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Balearic PAU Physics study materials
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Balearic PAU Physics practice questions

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Question 1Gravitational field

According to Newton's law of universal gravitation, the force between two point masses separated by a distance r is proportional to which quantity?

  1. The product of the masses divided by the square of the distance Correct answer.
  2. The sum of the masses divided by the square of the distance
  3. The product of the masses divided by the distance
  4. The sum of the masses divided by the distance

Explanation

Newton's law states F = G·m1·m2/r², so the force is directly proportional to the product of the masses and inversely proportional to the square of their separation. This inverse-square dependence on distance is the defining feature of the law.

Question 2Gravitational field

The gravitational field strength g at a point in space is defined as which of the following?

  1. The gravitational force per unit mass placed at that point Correct answer.
  2. The work needed to move a unit mass to that point
  3. The gravitational potential energy of a unit mass at that point
  4. The total gravitational force on a test mass at that point

Explanation

Field strength is force per unit mass, g = F/m, a vector quantity measured in N/kg (equivalent to m/s²). Potential energy and work per unit mass describe potential, not field strength, which is why those options are distinct concepts.

Question 3Gravitational field

Two spherical masses are moved so that the distance between their centres is doubled while the masses stay the same. The gravitational force between them becomes:

  1. Twice its original value
  2. Four times its original value
  3. One quarter of its original value Correct answer.
  4. One half of its original value

Explanation

Because F ∝ 1/r², doubling r multiplies the force by 1/2² = 1/4. The inverse-square law makes the force fall to a quarter of the initial value.

Question 4Gravitational field

Which statement correctly describes the gravitational potential at a point in the field of an isolated point mass?

  1. It is a vector pointing towards the mass
  2. It is positive and decreases with distance
  3. It is zero at the surface of the mass
  4. It is negative and tends to zero at infinity Correct answer.

Explanation

Gravitational potential is defined as V = -G·M/r, taking zero at infinity, so it is always negative and increases (becomes less negative) with distance. Potential is a scalar, so it has no direction.

Question 5Gravitational field

The escape velocity from the surface of a planet of mass M and radius R is given by which expression?

  1. √(2GM/R²)
  2. √(GM/R)
  3. 2GM/R
  4. √(2GM/R) Correct answer.

Explanation

Setting kinetic energy equal to the magnitude of gravitational potential energy, ½mv² = GMm/R, gives v = √(2GM/R). The factor 2 arises because potential energy is -GMm/R and must be fully overcome.

Question 6Gravitational field

For a satellite in a circular orbit of radius r around a planet of mass M, the orbital speed v satisfies:

  1. v² = GMr
  2. v² = GM/r²
  3. v² = GM/r Correct answer.
  4. v² = 2GM/r

Explanation

Equating gravitational force to centripetal force, GMm/r² = mv²/r, gives v² = GM/r. The orbital speed therefore decreases as the orbital radius increases.

Question 7Gravitational field

Kepler's third law relates the orbital period T of a planet to the semi-major axis a of its orbit. Which proportionality is correct?

  1. T² ∝ a³ Correct answer.
  2. T³ ∝ a²
  3. T² ∝ a
  4. T ∝ a²

Explanation

Kepler's third law states that the square of the period is proportional to the cube of the semi-major axis, T² ∝ a³. This follows from Newtonian gravitation for orbits around the same central mass.

Question 8Gravitational field

The gravitational field strength at the surface of a planet is 10 N/kg. What is the weight of an astronaut of mass 70 kg standing on that surface?

  1. 7 N
  2. 70 N
  3. 700 N Correct answer.
  4. 7000 N

Explanation

Weight is the gravitational force W = m·g = 70 × 10 = 700 N. Field strength in N/kg multiplied by mass in kg gives force in newtons.

Question 9Gravitational field

A satellite moves in a circular orbit around the Earth. Which statement about its gravitational potential energy U and kinetic energy K is correct?

  1. U is negative and K = -U
  2. U is negative and K = -U/2 Correct answer.
  3. U is positive and K = U
  4. U is zero and K equals the total energy

Explanation

For a circular orbit U = -GMm/r and K = GMm/2r, so K = -U/2 and the total energy is E = U/2, which is negative. This is the virial relation for inverse-square orbits.

Question 10Gravitational field

Two point masses, one of 2 kg and one of 8 kg, are separated by 4 m. Taking G = 6.67 × 10⁻¹¹ N·m²/kg², the gravitational force between them is approximately:

  1. 6.7 × 10⁻¹¹ N Correct answer.
  2. 5.3 × 10⁻¹⁰ N
  3. 2.7 × 10⁻¹⁰ N
  4. 1.3 × 10⁻¹⁰ N

Explanation

F = G·m1·m2/r² = (6.67 × 10⁻¹¹ × 2 × 8)/16 = (6.67 × 10⁻¹¹ × 16)/16 = 6.67 × 10⁻¹¹ N. The product of the masses divided by r² happens to equal 1 here.

Question 11Gravitational field

At what height above the Earth's surface does the gravitational field strength become one quarter of its surface value? Take the Earth's radius as R.

  1. h = 4R
  2. h = R/2
  3. h = R Correct answer.
  4. h = 2R

Explanation

Field strength varies as g ∝ 1/r², where r is measured from the Earth's centre. For g to fall to one quarter, r must double, so r = 2R and the height above the surface is h = R.

Question 12Gravitational field

A satellite is in a circular orbit at radius r around a planet. Which change would increase its orbital speed?

  1. Moving it to a larger orbital radius
  2. Increasing its own mass
  3. Increasing the mass of the satellite and decreasing the planet's mass
  4. Moving it to a smaller orbital radius Correct answer.

Explanation

Orbital speed is v = √(GM/r), which depends only on the central mass and the orbital radius. Reducing r increases v, while the satellite's own mass cancels out of the equation.

Question 13Gravitational field

The gravitational potential at a distance r from a point mass M is V = -GM/r. The potential difference between r = R and r = 2R is:

  1. -GM/(2R)
  2. GM/(2R) Correct answer.
  3. GM/R
  4. 3GM/(2R)

Explanation

V(2R) - V(R) = -GM/(2R) - (-GM/R) = GM/R - GM/(2R) = GM/(2R). The potential becomes less negative with distance, so the difference is positive.

Question 14Gravitational field

A body is projected vertically from the surface of the Earth with a speed equal to half the escape velocity. Ignoring air resistance, the maximum height reached, in terms of the Earth's radius R, is:

  1. R/3 Correct answer.
  2. R/4
  3. 2R/3
  4. R/2

Explanation

Energy conservation gives ½v² = GM(1/R - 1/r) with v² = ¼(2GM/R) = GM/(2R). Solving yields 1/r = 1/R - 1/(4R) = 3/(4R), so r = 4R/3 and h = R/3.

Question 15Gravitational field

Kepler's second law states that the line joining a planet to the Sun sweeps out equal areas in equal times. This is a direct consequence of which conservation law?

  1. Conservation of linear momentum
  2. Conservation of mass
  3. Conservation of mechanical energy
  4. Conservation of angular momentum Correct answer.

Explanation

The gravitational force is central, so it exerts no torque about the Sun and angular momentum L = m·r²·dθ/dt is constant. Constant areal velocity ½r²dθ/dt is exactly the statement of Kepler's second law.

Question 16Gravitational field

A planet has twice the mass of the Earth and twice its radius. The acceleration due to gravity at its surface, compared with the Earth's, is:

  1. Four times the Earth's value
  2. Twice the Earth's value
  3. Half the Earth's value Correct answer.
  4. The same as the Earth's value

Explanation

Surface gravity is g = GM/R². Doubling both M and R gives g' = G(2M)/(2R)² = (2/4)·GM/R² = g/2, so the surface gravity is halved.

Question 17Gravitational field

The gravitational potential energy of a mass m at the Earth's surface is often written as mgh for small heights h. This approximation is valid because:

  1. The gravitational constant G changes near the surface
  2. The potential energy is exactly linear in h at all heights
  3. The field is nearly uniform and h is much smaller than the Earth's radius Correct answer.
  4. The mass m becomes negligible at large heights

Explanation

Near the surface g is approximately constant, so the change in potential energy is mgh. Expanding -GMm/r about r = R gives mgh plus small correction terms of order h/R, negligible when h ≪ R.

Question 18Gravitational field

A geostationary satellite orbits the Earth above the equator. Which combination of properties must it have?

  1. A period of 12 hours and an equatorial orbit
  2. A period of 24 hours and any orbital radius
  3. A period of 24 hours and an orbit passing over the poles
  4. A period of 24 hours and an orbit in the plane of the equator Correct answer.

Explanation

A geostationary satellite must complete one orbit per sidereal day and remain fixed above one point of the equator, so its orbit must be equatorial with a period of about 24 hours. This fixes a unique orbital radius.

Question 19Gravitational field

The gravitational field inside a uniform solid sphere of density ρ at a distance r from its centre is:

  1. (4/3)πGρr, directed towards the centre Correct answer.
  2. Zero everywhere inside the sphere
  3. GM/r², directed towards the centre
  4. Constant and equal to the surface value

Explanation

Only the mass within radius r contributes, M(r) = (4/3)πρr³, so g = GM(r)/r² = (4/3)πGρr. The field grows linearly with r inside and falls as 1/r² outside.

Question 20Gravitational field

A rocket of mass 1000 kg is at a point where the gravitational potential is -6.0 × 10⁷ J/kg. What is its gravitational potential energy?

  1. -6.0 × 10¹³ J
  2. -6.0 × 10¹⁰ J Correct answer.
  3. -6.0 × 10⁴ J
  4. 6.0 × 10¹⁰ J

Explanation

Potential energy is U = m·V = 1000 × (-6.0 × 10⁷) = -6.0 × 10¹⁰ J. The negative sign reflects that the mass is bound in the field.

Question 21Gravitational field

Two satellites orbit the same planet with orbital radii r and 4r. What is the ratio of their orbital periods T₁/T₂?

  1. 1/8 Correct answer.
  2. 1/2
  3. 1/4
  4. 1/64

Explanation

Kepler's third law gives T² ∝ r³, so T₁/T₂ = (r/4r)^(3/2) = (1/4)^(3/2) = 1/8. The larger orbit has eight times the period of the smaller one.

Question 22Gravitational field

A body is launched from the Earth's surface with a speed equal to the escape velocity. Assuming no other forces act, its total mechanical energy is:

  1. Negative, so it enters a circular orbit
  2. Positive, so it accelerates away indefinitely
  3. Equal to its kinetic energy at the surface
  4. Zero, so it just reaches infinity with zero speed Correct answer.

Explanation

Escape velocity is defined by setting the total energy E = ½mv² - GMm/R to zero. The body therefore reaches infinity with zero kinetic energy, exactly the threshold between bound and unbound motion.

Question 23Gravitational field

A satellite of mass m orbits a planet of mass M in a circular orbit of radius r. To move it to a circular orbit of radius 2r, the minimum energy that must be supplied is:

  1. GMm/(4r) Correct answer.
  2. GMm/(2r)
  3. GMm/(8r)
  4. GMm/r

Explanation

Total orbital energy is E = -GMm/(2r). The change is ΔE = -GMm/(4r) - (-GMm/(2r)) = GMm/(4r), which is the energy that must be added to raise the orbit.

Question 24Gravitational field

A planet moves in an elliptical orbit around the Sun, with the Sun at one focus. At perihelion (closest approach) compared with aphelion (farthest point), the planet has:

  1. Smaller speed and greater kinetic energy
  2. Greater speed and greater kinetic energy Correct answer.
  3. Greater speed and greater potential energy
  4. Equal speed because angular momentum is conserved

Explanation

At perihelion the planet is closer to the Sun, so its gravitational potential energy is lower (more negative) and, by energy conservation, its kinetic energy and speed are greater. Angular momentum conservation also requires r·v to be larger when r is smaller.

Question 25Gravitational field

The gravitational potential due to two equal point masses M separated by a distance 2a is evaluated at the midpoint between them. The field and the potential there are:

  1. Field zero; potential -2GM/a Correct answer.
  2. Field zero; potential zero
  3. Field 2GM/a² towards one mass; potential -2GM/a
  4. Field zero; potential -GM/a

Explanation

At the midpoint the two field vectors are equal and opposite, so the resultant field is zero. Potentials are scalars and add, giving V = -GM/a - GM/a = -2GM/a.

Question 26Gravitational field

Which quantity is a scalar, and therefore has the same value at all points of a spherical surface centred on an isolated point mass?

  1. Gravitational acceleration of a test mass
  2. Gravitational field strength
  3. Gravitational potential Correct answer.
  4. Gravitational force on a test mass

Explanation

Gravitational potential V = -GM/r depends only on the distance r from the mass, so it takes a single value on any sphere centred on the mass; being a scalar it has no direction. Field strength, force and acceleration are vectors whose directions point towards the mass and therefore differ from point to point on that sphere.

Question 27Gravitational field

A satellite of mass 500 kg is in a circular orbit where the gravitational potential is -8.0 × 10⁶ J/kg. What is the total mechanical energy of the satellite in this orbit?

  1. -4.0 × 10⁹ J
  2. -8.0 × 10⁹ J
  3. -2.0 × 10⁹ J Correct answer.
  4. +4.0 × 10⁹ J

Explanation

For a circular orbit the total mechanical energy is E = ½ m V, because the kinetic energy equals minus one half of the potential energy. Substituting gives E = 0.5 × 500 × (-8.0 × 10⁶) = -2.0 × 10⁹ J. The negative sign reflects that the satellite is bound to the planet.

Question 28Gravitational field

A spherical planet has the same mean density as the Earth but twice its radius. How does the gravitational field strength at its surface compare with that at the Earth's surface?

  1. It is twice the Earth's value. Correct answer.
  2. It is the same as on the Earth.
  3. It is four times the Earth's value.
  4. It is one half of the Earth's value.

Explanation

For a uniform sphere, g = GM/R² and M = (4/3)πR³ρ, so g = (4/3)πGρR. With equal density, the surface field is proportional to the radius, so doubling R doubles g. The mass also increases eightfold, but the R² factor in the denominator reduces the effect to a factor of two.

Question 29Gravitational field

A comet moves in an elliptical orbit around the Sun. At the point where its distance from the Sun is 4.0 × 10¹¹ m, its speed is 6.0 × 10³ m/s and its velocity is perpendicular to the Sun–comet line. Using the angular-momentum (equal-area) condition, what is its speed when the Sun–comet distance is 1.0 × 10¹¹ m and the velocity is again perpendicular to that line?

  1. 2.4 × 10³ m/s
  2. 6.0 × 10³ m/s
  3. 1.5 × 10³ m/s
  4. 2.4 × 10⁴ m/s Correct answer.

Explanation

Kepler's second law requires the areal speed r·v/2 to stay constant when the velocity is perpendicular to the radius. Hence r₁v₁ = r₂v₂, giving v₂ = (4.0 × 10¹¹ × 6.0 × 10³)/(1.0 × 10¹¹) = 2.4 × 10⁴ m/s. The comet moves faster at the closer point, as expected near perihelion.

Question 30Gravitational field

Which statement correctly describes the escape velocity from the surface of a planet?

  1. It depends on the direction in which the body is launched.
  2. It is independent of the mass of the body launched. Correct answer.
  3. It equals the orbital speed of a satellite skimming the surface.
  4. It increases in proportion to the mass of the body launched.

Explanation

Escape velocity follows from equating ½mv² with GMm/R, so v = √(2GM/R): the mass m cancels and only the planet's mass and radius matter. The result is also independent of launch direction when air resistance is ignored. It is larger than the surface orbital speed by a factor of √2.

Question 31Electromagnetic field

Una carga puntual positiva se mueve con velocidad v paralela a un campo magnético uniforme B. ¿Qué fuerza magnética actúa sobre ella?

  1. Fuerza nula, porque el ángulo entre v y B es cero. Correct answer.
  2. Una fuerza qvB opuesta a B.
  3. Una fuerza qvB perpendicular a v.
  4. Una fuerza qvB en el mismo sentido que v.

Explanation

La fuerza magnética viene dada por F = qv×B, cuyo módulo es qvB·senθ. Si v es paralela a B, el ángulo θ es 0° y sen 0° = 0, por lo que la fuerza magnética se anula aunque la carga esté en movimiento.

Question 32Electromagnetic field

En el SI, ¿qué unidad corresponde al flujo magnético Φ a través de una superficie?

  1. El henrio (H).
  2. El weber (Wb). Correct answer.
  3. El tesla (T).
  4. El amperio por metro (A/m).

Explanation

El flujo magnético se define como Φ = ∫B·dS, producto de un campo en teslas por un área en metros cuadrados, y su unidad es el weber (1 Wb = 1 T·m²). El tesla mide la densidad de flujo, no el flujo total.

Question 33Electromagnetic field

Dos cargas eléctricas de signos opuestos se atraen con una fuerza F. Si se duplica la distancia que las separa manteniendo las cargas, ¿qué fuerza se ejerce ahora?

  1. 4F, de atracción.
  2. 2F, de atracción.
  3. F/4, de atracción. Correct answer.
  4. F/2, de repulsión.

Explanation

La ley de Coulomb establece F = kq₁q₂/r², de modo que la fuerza es inversamente proporcional al cuadrado de la distancia. Al duplicar r, el denominador se multiplica por cuatro y la fuerza queda en F/4, conservando el carácter atractivo propio de cargas de signo opuesto.

Question 34Electromagnetic field

¿Cuál es la dirección del campo eléctrico creado por una carga puntual negativa en un punto situado a su derecha?

  1. Perpendicular a la línea que une carga y punto.
  2. Hacia la izquierda, apuntando hacia la carga. Correct answer.
  3. Hacia la derecha, alejándose de la carga.
  4. El campo es nulo en ese punto.

Explanation

El campo eléctrico de una carga puntual apunta en el sentido en que se movería una carga de prueba positiva. Como la carga fuente es negativa, el campo apunta hacia ella, es decir, hacia la izquierda desde un punto situado a su derecha.

Question 35Electromagnetic field

Una espira circular se encuentra en el plano perpendicular a un campo magnético uniforme. ¿Cómo se calcula el flujo magnético que la atraviesa?

  1. Como el producto q·v·B sobre la espira.
  2. Como el producto B·A, siendo A el área de la espira. Correct answer.
  3. Como el producto B·L, siendo L el perímetro.
  4. Como B dividido por el área de la espira.

Explanation

Para un campo uniforme y una superficie plana perpendicular a él, el flujo vale Φ = B·A·cos0° = B·A. El coseno del ángulo entre B y la normal a la superficie vale uno en esta orientación.

Question 36Electromagnetic field

Una carga de 2 μC se mueve con velocidad de 3·10⁴ m/s perpendicularmente a un campo magnético de 0,5 T. ¿Cuál es el módulo de la fuerza magnética?

  1. 3·10⁻⁵ N
  2. 6·10⁻⁵ N
  3. 1,2·10⁻⁴ N
  4. 3·10⁻² N Correct answer.

Explanation

Con v perpendicular a B, el módulo es F = qvB = (2·10⁻⁶)(3·10⁴)(0,5). El producto da 2·10⁻⁶ · 1,5·10⁴ = 3·10⁻² N.

Question 37Electromagnetic field

¿Qué afirma la ley de Gauss para el campo eléctrico?

  1. Que el flujo eléctrico por una superficie cerrada es proporcional a la carga neta encerrada. Correct answer.
  2. Que el flujo eléctrico por una superficie cerrada depende del área de esa superficie.
  3. Que el campo eléctrico es siempre nulo en el interior de un conductor.
  4. Que la circulación del campo eléctrico entre dos puntos es siempre cero.

Explanation

La ley de Gauss establece que el flujo de E a través de cualquier superficie cerrada vale Q_encerrada/ε₀. Solo cuenta la carga interior; las cargas externas no contribuyen al flujo neto.

Question 38Electromagnetic field

Una partícula cargada entra con velocidad perpendicular a un campo magnético uniforme y describe una trayectoria circular. Si se duplica el módulo de la velocidad manteniendo el mismo campo, ¿qué ocurre con el radio de la trayectoria?

  1. Se multiplica por cuatro.
  2. Se duplica. Correct answer.
  3. Se reduce a la mitad.
  4. No cambia.

Explanation

La fuerza magnética actúa como fuerza centrípeta: qvB = mv²/r, de donde r = mv/(qB). El radio es directamente proporcional a la velocidad, por lo que al duplicar v se duplica r.

Question 39Electromagnetic field

Una carga positiva se mueve en el plano del papel hacia arriba mientras un campo magnético uniforme apunta hacia dentro del papel. ¿Hacia dónde se dirige la fuerza magnética?

  1. Hacia fuera del papel.
  2. Hacia la derecha del papel.
  3. Hacia la izquierda del papel. Correct answer.
  4. Hacia abajo, en sentido opuesto a la velocidad.

Explanation

Aplicando la regla de la mano derecha a F = qv×B: con v hacia arriba y B entrando en el plano, el producto vectorial apunta hacia la izquierda. Al ser la carga positiva, la fuerza conserva ese sentido.

Question 40Electromagnetic field

Un hilo recto muy largo transporta una corriente I. ¿Cómo varía el módulo del campo magnético que crea a una distancia r del hilo?

  1. Es proporcional a 1/r. Correct answer.
  2. Es proporcional a r.
  3. Es independiente de r.
  4. Es proporcional a 1/r².

Explanation

La ley de Ampère aplicada a un hilo recto indefinido da B = μ₀I/(2πr). El campo disminuye inversamente con la distancia al hilo, con líneas de campo circulares alrededor de él.

Question 41Electromagnetic field

Una carga puntual de 4 μC está en el vacío. ¿Cuál es el potencial eléctrico a 2 m de distancia? (k = 9·10⁹ N·m²/C²)

  1. 1,8·10⁷ V
  2. 3,6·10⁴ V
  3. 1,8·10⁴ V Correct answer.
  4. 9·10³ V

Explanation

El potencial de una carga puntual es V = kq/r = (9·10⁹)(4·10⁻⁶)/2. El cálculo da 3,6·10⁴/2 = 1,8·10⁴ V.

Question 42Electromagnetic field

Un electrón y un protón penetran con la misma velocidad perpendicularmente a un campo magnético uniforme. ¿Qué relación tienen los radios de sus trayectorias circulares?

  1. El radio del electrón es mayor, porque su carga tiene signo negativo.
  2. Son iguales, porque la carga tiene el mismo módulo.
  3. El radio del electrón es mayor, porque su masa es menor.
  4. El radio del protón es mayor, porque su masa es unas 1836 veces mayor. Correct answer.

Explanation

El radio vale r = mv/(|q|B). Ambos tienen el mismo |q| y la misma v, de modo que r es proporcional a la masa. Como la masa del protón es mucho mayor, su radio resulta aproximadamente 1836 veces mayor.

Question 43Electromagnetic field

Una bobina plana de 200 espiras y área 0,01 m² está sometida a un campo magnético perpendicular a su plano que aumenta de forma uniforme de 0 a 0,4 T en 0,2 s. ¿Cuál es la fem inducida media?

  1. 40 V
  2. 0,04 V
  3. 4 V Correct answer.
  4. 0,4 V

Explanation

La ley de Faraday da ε = N·ΔΦ/Δt = N·A·ΔB/Δt. Sustituyendo: 200 · 0,01 · (0,4/0,2) = 200 · 0,01 · 2 = 4 V.

Question 44Electromagnetic field

Una partícula cargada atraviesa una región con campos eléctrico y magnético uniformes, perpendiculares entre sí y a la velocidad. ¿Qué condición permite que la partícula no se desvíe?

  1. Que la carga sea negativa.
  2. Que qE sea igual a qvB, es decir, v = E/B. Correct answer.
  3. Que la velocidad sea igual a B/E.
  4. Que el campo eléctrico sea nulo.

Explanation

Cuando la fuerza eléctrica qE y la magnética qvB tienen igual módulo y sentidos opuestos, la fuerza neta es nula y la partícula sigue recta. Esto ocurre cuando v = E/B, base del selector de velocidades.

Question 45Electromagnetic field

Dos cargas iguales de 1 μC están separadas 0,3 m en el vacío. ¿Cuál es el módulo de la fuerza entre ellas? (k = 9·10⁹ N·m²/C²)

  1. 0,3 N
  2. 0,1 N Correct answer.
  3. 3·10⁻³ N
  4. 1 N

Explanation

Aplicando la ley de Coulomb: F = kq²/r² = (9·10⁹)(10⁻⁶)²/(0,3)² = 9·10⁻³/0,09 = 0,1 N. Al ser cargas del mismo signo, la fuerza es repulsiva.

Question 46Electromagnetic field

Una superficie esférica imaginaria encierra una carga neta nula, aunque en su interior hay un dipolo eléctrico. ¿Qué se puede afirmar sobre el flujo eléctrico neto a través de la superficie?

  1. Es negativo, porque hay una carga negativa.
  2. Depende del radio de la esfera y del momento dipolar.
  3. Es nulo, porque la carga neta encerrada es cero. Correct answer.
  4. Es positivo, porque el dipolo genera campo.

Explanation

La ley de Gauss relaciona el flujo neto únicamente con la carga neta encerrada: Φ = Q_enc/ε₀. Un dipolo tiene carga total cero, por lo que el flujo neto es nulo aunque el campo no sea cero en cada punto de la superficie.

Question 47Electromagnetic field

En un solenoide ideal largo con n espiras por unidad de longitud que transporta corriente I, ¿cómo es el campo magnético en su interior?

  1. Creciente linealmente desde el eje hacia las paredes.
  2. Uniforme, de módulo μ₀nI y dirigido según el eje. Correct answer.
  3. Uniforme, de módulo μ₀I/(2πr) y perpendicular al eje.
  4. Nulo, porque las espiras se compensan.

Explanation

Aplicando la ley de Ampère a un solenoide ideal se obtiene B = μ₀nI en el interior, constante y paralelo al eje. En el exterior ideal el campo es prácticamente nulo.

Question 48Electromagnetic field

Una espira conductora se aleja de un imán recto cuya cara norte mira hacia ella. ¿Qué efecto se observa según la ley de Lenz?

  1. Se induce corriente solo si la espira gira.
  2. Se induce una corriente que crea un polo norte en la cara próxima al imán, oponiéndose al alejamiento. Correct answer.
  3. Se induce una corriente que crea un polo sur en la cara próxima, atrayendo el imán.
  4. No se induce corriente porque el imán no se mueve.

Explanation

Al alejarse la espira, el flujo magnético a través de ella disminuye. La ley de Lenz indica que la corriente inducida se opone a esa disminución generando un campo en el mismo sentido que el del imán, es decir, un polo norte en la cara próxima que atrae al imán y frena el alejamiento.

Question 49Electromagnetic field

Un condensador de placas paralelas tiene capacidad C. Si se duplica el área de las placas y se duplica la separación entre ellas, ¿qué ocurre con la capacidad?

  1. Se reduce a la mitad.
  2. Se multiplica por cuatro.
  3. No cambia. Correct answer.
  4. Se duplica.

Explanation

La capacidad de un condensador plano es C = ε₀A/d. Al duplicar A y duplicar d simultáneamente, el factor 2 del numerador y el factor 2 del denominador se cancelan, de modo que C permanece igual.

Question 50Electromagnetic field

Una carga positiva se lanza con velocidad horizontal en una región donde solo existe un campo eléctrico uniforme vertical hacia abajo. ¿Qué tipo de trayectoria describe?

  1. Una recta horizontal sin aceleración.
  2. Una circunferencia de radio constante.
  3. Una parábola con aceleración vertical hacia abajo. Correct answer.
  4. Una hélice alrededor de la dirección del campo.

Explanation

La fuerza eléctrica qE es constante y vertical, por lo que la aceleración también lo es mientras la componente horizontal de la velocidad permanece constante. El resultado es una trayectoria parabólica análoga a la de un proyectil.