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Extremadura PAU Physics practice questions

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Question 1Gravity

Two point masses of m1 = 500 kg and m2 = 800 kg are separated by a distance of r = 2.0 m. Using Newton's law of universal gravitation with G = 6.674 x 10^-11 N*m^2/kg^2, what is the magnitude of the gravitational force between them?

  1. 6.67 x 10^-6 N Correct answer.
  2. 1.33 x 10^-5 N
  3. 3.34 x 10^-6 N
  4. 2.67 x 10^-5 N

Explanation

According to Newton's law of universal gravitation, F = G * m1 * m2 / r^2. Substituting the values: F = (6.674 x 10^-11) * (500) * (800) / (2.0)^2 = (6.674 x 10^-11) * 400,000 / 4 = (6.674 x 10^-11) * 100,000 = 6.674 x 10^-6 N.

Question 2Gravity

Planet A orbits a star at a mean orbital radius rA, with an orbital period of TA = 8.0 years. Planet B orbits the same star at a mean orbital radius rB = 4 * rA. According to Kepler's Third Law (T^2 / r^3 = constant), what is the orbital period of Planet B?

  1. 32 years
  2. 64 years Correct answer.
  3. 16 years
  4. 128 years

Explanation

Kepler's Third Law states that T^2 / r^3 is constant. Therefore, (TB / TA)^2 = (rB / rA)^3 = (4)^3 = 64. Taking the square root gives TB / TA = 8. Thus, TB = 8 * TA = 8 * 8.0 years = 64 years.

Question 3Gravity

What is the gravitational field strength g at an altitude h = RE above the surface of the Earth, where RE is the Earth's radius and g0 = 9.80 m/s^2 is the surface gravitational acceleration?

  1. 4.90 m/s^2
  2. 1.23 m/s^2
  3. 2.45 m/s^2 Correct answer.
  4. 9.80 m/s^2

Explanation

The distance from Earth's center to the altitude h = RE is r = RE + h = 2 * RE. Gravitational field strength varies inversely with the square of distance: g(r) = G * ME / r^2 = G * ME / (2 * RE)^2 = (1/4) * (G * ME / RE^2) = g0 / 4 = 9.80 / 4 = 2.45 m/s^2.

Question 4Gravity

Which statement correctly describes the sign and physical interpretation of gravitational potential energy Ep = -G * M * m / r for a two-body bound system?

  1. Ep is always negative because energy must be added to the system to separate the masses to infinity where Ep = 0. Correct answer.
  2. Ep is always positive because gravity is an attractive force.
  3. Ep is zero at the surface of the primary body and becomes negative at infinity.
  4. Ep is positive for bound orbits and negative for unbound hyperbolic trajectories.

Explanation

Gravitational potential energy is conventionally set to zero at infinity (r -> infinity). Because gravity is attractive, work must be done against the gravitational field to move a mass from distance r to infinity. Hence, Ep(r) < 0 for any finite distance r.

Question 5Gravity

A satellite of mass m = 1000 kg orbits Earth in a circular path at distance r = 4 * RE from Earth's center. Given Earth's surface gravity g0 = 9.80 m/s^2 and RE = 6.37 x 10^6 m, what is the satellite's circular orbital speed?

  1. 7.90 km/s
  2. 5.59 km/s
  3. 3.95 km/s Correct answer.
  4. 11.2 km/s

Explanation

For a circular orbit, gravitational force provides centripetal acceleration: G * ME * m / r^2 = m * vorb^2 / r => vorb = sqrt(G * ME / r). Since g0 = G * ME / RE^2, we have G * ME = g0 * RE^2. Thus vorb = sqrt(g0 * RE^2 / (4 * RE)) = sqrt(g0 * RE / 4) = 0.5 * sqrt(g0 * RE) = 0.5 * sqrt(9.80 * 6.37 x 10^6) = 0.5 * 7901 m/s = 3950 m/s = 3.95 km/s.

Question 6Gravity

What is the escape velocity v_esc from the surface of Earth (RE = 6.37 x 10^6 m, g0 = 9.80 m/s^2)?

  1. 7.90 km/s
  2. 11.2 km/s Correct answer.
  3. 15.8 km/s
  4. 9.80 km/s

Explanation

Escape velocity is obtained by setting total mechanical energy E = E_k + E_p = 0: (1/2) * m * v_esc^2 - G * ME * m / RE = 0 => v_esc = sqrt(2 * G * ME / RE) = sqrt(2 * g0 * RE). Substituting RE = 6.37 x 10^6 m and g0 = 9.80 m/s^2: v_esc = sqrt(2 * 9.80 * 6.37 x 10^6) = sqrt(1.2485 x 10^8) = 11174 m/s approx 11.2 km/s.

Question 7Gravity

According to Kepler's Second Law (Law of Equal Areas), how does the speed of a planet at perihelion (closest distance rp) compare to its speed at aphelion (farthest distance ra)?

  1. Speed at perihelion is higher, satisfying vp * rp = va * ra due to angular momentum conservation. Correct answer.
  2. Speed at aphelion is higher because the planet is farther from the Sun.
  3. Speed is equal at perihelion and aphelion because mechanical energy is constant.
  4. Speed at perihelion is lower because gravitational potential energy is maximum.

Explanation

Kepler's Second Law is a direct consequence of the conservation of angular momentum (L = m * r * v * sin(theta) = constant). At perihelion and aphelion, velocity is perpendicular to the position vector, so L = m * rp * vp = m * ra * va => vp * rp = va * ra. Since rp < ra, vp > va.

Question 8Gravity

A comet in an elliptical orbit around the Sun has a perihelion distance rp = 0.50 AU with speed vp = 54 km/s. What is its orbital speed va at aphelion, located at ra = 4.5 AU?

  1. 6.0 km/s Correct answer.
  2. 12 km/s
  3. 27 km/s
  4. 3.0 km/s

Explanation

Using conservation of angular momentum at perihelion and aphelion: vp * rp = va * ra => va = vp * (rp / ra) = 54 km/s * (0.50 / 4.5) = 54 / 9 = 6.0 km/s.

Question 9Gravity

How much work must be done by an external agent to lift a m = 200 kg satellite from Earth's surface (r1 = RE) to an altitude h = RE (r2 = 2 * RE)? (Use RE = 6.37 x 10^6 m, g0 = 9.80 m/s^2).

  1. 1.25 x 10^10 J
  2. 6.24 x 10^9 J Correct answer.
  3. 3.12 x 10^9 J
  4. 2.49 x 10^10 J

Explanation

Work done W = Ep(r2) - Ep(r1) = -G * ME * m / (2 * RE) - (-G * ME * m / RE) = G * ME * m / (2 * RE) = (1/2) * m * g0 * RE. Substituting m = 200 kg, g0 = 9.80 m/s^2, RE = 6.37 x 10^6 m: W = 0.5 * 200 * 9.80 * 6.37 x 10^6 = 6.2426 x 10^9 J approx 6.24 x 10^9 J.

Question 10Gravity

Two point masses M1 = 16 * M and M2 = M are fixed at a distance d = 10 m apart. At what distance x from M1 along the line connecting them does the net gravitational field intensity equal zero?

  1. 8.0 m Correct answer.
  2. 6.0 m
  3. 5.0 m
  4. 9.0 m

Explanation

The gravitational fields due to M1 and M2 point in opposite directions between the masses. Setting g1 = g2: G * M1 / x^2 = G * M2 / (d - x)^2 => 16 * M / x^2 = M / (10 - x)^2. Taking square roots: 4 / x = 1 / (10 - x) => 4 * (10 - x) = x => 40 - 4x = x => 5x = 40 => x = 8.0 m.

Question 11Gravity

What is the orbital radius r of a geostationary satellite around Earth (ME = 5.97 x 10^24 kg, G = 6.674 x 10^-11 N*m^2/kg^2, T = 86400 s)?

  1. 4.22 x 10^7 m Correct answer.
  2. 3.58 x 10^7 m
  3. 6.37 x 10^6 m
  4. 2.11 x 10^7 m

Explanation

From Kepler's Third Law for circular orbit: T^2 = (4 * pi^2 / (G * ME)) * r^3 => r = (G * ME * T^2 / (4 * pi^2))^(1/3). Substituting G = 6.674 x 10^-11, ME = 5.97 x 10^24 kg, T = 86400 s: G * ME * T^2 = (3.984 x 10^14) * (7.465 x 10^9) = 2.974 x 10^24. Dividing by 39.478: r^3 = 7.533 x 10^22 m^3 => r = 4.22 x 10^7 m (approx 42,200 km).

Question 12Gravity

An exoplanet has mass MP = 3 * ME and radius RP = 2 * RE. If an object weighs W0 = 600 N on Earth's surface, what is its weight WP on the exoplanet's surface?

  1. 900 N
  2. 450 N Correct answer.
  3. 1350 N
  4. 300 N

Explanation

Surface gravity gP = G * MP / RP^2 = G * (3 * ME) / (2 * RE)^2 = (3 / 4) * (G * ME / RE^2) = (3 / 4) * g0. Therefore, WP = m * gP = (3 / 4) * m * g0 = (3 / 4) * 600 N = 450 N.

Question 13Gravity

What is the total mechanical energy E of a satellite of mass m = 500 kg in a stable circular orbit of radius r = 2 * RE around Earth (g0 = 9.80 m/s^2, RE = 6.37 x 10^6 m)?

  1. -7.80 x 10^9 J Correct answer.
  2. -1.56 x 10^10 J
  3. +7.80 x 10^9 J
  4. -3.90 x 10^9 J

Explanation

For a circular orbit, kinetic energy Ek = (1/2) * G * ME * m / r and potential energy Ep = -G * ME * m / r. Thus total energy E = Ek + Ep = -G * ME * m / (2 * r). Substituting r = 2 * RE and G * ME = g0 * RE^2: E = -m * g0 * RE / 4 = -(500 * 9.80 * 6.37 x 10^6) / 4 = -3.1213 x 10^10 / 4 = -7.803 x 10^9 J approx -7.80 x 10^9 J.

Question 14Gravity

What is the escape velocity from the surface of the Moon, given Moon mass MM = 7.35 x 10^22 kg, Moon radius RM = 1.74 x 10^6 m, and G = 6.674 x 10^-11 N*m^2/kg^2?

  1. 2.38 km/s Correct answer.
  2. 1.68 km/s
  3. 4.76 km/s
  4. 11.2 km/s

Explanation

v_esc = sqrt(2 * G * MM / RM) = sqrt(2 * 6.674 x 10^-11 * 7.35 x 10^22 / (1.74 x 10^6)) = sqrt(9.81 x 10^12 / 1.74 x 10^6) = sqrt(5.638 x 10^6) = 2374 m/s = 2.38 km/s.

Question 15Gravity

A space probe of mass m = 400 kg is in a circular orbit of radius r1 = 3 * RE. How much energy must be supplied by engines to transfer it to a higher circular orbit of radius r2 = 6 * RE?

  1. 1.04 x 10^9 J
  2. 2.08 x 10^9 J Correct answer.
  3. 4.16 x 10^9 J
  4. 5.20 x 10^8 J

Explanation

The mechanical energy in a circular orbit of radius r is E(r) = -G * ME * m / (2 * r) = -m * g0 * RE^2 / (2 * r). The energy required to transfer between orbits is Delta E = E(r2) - E(r1) = (-m * g0 * RE / 12) - (-m * g0 * RE / 6) = m * g0 * RE / 12. Substituting m = 400 kg, g0 = 9.80 m/s^2, and RE = 6.37 x 10^6 m yields Delta E = (400 * 9.80 * 6.37 x 10^6) / 12 = 2.08 x 10^9 J.

Question 16Gravity

What is the ratio of gravitational potential V(r) at distance r = 2 * RE from Earth's center to the gravitational field strength magnitude g(r) at the same point?

  1. -r = -2 * RE Correct answer.
  2. -RE
  3. -0.5 * RE
  4. -4 * RE

Explanation

Gravitational potential is V(r) = -G * ME / r. Gravitational field magnitude is g(r) = G * ME / r^2. Taking the ratio: V(r) / g(r) = (-G * ME / r) / (G * ME / r^2) = -r. At r = 2 * RE, V / g = -2 * RE.

Question 17Gravity

A rocket is launched vertically from Earth's surface with initial speed v0 = 0.80 * v_esc. Neglecting air resistance, what is the maximum altitude h reached by the rocket above Earth's surface?

  1. 0.78 * RE
  2. 1.78 * RE Correct answer.
  3. 0.64 * RE
  4. 2.56 * RE

Explanation

Conservation of mechanical energy: (1/2) * m * v0^2 - G * ME * m / RE = -G * ME * m / (RE + h). Since v0 = 0.80 * sqrt(2 * G * ME / RE), we have v0^2 = 0.64 * (2 * G * ME / RE) = 1.28 * G * ME / RE. Thus: 0.64 * (G * ME / RE) - G * ME / RE = -G * ME / (RE + h) => -0.36 * (G * ME / RE) = -G * ME / (RE + h) => RE + h = RE / 0.36 = 2.778 * RE => h = 1.78 * RE.

Question 18Gravity

For a satellite in an elliptical orbit with semi-major axis a and eccentricity e, what is the ratio of perihelion speed vp to aphelion speed va?

  1. (1 + e) / (1 - e) Correct answer.
  2. (1 - e) / (1 + e)
  3. sqrt((1 + e) / (1 - e))
  4. (1 + e)^2 / (1 - e)^2

Explanation

Perihelion distance is rp = a * (1 - e) and aphelion distance is ra = a * (1 + e). By conservation of angular momentum: vp * rp = va * ra => vp / va = ra / rp = (a * (1 + e)) / (a * (1 - e)) = (1 + e) / (1 - e).

Question 19Gravity

Two identical stars of mass M = 2.0 x 10^30 kg orbit their common center of mass in circular orbits of radius R = 1.0 x 10^11 m. What is their orbital period T? (G = 6.674 x 10^-11 N*m^2/kg^2).

  1. 2.43 x 10^7 s (approx 281 days) Correct answer.
  2. 1.72 x 10^7 s (approx 199 days)
  3. 3.44 x 10^7 s (approx 398 days)
  4. 4.86 x 10^7 s (approx 562 days)

Explanation

The distance between the two stars is d = 2 * R. The gravitational force on each star is F = G * M^2 / (4 * R^2), providing centripetal acceleration for circular radius R: G * M^2 / (4 * R^2) = M * (4 * pi^2 / T^2) * R. Solving for T yields T = 4 * pi * sqrt(R^3 / (G * M)). Substituting R = 1.0 x 10^11 m, M = 2.0 x 10^30 kg, and G = 6.674 x 10^-11 N*m^2/kg^2 gives T = 3.44 x 10^7 s.

Question 20Gravity

What is the Gauss's Law formulation for gravitational fields, and what does it imply about the field inside a thin uniform spherical shell of mass M and radius R?

  1. Flux = -4 * pi * G * Minclosed; the gravitational field inside a uniform spherical shell is zero. Correct answer.
  2. Flux = +4 * pi * G * Minclosed; the field inside increases linearly with distance from center.
  3. Flux = 0 always; the field inside is constant and equal to surface field g0.
  4. Flux = -G * Minclosed / R^2; the field inside is directed outward.

Explanation

Gauss's law for gravity states that the gravitational flux through any closed surface is Phi_g = closed_integral(g * dA) = -4 * pi * G * Minclosed. Inside a thin uniform spherical shell of radius R, for a concentric Gaussian sphere of radius r < R, Minclosed = 0, so g(r) = 0 everywhere inside.

Question 21Electromagnetic Fields

Two point charges q1 = +2.0 uC and q2 = -4.0 uC are located 0.30 m apart in vacuum (K = 8.99 x 10^9 N*m^2/C^2). What is the magnitude of the electrostatic force between them?

  1. 0.80 N Correct answer.
  2. 0.24 N
  3. 2.40 N
  4. 0.48 N

Explanation

According to Coulomb's Law, F = K * |q1 * q2| / r^2. Substituting values: F = (8.99 x 10^9) * (2.0 x 10^-6) * (4.0 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 approx 0.80 N.

Question 22Electromagnetic Fields

What is the electric field intensity E created by a point charge q = +5.0 nC at a distance r = 0.50 m in vacuum (K = 8.99 x 10^9 N*m^2/C^2)?

  1. 180 N/C Correct answer.
  2. 90 N/C
  3. 360 N/C
  4. 45 N/C

Explanation

E = K * |q| / r^2 = (8.99 x 10^9) * (5.0 x 10^-9) / (0.50)^2 = 44.95 / 0.25 = 179.8 N/C approx 180 N/C directed radially outward.

Question 23Electromagnetic Fields

What is the electric potential V created by a point charge q = -10 nC at a distance r = 2.0 m in vacuum (K = 8.99 x 10^9 N*m^2/C^2)?

  1. -45 V Correct answer.
  2. -22.5 V
  3. +45 V
  4. -90 V

Explanation

V = K * q / r = (8.99 x 10^9) * (-10 x 10^-9) / 2.0 = -89.9 / 2.0 = -44.95 V approx -45 V.

Question 24Electromagnetic Fields

How much work is done by the electrostatic field when moving a positive charge q = +2.0 uC from point A (VA = 100 V) to point B (VB = 40 V)?

  1. +1.2 x 10^-4 J Correct answer.
  2. -1.2 x 10^-4 J
  3. +2.0 x 10^-4 J
  4. +8.0 x 10^-5 J

Explanation

Work done by conservative electric field is W_field = -q * Delta V = -q * (VB - VA) = q * (VA - VB) = (2.0 x 10^-6 C) * (100 V - 40 V) = 2.0 x 10^-6 * 60 = +1.2 x 10^-4 J.

Question 25Electromagnetic Fields

A charge q = +3.0 uC moves with velocity v = 4.0 x 10^5 m/s perpendicular to a uniform magnetic field B = 0.50 T. What is the magnitude of the magnetic Lorentz force acting on the charge?

  1. 0.60 N Correct answer.
  2. 1.20 N
  3. 0.30 N
  4. 6.0 N

Explanation

The magnetic Lorentz force magnitude is FB = |q| * v * B * sin(theta). Since velocity is perpendicular to B, sin(90 deg) = 1: FB = (3.0 x 10^-6 C) * (4.0 x 10^5 m/s) * (0.50 T) = 1.2 * 0.50 = 0.60 N.

Question 26Electromagnetic Fields

An electron (m = 9.11 x 10^-31 kg, e = 1.60 x 10^-19 C) enters a uniform magnetic field B = 0.020 T perpendicularly with speed v = 2.0 x 10^6 m/s. What is the radius r of its circular orbit?

  1. 0.57 mm Correct answer.
  2. 1.14 mm
  3. 0.28 mm
  4. 2.28 mm

Explanation

Centripetal force is provided by magnetic Lorentz force: m * v^2 / r = e * v * B => r = m * v / (e * B). Substituting values: r = (9.11 x 10^-31 kg * 2.0 x 10^6 m/s) / (1.60 x 10^-19 C * 0.020 T) = (1.822 x 10^-24) / (3.20 x 10^-21) = 5.69 x 10^-4 m = 0.57 mm.

Question 27Electromagnetic Fields

What is the magnitude of the magnetic field B at distance r = 0.05 m from a long straight wire carrying a steady current I = 10 A in vacuum? (mu0 = 4*pi x 10^-7 T*m/A).

  1. 4.0 x 10^-5 T Correct answer.
  2. 2.0 x 10^-5 T
  3. 8.0 x 10^-5 T
  4. 1.0 x 10^-5 T

Explanation

According to the Biot-Savart / Ampere's Law for a long straight wire: B = mu0 * 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 28Electromagnetic Fields

Two long parallel straight 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 and its nature?

  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, repulsive

Explanation

Force per unit length between parallel currents: F/L = mu0 * 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. Currents in the SAME direction produce an ATTRACTIVE force.

Question 29Electromagnetic Fields

A planar loop of area A = 0.040 m^2 is placed in a uniform magnetic field B = 0.30 T. The normal vector to the loop makes an angle theta = 60 deg with the magnetic field. What is the magnetic flux Phi through the loop?

  1. 0.0060 Wb Correct answer.
  2. 0.0104 Wb
  3. 0.0120 Wb
  4. 0.0030 Wb

Explanation

Magnetic flux is Phi = B * A * cos(theta) = (0.30 T) * (0.040 m^2) * cos(60 deg) = 0.0120 * 0.50 = 0.0060 Wb (6.0 mWb).

Question 30Electromagnetic Fields

The magnetic flux through a single loop decreases uniformly from Phi1 = 0.50 Wb to Phi2 = 0.10 Wb in Delta t = 0.20 s. According to Faraday's Law, what is the magnitude of the induced electromotive force (EMF)?

  1. 2.0 V Correct answer.
  2. 1.0 V
  3. 0.80 V
  4. 4.0 V

Explanation

Faraday's Law states magnitude |EMF| = |Delta Phi / Delta t| = |(0.10 - 0.50) / 0.20| = |-0.40 / 0.20| = 2.0 V.

Question 31Electromagnetic Fields

Two positive charges q1 = +4.0 uC and q2 = +9.0 uC are fixed at positions x1 = 0 m and x2 = 1.0 m on the x-axis. At what position x on the x-axis is the net electric field intensity zero?

  1. x = 0.40 m Correct answer.
  2. x = 0.31 m
  3. x = 0.50 m
  4. x = 0.60 m

Explanation

Between two positive charges, net field is zero where E1 = E2: K * q1 / x^2 = K * q2 / (1.0 - x)^2 => 4 / x^2 = 9 / (1.0 - x)^2. Taking square roots: 2 / x = 3 / (1.0 - x) => 2 * (1.0 - x) = 3x => 2 - 2x = 3x => 5x = 2 => x = 0.40 m.

Question 32Electromagnetic Fields

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

  1. 500 eV (8.0 x 10^-17 J) and 1.33 x 10^7 m/s Correct answer.
  2. 250 eV (4.0 x 10^-17 J) and 9.38 x 10^6 m/s
  3. 500 eV (8.0 x 10^-17 J) and 1.77 x 10^6 m/s
  4. 1000 eV (1.6 x 10^-16 J) and 1.88 x 10^7 m/s

Explanation

Kinetic energy gained Ek = e * Delta V = (1.60 x 10^-19 C) * (500 V) = 8.0 x 10^-17 J = 500 eV. Using classical Ek = (1/2) * m * v^2 => v = sqrt(2 * Ek / m) = sqrt(2 * 8.0 x 10^-17 / (9.11 x 10^-31)) = sqrt(1.756 x 10^14) = 1.325 x 10^7 m/s approx 1.33 x 10^7 m/s.

Question 33Electromagnetic Fields

A velocity selector uses uniform electric E = 3.0 x 10^4 N/C and magnetic B = 0.15 T fields perpendicular to each other. What speed v must a charged particle have to pass through undeflected?

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

Explanation

For zero net force in a velocity selector, FE = FB => q * E = q * v * B => v = E / B = (3.0 x 10^4 N/C) / (0.15 T) = 2.0 x 10^5 m/s.

Question 34Electromagnetic Fields

What is the cyclotron frequency f of a proton (mp = 1.67 x 10^-27 kg, e = 1.60 x 10^-19 C) orbiting in a magnetic field B = 0.50 T?

  1. 7.63 MHz Correct answer.
  2. 15.3 MHz
  3. 3.82 MHz
  4. 47.9 MHz

Explanation

Cyclotron frequency f = q * B / (2 * pi * m) = (1.60 x 10^-19 C * 0.50 T) / (2 * pi * 1.67 x 10^-27 kg) = (8.0 x 10^-20) / (1.0493 x 10^-26) = 7.624 x 10^6 Hz = 7.63 MHz.

Question 35Electromagnetic Fields

What is the magnetic field intensity B at the center of a circular coil of N = 100 turns, radius R = 0.050 m, carrying a current I = 2.0 A in vacuum? (mu0 = 4*pi x 10^-7 T*m/A).

  1. 2.51 x 10^-3 T Correct answer.
  2. 5.03 x 10^-3 T
  3. 1.26 x 10^-3 T
  4. 7.54 x 10^-3 T

Explanation

Magnetic field at the center of a flat circular coil: B = N * mu0 * I / (2 * R) = (100 * 4*pi x 10^-7 * 2.0) / (2 * 0.050) = (2.513 x 10^-4) / 0.10 = 2.513 x 10^-3 T (2.51 mT).

Question 36Electromagnetic Fields

A conducting rod of length L = 0.50 m moves at constant speed v = 4.0 m/s perpendicular to a uniform magnetic field B = 0.30 T on parallel conducting rails. What is the induced EMF across the rod?

  1. 0.60 V Correct answer.
  2. 1.20 V
  3. 0.30 V
  4. 2.40 V

Explanation

Motional EMF for a rod moving perpendicular to B field: EMF = B * L * v = (0.30 T) * (0.50 m) * (4.0 m/s) = 0.60 V.

Question 37Electromagnetic Fields

A rectangular coil of N = 50 turns and area A = 0.020 m^2 rotates at a constant angular speed omega = 100 rad/s in a uniform magnetic field B = 0.40 T. What is the peak induced voltage (EMF0)?

  1. 40 V Correct answer.
  2. 400 V
  3. 4.0 V
  4. 80 V

Explanation

Magnetic flux is Phi(t) = N * B * A * cos(omega * t). By Faraday's Law, induced EMF(t) = -dPhi/dt = N * B * A * omega * sin(omega * t). The peak voltage EMF0 = N * B * A * omega = 50 * 0.40 * 0.020 * 100 = 40 V.

Question 38Electromagnetic Fields

An inductor with self-inductance L = 0.040 H carries a current decreasing at a constant rate dI/dt = -50 A/s. What is the self-induced EMF across the inductor?

  1. +2.0 V Correct answer.
  2. -2.0 V
  3. +0.80 V
  4. +1.25 V

Explanation

Self-induced EMF is given by Faraday/Lenz law: EMF = -L * (dI/dt) = -(0.040 H) * (-50 A/s) = +2.0 V. The positive sign indicates EMF opposes current decrease.

Question 39Electromagnetic Fields

Three identical positive point charges q = +3.0 uC are located at the vertices of an equilateral triangle of side length a = 0.50 m in vacuum (K = 8.99 x 10^9 N*m^2/C^2). What is the total electrostatic potential energy U of the system?

  1. 0.49 N*m (J) Correct answer.
  2. 0.16 J
  3. 0.32 J
  4. 0.97 J

Explanation

Total potential energy of a system of 3 charges is U = U12 + U13 + U23 = 3 * (K * q^2 / a) = 3 * (8.99 x 10^9 * (3.0 x 10^-6)^2 / 0.50) = 3 * (8.99 x 10^9 * 9.0 x 10^-12 / 0.50) = 3 * (0.08091 / 0.50) = 3 * 0.1618 = 0.4854 J approx 0.49 J.

Question 40Electromagnetic Fields

A square loop of side L = 0.10 m carrying current I = 2.0 A is placed in a uniform magnetic field B = 0.50 T. What is the maximum torque tau_max exerted on the loop?

  1. 0.010 N*m Correct answer.
  2. 0.10 N*m
  3. 0.005 N*m
  4. 0.020 N*m

Explanation

Magnetic dipole moment mu = I * A = (2.0 A) * (0.10 m)^2 = 0.020 A*m^2. Maximum torque tau_max = mu * B = (0.020 A*m^2) * (0.50 T) = 0.010 N*m.

Question 41Electromagnetic Fields

What is Gauss's Law for magnetic fields (closed_integral B * dA = 0) and what physical principle does it express?

  1. Magnetic monopoles do not exist; net magnetic flux through any closed surface is always zero. Correct answer.
  2. Magnetic fields are created by static electric charges.
  3. Magnetic flux through a closed loop changes whenever current flows.
  4. Magnetic fields are strictly conservative scalar fields.

Explanation

Gauss's law for magnetism states that the net magnetic flux through any closed surface is zero. This implies that magnetic field lines are continuous loops without sources or sinks, meaning isolated magnetic monopoles do not exist in nature.

Question 42Electromagnetic Fields

A charged particle (+q, mass m) enters a region with uniform electric E and magnetic B fields aligned in parallel along the +z axis with initial velocity v in the +x direction. What trajectory does it follow?

  1. Helical trajectory with increasing pitch along the +z axis. Correct answer.
  2. Uniform circular motion in the xy plane.
  3. Straight line trajectory along the +x axis.
  4. Parabolic path confined to the xz plane.

Explanation

Magnetic field B along +z produces Lorentz force FB = q * v_perp x B in xy plane, causing uniform circular motion in xy plane. Electric field E along +z exerts force FE = q*E along +z, causing constant acceleration a_z = qE/m in z direction. Combined, the trajectory is a helix with continuously increasing pitch along z.

Question 43Electromagnetic Fields

In a mass spectrometer, singly ionized isotopes 35Cl+ (m1 = 35 u) and 37Cl+ (m2 = 37 u) are accelerated through potential difference V and enter magnetic field B perpendicularly. What is the ratio of their orbital radii r2 / r1?

  1. sqrt(37 / 35) approx 1.028 Correct answer.
  2. 37 / 35 approx 1.057
  3. (37 / 35)^2 approx 1.118
  4. 1.000

Explanation

Ion speed after acceleration V is v = sqrt(2 * q * V / m). Radius in magnetic field is r = m * v / (q * B) = (1/B) * sqrt(2 * m * V / q). Therefore, radius is proportional to sqrt(m). The ratio r2 / r1 = sqrt(m2 / m1) = sqrt(37 / 35) approx 1.028.

Question 44Electromagnetic Fields

A circular loop of radius r0 = 0.10 m is placed in a spatially uniform magnetic field pointing perpendicular to the loop plane. The magnetic field magnitude varies in time as B(t) = B0 * cos(omega * t), with B0 = 0.50 T and omega = 100 pi rad/s. What is the induced electric field amplitude E0 along the loop?

  1. 7.85 V/m Correct answer.
  2. 15.7 V/m
  3. 2.50 V/m
  4. 5.00 V/m

Explanation

By Faraday's law in integral form: closed_integral(E * dl) = -dPhi/dt => E * (2 * pi * r0) = -A * dB/dt = -(pi * r0^2) * (-B0 * omega * sin(omega * t)). Thus E(t) = (r0 / 2) * B0 * omega * sin(omega * t). Amplitude E0 = (r0 / 2) * B0 * omega = (0.10 / 2) * 0.50 * (100 pi) = 0.05 * 0.50 * 314.16 = 7.854 V/m approx 7.85 V/m.

Question 45Electromagnetic Fields

Two long parallel wires separated by distance d carry equal currents I in opposite directions. What is the magnitude of the net magnetic field B at a point midway between them?

  1. mu0 * I / (pi * d) Correct answer.
  2. 2 * mu0 * I / (pi * d)
  3. 0
  4. mu0 * I / (2 * pi * d)

Explanation

Midway between the wires, the distance to each wire is r = d / 2. By the right-hand rule, opposite currents produce magnetic fields in the same direction at the midpoint. The magnetic field of a single wire at distance d/2 is B1 = mu0 * I / (2 * pi * (d/2)) = mu0 * I / (pi * d). Adding the two fields gives B_net = B1 + B2 = 2 * mu0 * I / (pi * d).

Question 46Waves Optics

A transverse harmonic wave has frequency f = 400 Hz and propagates with speed v = 320 m/s. What is its wavelength lambda?

  1. 0.80 m Correct answer.
  2. 1.25 m
  3. 0.64 m
  4. 1.60 m

Explanation

Using wave equation v = f * lambda => lambda = v / f = (320 m/s) / (400 Hz) = 0.80 m.

Question 47Waves Optics

What is the phase difference Delta phi (in radians) between two points separated by distance Delta x = 0.25 * lambda along the direction of propagation of a harmonic wave?

  1. pi / 2 rad (90 deg) Correct answer.
  2. pi rad (180 deg)
  3. pi / 4 rad (45 deg)
  4. 2 * pi rad (360 deg)

Explanation

Phase difference Delta phi = k * Delta x = (2 * pi / lambda) * (0.25 * lambda) = 2 * pi * 0.25 = pi / 2 rad (90 degrees).

Question 48Waves Optics

What is the sound intensity level beta (in decibels) for a sound wave of intensity I = 1.0 x 10^-5 W/m^2? (Reference threshold I0 = 1.0 x 10^-12 W/m^2).

  1. 70 dB Correct answer.
  2. 50 dB
  3. 120 dB
  4. 90 dB

Explanation

beta = 10 * log10(I / I0) = 10 * log10((1.0 x 10^-5) / (1.0 x 10^-12)) = 10 * log10(10^7) = 10 * 7 = 70 dB.

Question 49Waves Optics

Light travels from air (n1 = 1.00) into glass (n2 = 1.50) at an incident angle theta1 = 30.0 deg. According to Snell's law, what is the angle of refraction theta2?

  1. 19.5 deg Correct answer.
  2. 48.6 deg
  3. 22.5 deg
  4. 15.0 deg

Explanation

Snell's Law: n1 * sin(theta1) = n2 * sin(theta2) => 1.00 * sin(30.0 deg) = 1.50 * sin(theta2) => 0.500 = 1.50 * sin(theta2) => sin(theta2) = 0.500 / 1.50 = 0.3333. Taking arcsin(0.3333) = 19.47 deg approx 19.5 deg.

Question 50Waves Optics

What is the critical angle theta_c for total internal reflection at an interface between water (n1 = 1.33) and air (n2 = 1.00)?

  1. 48.8 deg Correct answer.
  2. 41.8 deg
  3. 53.1 deg
  4. 30.0 deg

Explanation

Critical angle sin(theta_c) = n2 / n1 = 1.00 / 1.33 = 0.7519 => theta_c = arcsin(0.7519) = 48.75 deg approx 48.8 deg.