^-3 T

Explanation

Solenoid interior field: B = mu_0 (N / L) I = (4 pi x 10^-7) * (1000 / 0.50) * 4.0 = (1.2566 x 10^-6) * 2000 * 4.0 = 1.0053 x 10^-2 T ≈ 1.01 x 10^-2 T (10.1 mT).

Question 50Electromagnetic Fields

A velocity selector uses perpendicular electric field E = 3.0 x 10^4 V/m and magnetic field B = 0.15 T. What speed v must a charged particle have to pass undeflected?

  1. 4.5 x 10^3 m/s
  2. 5.0 x 10^-6 m/s
  3. 2.0 x 10^5 m/s Correct answer.
  4. 1.0 x 10^5 m/s

Explanation

For zero deflection, electric force equals magnetic force: q E = q v B => v = E / B = (3.0 x 10^4 V/m) / (0.15 T) = 2.0 x 10^5 m/s.