Physics II Practice

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1.1Compute Force from an Electric Field

Exam I | Problem 1.1 |

A $3.0\,\mu\text{C}$ charge is placed in a uniform electric field of $250\ \text{N/C}$. What is the force magnitude?

Solution

Use

$$ F = qE $$

F = (3.0 \times 10^{-6})(250) = 7.5 \times 10^{-4}\ \text{N}

The force magnitude is $7.5 \times 10^{-4}\ \text{N}$.

1.2Find the Force Between Two Point Charges

Exam I | Problem 1.2 |

Coulomb's law, Charge

A $+2.0\,\mu\text{C}$ charge and a $-3.0\,\mu\text{C}$ charge are $0.40\ \text{m}$ apart. What is the magnitude of the force between them? State whether the force is attractive or repulsive.

Solution

Use Coulomb's law:

F = k \frac{|q_1 q_2|}{r^2}

Substitute the values:

F = (9.0 \times 10^9)\frac{(2.0 \times 10^{-6})(3.0 \times 10^{-6})}{(0.40)^2}

F = 0.34\ \text{N}

Because the charges have opposite signs, the force is attractive.

1.3Electric Field of a Point Charge

Exam I | Problem 1.3 |

Electric field

Find the magnitude of the electric field $0.30\ \text{m}$ from a $+4.0\,\text{nC}$ point charge.

Solution

For a point charge,

\mathbf{E} = k \frac{q}{r^2}

E = (9.0 \times 10^9)\frac{4.0 \times 10^{-9}}{(0.30)^2}

E = 400\ \text{N/C}

The field points away from the positive charge.

1.4Calculate Electric Flux

Exam I | Problem 1.4 |

Electric flux

A uniform field of $150\ \text{N/C}$ passes through a flat $0.20\ \text{m}^2$ surface. The field makes a $60^\circ$ angle with the surface normal. What is the electric flux?

Solution

For a uniform field through a flat surface,

\Phi_E = EA\cos\theta

\Phi_E = (150)(0.20)\cos 60^\circ

\Phi_E = 15\ \text{N}\cdot\text{m}^2/\text{C}

1.5Use Gauss's Law to Find Enclosed Charge

Exam I | Problem 1.5 |

Gauss's law

A closed surface has electric flux $1.13 \times 10^4\ \text{N}\cdot\text{m}^2/\text{C}$. What charge is enclosed?

Solution

Gauss's law gives

\oint \mathbf{E}\cdot d\mathbf{A} = \frac{Q_{enc}}{\epsilon_0}

Q_{enc} = \epsilon_0 \Phi_E

Substitute:

Q_{enc} = (8.85 \times 10^{-12})(1.13 \times 10^4)

Q_{enc} \approx 1.0 \times 10^{-7}\ \text{C}

The enclosed charge is positive because the flux is positive.

1.6Potential of a Point Charge

Exam I | Problem 1.6 |

Electric potential

What is the electric potential $0.20\ \text{m}$ from a $-5.0\,\text{nC}$ point charge?

Solution

Use the point-charge potential formula:

V = k\frac{q}{r}

V = (9.0 \times 10^9)\frac{-5.0 \times 10^{-9}}{0.20}

V = -225\ \text{V}

1.7Find a Capacitor's Capacitance

Exam I | Problem 1.7 |

Capacitance

A capacitor stores $12\,\mu\text{C}$ when the potential difference across it is $6\ \text{V}$. What is its capacitance?

Solution

Use the definition of capacitance:

C = \frac{Q}{\Delta V}

C = \frac{12 \times 10^{-6}}{6} = 2.0 \times 10^{-6}\ \text{F}

Therefore,

C = 2.0\ \mu\text{F}

1.8Current and Power in a Resistor

Exam I | Problem 1.8 |

Ohm's law, Power

A $12\ \text{V}$ battery is connected across a $6\ \Omega$ resistor. What current flows, and what power is dissipated?

Solution

First use Ohm's law:

I = \frac{V}{R} = \frac{12}{6} = 2\ \text{A}

Then use the power formula:

P = IV = (2)(12) = 24\ \text{W}

The current is $2\ \text{A}$ and the power is $24\ \text{W}$.

1.9Magnetic Force on a Moving Charge

Exam I | Problem 1.9 |

Magnetic force

A $+2.0\,\mu\text{C}$ charge moves at $300\ \text{m/s}$ perpendicular to a $0.50\ \text{T}$ magnetic field. What is the magnetic force magnitude?

Solution

For a charge moving perpendicular to the field,

$$ F = qvB $$

F = (2.0 \times 10^{-6})(300)(0.50)

F = 3.0 \times 10^{-4}\ \text{N}

1.10Electric Field Amplitude of an EM Wave

Exam I | Problem 1.10 |

EM waves

An electromagnetic wave in vacuum has magnetic-field amplitude $2.0 \times 10^{-6}\ \text{T}$. What is the electric-field amplitude?

Solution

In vacuum,

$$ E = cB $$

E = (3.0 \times 10^8)(2.0 \times 10^{-6})

E = 600\ \text{V/m}

Difficulty

Integrated

2.1Equivalent Resistance in Series and Parallel

Exam II | Problem 2.1 |

A $4\ \Omega$ resistor is in series with a parallel pair of $6\ \Omega$ and $3\ \Omega$ resistors. If the combination is connected to a $12\ \text{V}$ source, what total current flows?

Solution

First combine the parallel resistors:

\frac{1}{R_p} = \frac{1}{6} + \frac{1}{3} = \frac{1}{2}

R_p = 2\ \Omega

Now add the series resistor:

R_{eq} = 4 + 2 = 6\ \Omega

Use Ohm's law for the whole circuit:

I = \frac{V}{R_{eq}} = \frac{12}{6} = 2\ \text{A}

2.2RC Time Constant and Charging Charge

Exam II | Problem 2.2 |

A $20\,\mu\text{F}$ capacitor charges through a $150\ \text{k}\Omega$ resistor from a $9\ \text{V}$ battery. Find the time constant and the charge after one time constant.

Solution

The time constant is

\tau = RC = (150 \times 10^3)(20 \times 10^{-6}) = 3.0\ \text{s}

The final charge is

Q_f = CV = (20 \times 10^{-6})(9) = 180 \times 10^{-6}\ \text{C}

After one time constant,

Q(\tau) = Q_f\left(1-e^{-1}\right)

Q(\tau) \approx 180\ \mu\text{C}(0.632) \approx 114\ \mu\text{C}

2.3Currents in Parallel Branches

Exam II | Problem 2.3 |

A $12\ \text{V}$ source is connected across a $6\ \Omega$ resistor and a $3\ \Omega$ resistor in parallel. What is the current in each branch and the total current?

Solution

In parallel, each branch has the full $12\ \text{V}$ across it.

For the $6\ \Omega$ branch:

I_1 = \frac{12}{6} = 2\ \text{A}

For the $3\ \Omega$ branch:

I_2 = \frac{12}{3} = 4\ \text{A}

The total current is

I_{tot} = I_1 + I_2 = 6\ \text{A}

2.4Field from a Long Straight Wire

Exam II | Problem 2.4 |

Magnetic field of a wire

A long straight wire carries $10\ \text{A}$. What is the magnetic field magnitude $5.0\ \text{cm}$ from the wire?

Solution

For a long straight wire,

B = \frac{\mu_0 I}{2\pi r}

Substitute the values:

B = \frac{(4\pi \times 10^{-7})(10)}{2\pi(0.050)}

B = 4.0 \times 10^{-5}\ \text{T}

2.5Force on a Current-Carrying Wire

Exam II | Problem 2.5 |

Magnetic force on a wire

A $0.40\ \text{m}$ wire carries $5.0\ \text{A}$ in a $0.20\ \text{T}$ field. The wire is perpendicular to the field. What is the magnetic force?

Solution

Use the wire-force formula:

F = ILB\sin\theta

With $\theta = 90^\circ$,

F = (5.0)(0.40)(0.20) = 0.40\ \text{N}

2.6Induced emf in a Coil

Exam II | Problem 2.6 |

Faraday's law, Lenz's law

A $25$-turn coil of area $0.020\ \text{m}^2$ sits perpendicular to a magnetic field. The field drops from $0.60\ \text{T}$ to $0.10\ \text{T}$ in $0.10\ \text{s}$. The coil resistance is $5.0\ \Omega$. Find the induced emf and current magnitude.

Solution

The change in field is

\Delta B = 0.60 - 0.10 = 0.50\ \text{T}

The flux change per turn is

\Delta \Phi_B = A\Delta B = (0.020)(0.50) = 0.010\ \text{Wb}

So the induced emf magnitude is

\mathcal{E} = N\frac{\Delta \Phi_B}{\Delta t} = 25\cdot\frac{0.010}{0.10} = 2.5\ \text{V}

Then

I = \frac{\mathcal{E}}{R} = \frac{2.5}{5.0} = 0.50\ \text{A}

2.7Current Growth in an RL Circuit

Exam II | Problem 2.7 |

RL circuits

An RL circuit has $L = 2.0\ \text{H}$ and $R = 10\ \Omega$. If the current is growing toward a maximum of $4.0\ \text{A}$, what is the time constant and the current after one time constant?

Solution

The time constant is

\tau = \frac{L}{R} = \frac{2.0}{10} = 0.20\ \text{s}

For current growth,

I(t) = I_{\max}\left(1-e^{-tR/L}\right)

At $t=\tau$,

I(\tau) = 4.0(1-e^{-1}) \approx 4.0(0.632) = 2.53\ \text{A}

2.8Resonance in a Series RLC Circuit

Exam II | Problem 2.8 |

A series RLC circuit has $R = 15\ \Omega$, $L = 0.25\ \text{H}$, and $C = 0.040\ \text{F}$. What is the resonant angular frequency, and what current amplitude flows at resonance if the source has amplitude $30\ \text{V}$?

Solution

At resonance,

\omega_0 = \frac{1}{\sqrt{LC}}

\omega_0 = \frac{1}{\sqrt{(0.25)(0.040)}} = \frac{1}{\sqrt{0.010}} = 10\ \text{rad/s}

At resonance, $X_L = X_C$, so the impedance is just

Z = R = 15\ \Omega

Thus the current amplitude is

I_0 = \frac{V_0}{Z} = \frac{30}{15} = 2\ \text{A}

Difficulty

Applied

3.1Field from an Infinite Line Charge

Final | Problem 3.1 |

An infinite line has linear charge density $4.0\,\mu\text{C/m}$. What is the electric field magnitude $0.10\ \text{m}$ away?

Solution

Use a cylindrical Gaussian surface. Gauss's law gives

E(2\pi rL) = \frac{\lambda L}{\epsilon_0}

E = \frac{\lambda}{2\pi \epsilon_0 r}

Substitute:

E = \frac{4.0 \times 10^{-6}}{2\pi(8.85 \times 10^{-12})(0.10)}

E \approx 7.2 \times 10^5\ \text{N/C}

The field points outward from the line if the charge density is positive.

3.2Infer Resistivity from Geometry

Final | Problem 3.2 |

Resistivity

A wire is $2.0\ \text{m}$ long, has resistance $0.80\ \Omega$, and cross-sectional area $2.0 \times 10^{-6}\ \text{m}^2$. What is its resistivity?

Solution

Use

R = \rho \frac{L}{A}

\rho = \frac{RA}{L}

Substitute:

\rho = \frac{(0.80)(2.0 \times 10^{-6})}{2.0}

\rho = 8.0 \times 10^{-7}\ \Omega\cdot\text{m}

3.3Radius of Circular Motion in a Magnetic Field

Final | Problem 3.3 |

A particle with mass $3.2 \times 10^{-26}\ \text{kg}$ and charge $1.6 \times 10^{-19}\ \text{C}$ moves at $2.0 \times 10^5\ \text{m/s}$ perpendicular to a $0.50\ \text{T}$ field. What is the radius of its path?

Solution

When velocity is perpendicular to the magnetic field, the magnetic force provides centripetal force. The radius is

r = \frac{mv}{|q|B}

Substitute:

r = \frac{(3.2 \times 10^{-26})(2.0 \times 10^5)}{(1.6 \times 10^{-19})(0.50)}

r = 0.080\ \text{m}

3.4Image Formation with a Thin Lens

Final | Problem 3.4 |

Thin lenses

A converging lens has focal length $15\ \text{cm}$. An object is placed $30\ \text{cm}$ from the lens. Find the image distance and magnification.

Solution

Use the thin-lens equation:

\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}

Substitute the values:

\frac{1}{15} = \frac{1}{30} + \frac{1}{d_i}

\frac{1}{d_i} = \frac{1}{30}

and

d_i = 30\ \text{cm}

The magnification is

m = -\frac{d_i}{d_o} = -\frac{30}{30} = -1

The image is real, inverted, and the same size as the object.

3.5Double-Slit Fringe Position

Final | Problem 3.5 |

Interference

A double slit has slit separation $0.20\ \text{mm}$. A screen is $2.0\ \text{m}$ away. Light of wavelength $500\ \text{nm}$ is used. Where is the second bright fringe?

Solution

For small angles, bright fringes occur at

y_m \approx \frac{m\lambda L}{d}

For the second bright fringe, $m=2$:

y_2 = \frac{2(500 \times 10^{-9})(2.0)}{0.20 \times 10^{-3}}

y_2 = 1.0 \times 10^{-2}\ \text{m}

So the second bright fringe is $1.0\ \text{cm}$ from the center.

Difficulty

Challenge

4.1Induction with Lenz's Law

Final | Problem 4.1 |

Faraday's law, Lenz's law

A square single-turn loop with side $0.20\ \text{m}$ lies in a magnetic field pointing out of the page. The field increases from $0.10\ \text{T}$ to $0.30\ \text{T}$ in $0.50\ \text{s}$. If the loop resistance is $0.40\ \Omega$, what are the induced current magnitude and direction?

Solution

The area of the loop is

A = (0.20)^2 = 0.040\ \text{m}^2

The change in field is

\Delta B = 0.30 - 0.10 = 0.20\ \text{T}

So the flux change is

\Delta \Phi_B = A\Delta B = (0.040)(0.20) = 0.008\ \text{Wb}

The induced emf magnitude is

\mathcal{E} = \frac{\Delta \Phi_B}{\Delta t} = \frac{0.008}{0.50} = 0.016\ \text{V}

Thus the induced current is

I = \frac{\mathcal{E}}{R} = \frac{0.016}{0.40} = 0.040\ \text{A}

Because the out-of-page flux is increasing, the induced field must point into the page, so the current is clockwise.

4.2Exponential Charging in an RC Circuit

Final | Problem 4.2 |

A $50\,\mu\text{F}$ capacitor charges through a $40\ \text{k}\Omega$ resistor from a $24\ \text{V}$ battery. Find the time constant, the charge after two time constants, and the current after two time constants.

Solution

The time constant is

\tau = RC = (40 \times 10^3)(50 \times 10^{-6}) = 2.0\ \text{s}

The final charge is

Q_f = CV = (50 \times 10^{-6})(24) = 1.2 \times 10^{-3}\ \text{C}

After two time constants,

Q(2\tau) = Q_f(1-e^{-2})

Q(2\tau) \approx (1.2 \times 10^{-3})(0.865) \approx 1.04 \times 10^{-3}\ \text{C}

For the current,

I(t) = \frac{V}{R}e^{-t/RC}

so at $t = 2\tau$,

I(2\tau) = \frac{24}{40\,000}e^{-2} \approx 8.1 \times 10^{-5}\ \text{A}

That is about $81\ \mu\text{A}$.

4.3Series RLC Behavior Away from Resonance

Final | Problem 4.3 |

A series circuit has $R = 30\ \Omega$, $L = 0.20\ \text{H}$, and $C = 100\ \mu\text{F}$. It is driven by a $120\ \text{V}$ source at $50\ \text{Hz}$. Find $X_L$, $X_C$, the impedance, and whether the current leads or lags the source voltage.

Solution

First find the angular frequency:

\omega = 2\pi f = 2\pi(50) \approx 314\ \text{rad/s}

Then compute the reactances:

X_L = \omega L \approx (314)(0.20) = 62.8\ \Omega

X_C = \frac{1}{\omega C} = \frac{1}{(314)(100 \times 10^{-6})} \approx 31.8\ \Omega

Now find the impedance:

Z = \sqrt{R^2 + (X_L - X_C)^2}

Z = \sqrt{30^2 + (62.8 - 31.8)^2} \approx 43.1\ \Omega

The current amplitude is

I_0 = \frac{V_0}{Z} = \frac{120}{43.1} \approx 2.8\ \text{A}

Because $X_L > X_C$, the circuit is net inductive, so the current lags the source voltage.

4.4Total Internal Reflection

Final | Problem 4.4 |

Light goes from glass with index $1.5$ into air with index $1.0$. What is the critical angle? If the incident angle is $45^\circ$, what happens?

Solution

Use the critical-angle condition:

\sin \theta_c = \frac{n_2}{n_1}

\sin \theta_c = \frac{1.0}{1.5} = \frac{2}{3}

which gives

\theta_c \approx 41.8^\circ

Since $45^\circ > 41.8^\circ$, the light undergoes total internal reflection.