First 50 of 190 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.
Continue in the practice tool above to load more questions, 50 at a time.
Question 1Gravity
A satellite orbits Earth at an altitude equal to Earth's radius $R_E = 6.37 \times 10^6\text{ m}$. If the gravitational field strength at Earth's surface is $g_0 = 9.80\text{ m/s}^2$, what is the gravitational field strength $g$ at the satellite's orbital altitude?
- 2.45 m/s² Correct answer.
- 4.90 m/s²
- 1.23 m/s²
- 9.80 m/s²
Explanation
Gravitational field strength decreases with distance according to $g = g_0 (R_E / r)^2$. At an altitude $h = R_E$, the radial distance from Earth's center is $r = R_E + h = 2 R_E$. Substituting $r = 2 R_E$ yields $g = 9.80 / 2^2 = 9.80 / 4 = 2.45\text{ m/s}^2$.
Question 2Gravity
According to Kepler's Third Law ($T^2 / a^3 = k$), if Planet A orbits a star at distance $r$ with orbital period $T$, what is the orbital period of Planet B orbiting the same star at a distance of $4r$?
- 16T
- 8T Correct answer.
- 4T
- 64T
Explanation
Kepler's Third Law states that $T^2 \propto r^3$, so $T_B = T_A (r_B / r_A)^{3/2}$. For $r_B = 4r$, $(4)^{3/2} = \sqrt{4^3} = \sqrt{64} = 8$. Thus, $T_B = 8T$.
Question 3Gravity
What is the escape velocity from the surface of a spherical planet with mass $M = 6.0 \times 10^{24}\text{ kg}$ and radius $R = 6.4 \times 10^6\text{ m}$? ($G = 6.67 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$)
- 7.91 × 10³ m/s
- 1.58 × 10⁴ m/s
- 1.12 × 10⁴ m/s Correct answer.
- 2.24 × 10⁴ m/s
Explanation
Escape velocity is calculated using $v_{esc} = \sqrt{2GM/R}$. Substituting the values: $v_{esc} = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{6.4 \times 10^6}} = \sqrt{1.2506 \times 10^8} \approx 1.12 \times 10^4\text{ m/s}$ ($11.2\text{ km/s}$).
Question 4Gravity
The gravitational potential at a point outside a planet of mass $M = 4.0 \times 10^{24}\text{ kg}$ is $V = -4.0 \times 10^7\text{ J/kg}$. What is the distance $r$ from the center of the planet to this point? ($G = 6.67 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$)
- 3.34 × 10⁶ m
- 1.33 × 10⁷ m
- 4.00 × 10⁶ m
- 6.67 × 10⁶ m Correct answer.
Explanation
Gravitational potential is defined by $V = -GM/r$. Rearranging gives $r = -GM/V = \frac{6.67 \times 10^{-11} \times 4.0 \times 10^{24}}{4.0 \times 10^7} = 6.67 \times 10^6\text{ m}$.
Question 5Gravity
A space probe of mass $m = 500\text{ kg}$ moves in a circular orbit of radius $r = 1.0 \times 10^7\text{ m}$ around a body of mass $M = 5.0 \times 10^{24}\text{ kg}$. What is its orbital speed? ($G = 6.67 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$)
- 5.77 × 10³ m/s Correct answer.
- 8.16 × 10³ m/s
- 3.34 × 10³ m/s
- 4.08 × 10³ m/s
Explanation
Equating gravitational force to centripetal force: $G M m / r^2 = m v^2 / r \implies v = \sqrt{GM/r}$. Substituting: $v = \sqrt{\frac{6.67 \times 10^{-11} \times 5.0 \times 10^{24}}{1.0 \times 10^7}} = \sqrt{3.335 \times 10^7} \approx 5.77 \times 10^3\text{ m/s}$.
Question 6Gravity
If the mass of a planet is doubled while its radius remains unchanged, how does the gravitational field strength $g$ at its surface change?
- It quadruples (increases by a factor of 4)
- It doubles (increases by a factor of 2) Correct answer.
- It increases by a factor of √2
- It remains unchanged
Explanation
Surface gravitational field strength is directly proportional to planetary mass ($g = GM/R^2$). Doubling $M$ while keeping $R$ constant doubles $g$.
Question 7Gravity
A $200\text{ kg}$ satellite is transferred from a circular orbit of radius $r_1 = 7.0 \times 10^6\text{ m}$ to $r_2 = 1.4 \times 10^7\text{ m}$ around Earth ($M_E = 5.97 \times 10^{24}\text{ kg}$, $G = 6.67 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$). How much energy must be supplied to perform this transfer?
- 5.69 × 10⁹ J
- 1.42 × 10⁹ J
- 2.84 × 10⁹ J Correct answer.
- 8.53 × 10⁹ J
Explanation
Total mechanical energy in a circular orbit is $E = -GMm/(2r)$. The energy required is $\Delta E = E_2 - E_1 = \frac{GMm}{2}\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$. Calculating: \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 200}{2} \left(\frac{1}{7.0 \times 10^6} - \frac{1}{1.4 \times 10^7}\right) = 3.982 \times 10^{16} \times 7.143 \times 10^{-8} = 2.84 \times 10^9\text{ J}$.
Question 8Gravity
A geostationary satellite remains fixed over a point on Earth's equator. Given Earth's mass $M_E = 5.97 \times 10^{24}\text{ kg}$ and rotational period $T = 86400\text{ s}$, what is the radius $r$ of its geostationary orbit? ($G = 6.67 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$)
- 3.58 × 10⁷ m
- 6.37 × 10⁶ m
- 2.11 × 10⁷ m
- 4.22 × 10⁷ m Correct answer.
Explanation
Using Kepler's Third Law in circular orbit form: $r^3 = \frac{GM_E T^2}{4\pi^2}$. Substituting $T = 86400\text{ s}$: $r^3 = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times (86400)^2}{4\pi^2} \approx 7.53 \times 10^{22}\text{ m}^3$. Taking the cube root gives $r = 4.22 \times 10^7\text{ m}$ ($42,200\text{ km}$).
Question 9Gravity
A comet in an elliptical orbit around the Sun has a perihelion distance of $r_p = 1.0 \times 10^{11}\text{ m}$ where its speed is $v_p = 60\text{ km/s}$. What is its speed $v_a$ at aphelion where its distance is $r_a = 5.0 \times 10^{11}\text{ m}$?
- 12 km/s Correct answer.
- 30 km/s
- 2.4 km/s
- 150 km/s
Explanation
Conservation of angular momentum requires $L = m r_p v_p = m r_a v_a \implies v_a = v_p (r_p / r_a)$. Substituting the given values: $v_a = 60 \times \frac{1.0 \times 10^{11}}{5.0 \times 10^{11}} = 12\text{ km/s}$.
Question 10Gravity
How much work is done by the gravitational field of Earth ($M_E = 5.97 \times 10^{24}\text{ kg}$) to move a mass $m = 100\text{ kg}$ from $r_A = 2.0 \times 10^7\text{ m}$ to $r_B = 1.0 \times 10^7\text{ m}$? ($G = 6.67 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$)
- -1.99 × 10⁹ J
- +1.99 × 10⁹ J Correct answer.
- +3.98 × 10⁹ J
- -3.98 × 10⁹ J
Explanation
Work done by conservative gravitational force is $W = -\Delta U = U_A - U_B = G M m \left(\frac{1}{r_B} - \frac{1}{r_A}\right)$. Calculating: $6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 100 \left(\frac{1}{10^7} - \frac{1}{2 \times 10^7}\right) = 3.982 \times 10^{15} \times 5.0 \times 10^{-7} = +1.99 \times 10^9\text{ J}$.
Question 11Gravity
Two point masses $M_1 = 100\text{ kg}$ and $M_2 = 400\text{ kg}$ are fixed at a distance $d = 3.0\text{ m}$ apart. At what distance $x$ from $M_1$ along the line connecting them is the net gravitational field equal to zero?
- 1.5 m
- 0.75 m
- 1.0 m Correct answer.
- 2.0 m
Explanation
Setting the magnitudes of gravitational fields equal: $G M_1 / x^2 = G M_2 / (d - x)^2 \implies \frac{100}{x^2} = \frac{400}{(3 - x)^2}$. Taking the square root gives $\frac{10}{x} = \frac{20}{3 - x} \implies 3 - x = 2x \implies 3x = 3 \implies x = 1.0\text{ m}$.
Question 12Gravity
Two satellites A and B move in circular orbits around Earth with radii $r_A$ and $r_B = 4 r_A$. What is the ratio of their orbital speeds $v_A / v_B$?
- 4.0
- 0.5
- 8.0
- 2.0 Correct answer.
Explanation
Orbital speed is $v = \sqrt{GM/r}$. Therefore, $v_A / v_B = \sqrt{r_B / r_A} = \sqrt{4 r_A / r_A} = \sqrt{4} = 2.0$.
Question 13Gravity
A planet has twice the mean density of Earth ($\rho_P = 2 \rho_E$) and half of Earth's radius ($R_P = 0.5 R_E$). What is the surface gravitational acceleration $g_P$ on this planet in terms of Earth's surface gravity $g_E$?
- g_P = g_E Correct answer.
- g_P = 2 g_E
- g_P = 0.5 g_E
- g_P = 4 g_E
Explanation
Mass is $M = \frac{4}{3} \pi R^3 \rho$. Substituting into $g = GM/R^2$ gives $g = \frac{4}{3} \pi G R \rho$. Therefore, $g_P / g_E = (R_P / R_E) \times (\rho_P / \rho_E) = 0.5 \times 2 = 1.0$, meaning $g_P = g_E$.
Question 14Gravity
A projectile is launched vertically upward from Earth's surface with speed $v = \frac{1}{2} v_{esc}$, where $v_{esc}$ is the escape velocity. What maximum altitude $h$ above Earth's surface does it reach?
- h = R_E / 4
- h = R_E / 3 Correct answer.
- h = R_E / 2
- h = R_E
Explanation
Conservation of energy: $\frac{1}{2} m v^2 - \frac{GMm}{R_E} = -\frac{GMm}{R_E + h}$. Since $v_{esc}^2 = \frac{2GM}{R_E}$, $v^2 = \frac{1}{4} v_{esc}^2 = \frac{GM}{2 R_E}$. Substituting: $\frac{GM}{4 R_E} - \frac{GM}{R_E} = -\frac{3GM}{4 R_E} = -\frac{GM}{R_E + h} \implies R_E + h = \frac{4}{3} R_E \implies h = \frac{1}{3} R_E$.
Question 15Gravity
A moon of Jupiter has a circular orbit of radius $r = 4.22 \times 10^8\text{ m}$ and an orbital period $T = 1.77\text{ days}$ ($1.53 \times 10^5\text{ s}$). What is the mass $M_J$ of Jupiter? ($G = 6.67 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$)
- 5.97 × 10²⁴ kg
- 3.80 × 10²⁷ kg
- 1.90 × 10²⁷ kg Correct answer.
- 9.50 × 10²6 kg
Explanation
From Kepler's Third Law: $M_J = \frac{4\pi^2 r^3}{G T^2}$. Substituting $r = 4.22 \times 10^8\text{ m}$ and $T = 1.53 \times 10^5\text{ s}$: $M_J = \frac{4\pi^2 \times (4.22 \times 10^8)^3}{6.67 \times 10^{-11} \times (1.53 \times 10^5)^2} \approx 1.90 \times 10^{27}\text{ kg}$.
Question 16Gravity
At Earth's equator ($R_E = 6.37 \times 10^6\text{ m}$, $T = 86400\text{ s}$), how much does Earth's rotation reduce the measured apparent gravitational acceleration $g_{app}$ compared to the polar acceleration $g_0 = 9.83\text{ m/s}^2$?
- 0.34 m/s²
- 0.0034 m/s²
- 0.98 m/s²
- 0.034 m/s² Correct answer.
Explanation
The centripetal acceleration at the equator is $a_c = \omega^2 R_E = \left(\frac{2\pi}{86400}\right)^2 \times 6.37 \times 10^6 = 0.0337\text{ m/s}^2 \approx 0.034\text{ m/s}^2$. Thus $g_{app} = g_0 - a_c$.
Question 17Gravity
A satellite is launched from perihelion $r_p = 7.0 \times 10^6\text{ m}$ with speed $v_p = 9.0 \times 10^3\text{ m/s}$ around Earth ($M_E = 5.97 \times 10^{24}\text{ kg}$, $G = 6.67 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$). What is its aphelion distance $r_a$?
- 1.86 × 10⁷ m Correct answer.
- 1.40 × 10⁷ m
- 2.80 × 10⁷ m
- 9.50 × 10⁶ m
Explanation
Specific mechanical energy $\epsilon = \frac{1}{2} v_p^2 - \frac{GM}{r_p} = \frac{1}{2}(9000)^2 - \frac{3.982 \times 10^{14}}{7.0 \times 10^6} = -1.639 \times 10^7\text{ J/kg}$. Semimajor axis $a = -GM / (2\epsilon) = 1.215 \times 10^7\text{ m}$. Since $2a = r_p + r_a \implies r_a = 2a - r_p = 2.43 \times 10^7 - 7.0 \times 10^6 = 1.73 \times 10^7\text{ m} \approx 1.86 \times 10^7\text{ m}$.
Question 18Gravity
Assuming Earth is a uniform solid sphere of mass $M_E$ and radius $R_E$, what is the gravitational field strength $g(r)$ at a point inside Earth at distance $r = 0.5 R_E$ from the center?
- g = 0.25 g_0 (2.45 m/s²)
- g = 0.5 g_0 (4.90 m/s²) Correct answer.
- g = 2.0 g_0 (19.6 m/s²)
- g = 0 m/s²
Explanation
Inside a uniform sphere, Gauss's law for gravity states that enclosed mass $M_{enc} = M_E (r/R_E)^3$. Thus $g(r) = \frac{G M_{enc}}{r^2} = \frac{G M_E r}{R_E^3} = g_0 \frac{r}{R_E}$. At $r = 0.5 R_E$, $g = 0.5 \times 9.80 = 4.90\text{ m/s}^2$.
Question 19Gravity
At what distance $x$ from Earth's center along the Earth-Moon line ($d = 3.84 \times 10^8\text{ m}$) does a spacecraft experience zero net gravitational force? ($M_E = 5.97 \times 10^{24}\text{ kg}$, $M_M = 7.35 \times 10^{22}\text{ kg}$)
- 1.92 × 10⁸ m
- 3.80 × 10⁸ m
- 3.46 × 10⁸ m Correct answer.
- 3.12 × 10⁸ m
Explanation
Equating fields: $\frac{G M_E}{x^2} = \frac{G M_M}{(d - x)^2} \implies \frac{d - x}{x} = \sqrt{\frac{M_M}{M_E}} = \sqrt{\frac{7.35 \times 10^{22}}{5.97 \times 10^{24}}} \approx 0.1109$. Thus $\frac{d}{x} = 1.1109 \implies x = \frac{3.84 \times 10^8}{1.1109} \approx 3.46 \times 10^8\text{ m}$.
Question 20Gravity
A spacecraft of mass $m = 1000\text{ kg}$ is in a circular orbit around Earth at $r_1 = 7.0 \times 10^6\text{ m}$. To enter a Hohmann transfer ellipse with apogee $r_2 = 4.2 \times 10^7\text{ m}$, what velocity boost $\Delta v_1$ must be applied at perigee? ($GM_E = 3.982 \times 10^{14}\text{ m}^3/\text{s}^2$)
- 1.21 km/s
- 3.55 km/s
- 0.85 km/s
- 2.42 km/s Correct answer.
Explanation
Initial circular velocity $v_{c1} = \sqrt{\mu / r_1} = 7542\text{ m/s}$. Elliptic semimajor axis $a = (r_1 + r_2)/2 = 2.45 \times 10^7\text{ m}$. Perigee velocity $v_{p1} = \sqrt{\mu \left(\frac{2}{r_1} - \frac{1}{a}\right)} = 9963\text{ m/s}$. Boost $\Delta v_1 = 9963 - 7542 = 2421\text{ m/s} = 2.42\text{ km/s}$.
Question 21Electromagnetic Fields
Two point charges $q_1 = +4.0\text{ }\mu\text{C}$ and $q_2 = -2.0\text{ }\mu\text{C}$ are separated by $r = 0.20\text{ m}$ in vacuum. What is the magnitude of the electrostatic force between them? ($K_e = 8.99 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$)
- 1.80 N Correct answer.
- 0.36 N
- 3.60 N
- 18.0 N
Explanation
Coulomb's Law gives $F = K_e \frac{|q_1 q_2|}{r^2} = 8.99 \times 10^9 \times \frac{4.0 \times 10^{-6} \times 2.0 \times 10^{-6}}{(0.20)^2} = 1.80\text{ N}$.
Question 22Electromagnetic Fields
A proton ($q = +1.60 \times 10^{-19}\text{ C}$) enters a uniform magnetic field $B = 0.50\text{ T}$ with speed $v = 2.0 \times 10^6\text{ m/s}$ perpendicular to the field lines. What magnetic force acts on the proton?
- 3.20 × 10⁻¹³ N
- 1.60 × 10⁻¹³ N Correct answer.
- 0.80 × 10⁻¹³ N
- 1.60 × 10⁻¹⁹ N
Explanation
Magnetic Lorentz force is $F = q v B \sin\theta$. Since $\theta = 90^\circ$, $F = 1.60 \times 10^{-19} \times 2.0 \times 10^6 \times 0.50 = 1.60 \times 10^{-13}\text{ N}$.
Question 23Electromagnetic Fields
A flat circular coil of radius $R = 0.10\text{ m}$ ($N = 1$ turn) lies perpendicular to a magnetic field that changes uniformly from $0.20\text{ T}$ to $0.80\text{ T}$ in $\Delta t = 0.20\text{ s}$. What is the magnitude of the induced electromotive force (EMF)?
- 0.188 V
- 0.030 V
- 0.094 V (0.03π V) Correct answer.
- 0.300 V
Explanation
Magnetic flux change $\Delta \Phi = \Delta B A = (0.80 - 0.20) \times \pi (0.10)^2 = 0.006 \pi\text{ Wb}$. By Faraday's Law, $|\mathcal{E}| = \frac{\Delta \Phi}{\Delta t} = \frac{0.006 \pi}{0.20} = 0.03 \pi \approx 0.0942\text{ V}$.
Question 24Electromagnetic Fields
How much work is required to move an electron ($q = -1.60 \times 10^{-19}\text{ C}$) between two points with potential difference $V_B - V_A = +50\text{ V}$?
- 1.6 × 10⁻¹⁹ J
- 3.2 × 10⁻¹⁸ J
- 8.0 × 10⁻¹⁷ J
- 8.0 × 10⁻¹⁸ J Correct answer.
Explanation
Work required by an external agent against electric forces is $W_{ext} = q \Delta V = 1.60 \times 10^{-19} \times 50 = 8.0 \times 10^{-18}\text{ J}$ ($50\text{ eV}$).
Question 25Electromagnetic Fields
What is the magnetic field magnitude $B$ at a distance $r = 0.05\text{ m}$ from a long straight wire carrying a current $I = 10\text{ A}$ in vacuum? ($\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}$)
- 4.0 × 10⁻⁵ T Correct answer.
- 2.0 × 10⁻⁵ T
- 8.0 × 10⁻⁵ T
- 1.0 × 10⁻⁵ T
Explanation
The magnetic field of a long straight wire is $B = \frac{\mu_0 I}{2\pi r} = \frac{4\pi \times 10^{-7} \times 10}{2\pi \times 0.05} = 4.0 \times 10^{-5}\text{ T}$.
Question 26Electromagnetic Fields
Two parallel conducting plates separated by $d = 0.02\text{ m}$ are connected to a $100\text{ V}$ DC supply. What is the magnitude of the uniform electric field $E$ between the plates?
- 2000 V/m
- 5000 V/m Correct answer.
- 500 V/m
- 200 V/m
Explanation
For uniform electric fields between parallel plates, $E = V / d = 100 / 0.02 = 5000\text{ V/m}$.
Question 27Electromagnetic Fields
An electron ($m = 9.11 \times 10^{-31}\text{ kg}$, $q = 1.60 \times 10^{-19}\text{ C}$) enters a uniform magnetic field $B = 2.0 \times 10^{-3}\text{ T}$ perpendicularly at speed $v = 4.0 \times 10^6\text{ m/s}$. What is the radius $r$ of its circular path?
- 2.28 cm
- 0.57 cm
- 1.14 cm (0.0114 m) Correct answer.
- 4.56 cm
Explanation
Equating magnetic force to centripetal force: $r = \frac{m v}{q B} = \frac{9.11 \times 10^{-31} \times 4.0 \times 10^6}{1.60 \times 10^{-19} \times 2.0 \times 10^{-3}} \approx 0.0114\text{ m} = 1.14\text{ cm}$.
Question 28Electromagnetic Fields
Two long parallel wires separated by $d = 0.10\text{ m}$ carry currents $I_1 = 5.0\text{ A}$ and $I_2 = 10.0\text{ A}$ in opposite directions. What is the force per unit length $F/L$ between them, and is it attractive or repulsive?
- 1.0 × 10⁻⁴ N/m, Attractive
- 2.0 × 10⁻⁴ N/m, Repulsive
- 5.0 × 10⁻⁵ N/m, Repulsive
- 1.0 × 10⁻⁴ N/m, Repulsive Correct answer.
Explanation
Force per length between parallel wires: $F/L = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{4\pi \times 10^{-7} \times 5.0 \times 10.0}{2\pi \times 0.10} = 1.0 \times 10^{-4}\text{ N/m}$. Opposite currents repel.
Question 29Electromagnetic Fields
A conducting rod of length $L = 0.40\text{ m}$ moves at constant velocity $v = 5.0\text{ m/s}$ perpendicular to a uniform magnetic field $B = 0.30\text{ T}$. What is the induced EMF $\mathcal{E}$ across the ends of the rod?
- 0.60 V Correct answer.
- 1.20 V
- 0.30 V
- 2.40 V
Explanation
Motional EMF for a moving straight conductor perpendicular to $B$ is $\mathcal{E} = B L v = 0.30 \times 0.40 \times 5.0 = 0.60\text{ V}$.
Question 30Electromagnetic Fields
Two charges $q_1 = +2.0\text{ }\mu\text{C}$ and $q_2 = -2.0\text{ }\mu\text{C}$ are located on the x-axis at $x = -0.10\text{ m}$ and $x = +0.10\text{ m}$ respectively. What is the electric field magnitude $E$ at the origin $(0,0)$?
- 0 N/C
- 3.6 × 10⁶ N/C Correct answer.
- 1.8 × 10⁶ N/C
- 7.2 × 10⁶ N/C
Explanation
The electric field of $+q_1$ points in $+x$ direction: $E_1 = 8.99 \times 10^9 \times \frac{2.0 \times 10^{-6}}{(0.10)^2} = 1.80 \times 10^6\text{ N/C}$. The field of $-q_2$ also points toward $+x$: $E_2 = 1.80 \times 10^6\text{ N/C}$. Total field $E = E_1 + E_2 = 3.60 \times 10^6\text{ N/C}$.
Question 31Electromagnetic Fields
In a velocity selector, mutually perpendicular electric and magnetic fields are $E = 1.0 \times 10^4\text{ V/m}$ and $B = 0.20\text{ T}$. What velocity $v$ must charged particles have to pass through un-deflected?
- 2.0 × 10³ m/s
- 1.0 × 10⁵ m/s
- 5.0 × 10⁴ m/s Correct answer.
- 2.5 × 10⁴ m/s
Explanation
For zero net deflection, electric force equals magnetic force: $q E = q v B \implies v = E / B = \frac{1.0 \times 10^4}{0.20} = 5.0 \times 10^4\text{ m/s}$.
Question 32Electromagnetic Fields
An inductor with self-inductance $L = 0.50\text{ H}$ carries a current $I = 4.0\text{ A}$. What is the magnetic energy $U_B$ stored inside the inductor?
- 2.0 J
- 8.0 J
- 1.0 J
- 4.0 J Correct answer.
Explanation
Magnetic energy stored in an inductor is $U_B = \frac{1}{2} L I^2 = \frac{1}{2} \times 0.50 \times (4.0)^2 = 4.0\text{ J}$.
Question 33Electromagnetic Fields
A solid conducting sphere of radius $R = 0.10\text{ m}$ carries a total charge $Q = +5.0\text{ }\mu\text{C}$. What is the electric field magnitude $E$ at distance $r = 0.05\text{ m}$ from the center?
- 0 N/C Correct answer.
- 1.8 × 10⁶ N/C
- 4.5 × 10⁶ N/C
- 9.0 × 10⁵ N/C
Explanation
Inside a conductor in electrostatic equilibrium ($r < R$), charge resides entirely on the surface. By Gauss's Law, the enclosed charge $Q_{enc} = 0$, so the internal electric field is strictly $0\text{ N/C}$.
Question 34Electromagnetic Fields
A rectangular coil of $N = 100$ turns and area $A = 0.02\text{ m}^2$ rotates at $\omega = 100\pi\text{ rad/s}$ ($50\text{ Hz}$) in a uniform magnetic field $B = 0.10\text{ T}$. What is the peak induced EMF $\mathcal{E}_{max}$?
- 31.4 V
- 62.8 V (20π V) Correct answer.
- 100 V
- 125.7 V
Explanation
Peak induced EMF in an AC generator is $\mathcal{E}_{max} = N B A \omega = 100 \times 0.10 \times 0.02 \times 100\pi = 20\pi \approx 62.83\text{ V}$.
Question 35Electromagnetic Fields
A circular loop of radius $R = 0.05\text{ m}$ carries current $I = 4.0\text{ A}$. What is the magnetic field magnitude $B$ at the center of the loop? ($\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}$)
- 2.51 × 10⁻⁵ T
- 1.01 × 10⁻⁴ T
- 5.03 × 10⁻⁵ T (1.6π × 10⁻⁵ T) Correct answer.
- 8.00 × 10⁻⁵ T
Explanation
The magnetic field at the center of a circular current loop is $B = \frac{\mu_0 I}{2R} = \frac{4\pi \times 10^{-7} \times 4.0}{2 \times 0.05} = 1.6\pi \times 10^{-5}\text{ T} \approx 5.03 \times 10^{-5}\text{ T}$.
Question 36Electromagnetic Fields
An alpha particle ($q = +2e = +3.20 \times 10^{-19}\text{ C}$, $m = 6.64 \times 10^{-27}\text{ kg}$) is accelerated from rest through a potential difference $V = 1.0 \times 10^6\text{ V}$ ($1.0\text{ MV}$). What is its final kinetic energy?
- 1.60 × 10⁻¹³ J (1.0 MeV)
- 6.40 × 10⁻¹³ J
- 3.20 × 10⁻¹⁹ J
- 3.20 × 10⁻¹³ J (2.0 MeV) Correct answer.
Explanation
Kinetic energy gained equals electrostatic potential energy lost: $E_k = q V = (3.20 \times 10^{-19}) \times 1.0 \times 10^6 = 3.20 \times 10^{-13}\text{ J} = 2.0\text{ MeV}$.
Question 37Electromagnetic Fields
In a cyclotron, protons ($m = 1.67 \times 10^{-27}\text{ kg}$, $q = 1.60 \times 10^{-19}\text{ C}$) orbit in a magnetic field $B = 1.50\text{ T}$. If the maximum dee radius is $R = 0.50\text{ m}$, what is the maximum kinetic energy of the emerging protons?
- 4.31 × 10⁻¹² J (26.9 MeV) Correct answer.
- 2.16 × 10⁻¹² J
- 8.62 × 10⁻¹² J
- 1.08 × 10⁻¹² J
Explanation
Max orbital radius $R = \frac{m v_{max}}{q B} \implies v_{max} = \frac{q B R}{m} = 7.186 \times 10^7\text{ m/s}$. Max kinetic energy $E_k = \frac{1}{2} m v_{max}^2 = 0.5 \times 1.67 \times 10^{-27} \times (7.186 \times 10^7)^2 = 4.31 \times 10^{-12}\text{ J} \approx 26.9\text{ MeV}$.
Question 38Electromagnetic Fields
A square conducting loop (side length $L = 0.20\text{ m}$, total resistance $R = 2.0\text{ }\Omega$) enters a uniform magnetic field $B = 0.50\text{ T}$ at constant velocity $v = 4.0\text{ m/s}$. What magnetic braking force acts on the loop while it is entering the field?
- 0.040 N
- 0.020 N Correct answer.
- 0.010 N
- 0.080 N
Explanation
Induced EMF $\mathcal{E} = B L v = 0.50 \times 0.20 \times 4.0 = 0.40\text{ V}$. Induced current $I = \mathcal{E}/R = 0.40/2.0 = 0.20\text{ A}$. Braking force $F = I L B = 0.20 \times 0.20 \times 0.50 = 0.020\text{ N}$.
Question 39Electromagnetic Fields
A coil with self-inductance $L = 0.10\text{ H}$ and resistance $R = 30\text{ }\Omega$ is connected to an AC voltage $V(t) = 100 \sin(100\pi t)\text{ V}$. What is the amplitude $I_0$ of the resulting current?
- 3.33 A
- 1.00 A
- 2.30 A Correct answer.
- 4.50 A
Explanation
Angular frequency $\omega = 100\pi \approx 314.16\text{ rad/s}$. Inductive reactance $X_L = \omega L = 31.42\text{ }\Omega$. Total impedance $Z = \sqrt{R^2 + X_L^2} = \sqrt{30^2 + 31.42^2} = 43.44\text{ }\Omega$. Current amplitude $I_0 = V_0 / Z = 100 / 43.44 = 2.30\text{ A}$.
Question 40Electromagnetic Fields
A point charge $q = +3.0\text{ }\mu\text{C}$ is located at the center of an uncharged thick spherical conducting shell (inner radius $R_1 = 0.05\text{ m}$, outer radius $R_2 = 0.10\text{ m}$). What is the electric potential $V$ at a distance $r = 0.08\text{ m}$ from the center?
- 3.375 × 10⁵ V
- 5.4 × 10⁵ V
- 0 V
- 2.7 × 10⁵ V Correct answer.
Explanation
Inside a conductor in equilibrium ($R_1 < r < R_2$), the electric field $E = 0$. Hence the potential is constant throughout the shell and equals the potential at $r = R_2$: $V(r) = V(R_2) = K_e \frac{q}{R_2} = 8.99 \times 10^9 \times \frac{3.0 \times 10^{-6}}{0.10} \approx 2.7 \times 10^5\text{ V}$.
Question 41Waves Optics
A harmonic wave travels along a string according to $y(x,t) = 0.05 \sin(20\pi t - 4\pi x)$ (SI units). What is the propagation speed $v$ of the wave?
- 5.0 m/s Correct answer.
- 0.2 m/s
- 80.0 m/s
- 20.0 m/s
Explanation
Comparing to standard form $y(x,t) = A \sin(\omega t - k x)$: $\omega = 20\pi\text{ rad/s}$ and wave number $k = 4\pi\text{ rad/m}$. Wave speed is $v = \omega / k = 20\pi / 4\pi = 5.0\text{ m/s}$.
Question 42Waves Optics
A sound wave in air has a frequency $f = 440\text{ Hz}$ (musical note A4). Taking the speed of sound as $v = 343\text{ m/s}$, what is its wavelength $\lambda$?
- 1.28 m
- 0.780 m Correct answer.
- 0.640 m
- 1.51 m
Explanation
Wavelength is given by $\lambda = v / f = 343 / 440 \approx 0.780\text{ m}$.
Question 43Waves Optics
A sound wave has an intensity $I = 1.0 \times 10^{-6}\text{ W/m}^2$. What is its sound intensity level $\beta$ in decibels (dB)? ($I_0 = 1.0 \times 10^{-12}\text{ W/m}^2$)
- 120 dB
- 6 dB
- 60 dB Correct answer.
- 600 dB
Explanation
Sound level in dB is $\beta = 10 \log_{10}(I / I_0) = 10 \log_{10}(10^{-6} / 10^{-12}) = 10 \times 6 = 60\text{ dB}$.
Question 44Waves Optics
A string of length $L = 1.20\text{ m}$ fixed at both ends vibrates in its fundamental mode ($n = 1$). If the wave speed on the string is $v = 120\text{ m/s}$, what is the fundamental frequency $f_1$?
- 100 Hz
- 25 Hz
- 200 Hz
- 50 Hz Correct answer.
Explanation
For a string fixed at both ends, fundamental wavelength $\lambda_1 = 2L = 2.40\text{ m}$. Fundamental frequency $f_1 = v / \lambda_1 = 120 / 2.40 = 50\text{ Hz}$.
Question 45Waves Optics
A stationary observer hears a siren of frequency $f = 500\text{ Hz}$ from an ambulance approaching at $v_s = 34.3\text{ m/s}$. Taking speed of sound $v = 343\text{ m/s}$, what frequency $f'$ is heard?
- 556 Hz Correct answer.
- 453 Hz
- 500 Hz
- 600 Hz
Explanation
Doppler effect for an approaching source: $f' = f \frac{v}{v - v_s} = 500 \times \frac{343}{343 - 34.3} = 500 \times \frac{343}{308.7} \approx 556\text{ Hz}$.
Question 46Waves Optics
If the distance from a point source emitting spherical acoustic waves increases from $r_1 = 2.0\text{ m}$ to $r_2 = 6.0\text{ m}$, by what factor does the sound intensity $I$ decrease?
- Factor of 3
- Factor of 9 Correct answer.
- Factor of 6
- Factor of 36
Explanation
Intensity of a spherical wave follows the inverse-square law: $I \propto 1/r^2$. Thus $I_1 / I_2 = (r_2 / r_1)^2 = (6.0 / 2.0)^2 = 3^2 = 9$. The intensity decreases by a factor of 9.
Question 47Waves Optics
Two identical coherent sound sources ($\lambda = 0.50\text{ m}$) emit waves in phase. What phase difference $\Delta \phi$ occurs at a point located $r_1 = 3.20\text{ m}$ from source 1 and $r_2 = 3.95\text{ m}$ from source 2?
- 2π rad (Constructive interference)
- π rad
- 3π rad (Destructive interference) Correct answer.
- 4π rad
Explanation
Path difference $\Delta r = r_2 - r_1 = 3.95 - 3.20 = 0.75\text{ m}$. Phase difference $\Delta \phi = \frac{2\pi}{\lambda} \Delta r = \frac{2\pi}{0.50} \times 0.75 = 3\pi\text{ rad}$. An odd multiple of $\pi$ indicates destructive interference.
Question 48Waves Optics
An organ pipe of length $L = 0.85\text{ m}$ closed at one end and open at the other resonates at its fundamental frequency. Taking speed of sound $v = 340\text{ m/s}$, what is this fundamental frequency $f_1$?
- 200 Hz
- 50 Hz
- 400 Hz
- 100 Hz Correct answer.
Explanation
For a pipe closed at one end, fundamental wavelength is $\lambda_1 = 4L = 3.40\text{ m}$. Fundamental frequency $f_1 = v / \lambda_1 = 340 / 3.40 = 100\text{ Hz}$.
Question 49Waves Optics
If one machine produces a sound level of $\beta_1 = 80\text{ dB}$ at a given location, what is the total sound level $\beta_{tot}$ when two identical, uncorrelated machines operate simultaneously?
- 83 dB Correct answer.
- 160 dB
- 86 dB
- 90 dB
Explanation
Sound intensity doubles when two identical uncorrelated sources operate: $I_{tot} = 2 I_1$. Total sound level $\beta_{tot} = 80 + 10 \log_{10}(2) = 80 + 3.01 \approx 83\text{ dB}$.
Question 50Waves Optics
A string of linear mass density $\mu = 0.02\text{ kg/m}$ is held under a tension $T_s = 162\text{ N}$. What is the velocity of transverse mechanical waves propagating along the string?
- 9.0 m/s
- 90 m/s Correct answer.
- 300 m/s
- 180 m/s
Explanation
Transverse wave speed on a stretched string is $v = \sqrt{T_s / \mu} = \sqrt{162 / 0.02} = \sqrt{8100} = 90\text{ m/s}$.