Pomodoro
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Difficulty
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
so
The force magnitude is $7.5 \times 10^{-4}\ \text{N}$.
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:
Substitute the values:
Because the charges have opposite signs, the force is attractive.
Find the magnitude of the electric field $0.30\ \text{m}$ from a $+4.0\,\text{nC}$ point charge.
Solution
For a point charge,
so
The field points away from the positive charge.
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,
so
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
so
Substitute:
The enclosed charge is positive because the flux is positive.
What is the electric potential $0.20\ \text{m}$ from a $-5.0\,\text{nC}$ point charge?
Solution
Use the point-charge potential formula:
So
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:
So
Therefore,
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:
Then use the power formula:
The current is $2\ \text{A}$ and the power is $24\ \text{W}$.
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,
so
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,
so
Difficulty
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:
so
Now add the series resistor:
Use Ohm's law for the whole circuit:
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
The final charge is
After one time constant,
so
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:
For the $3\ \Omega$ branch:
The total current is
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,
Substitute the values:
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:
With $\theta = 90^\circ$,
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
The flux change per turn is
So the induced emf magnitude is
Then
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
For current growth,
At $t=\tau$,
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,
so
At resonance, $X_L = X_C$, so the impedance is just
Thus the current amplitude is
Difficulty
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
so
Substitute:
The field points outward from the line if the charge density is positive.
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
so
Substitute:
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
Substitute:
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:
Substitute the values:
so
and
The magnification is
The image is real, inverted, and the same size as the object.
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
For the second bright fringe, $m=2$:
So the second bright fringe is $1.0\ \text{cm}$ from the center.
Difficulty
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
The change in field is
So the flux change is
The induced emf magnitude is
Thus the induced current is
Because the out-of-page flux is increasing, the induced field must point into the page, so the current is clockwise.
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
The final charge is
After two time constants,
so
For the current,
so at $t = 2\tau$,
That is about $81\ \mu\text{A}$.
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:
Then compute the reactances:
Now find the impedance:
The current amplitude is
Because $X_L > X_C$, the circuit is net inductive, so the current lags the source voltage.
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:
So
which gives
Since $45^\circ > 41.8^\circ$, the light undergoes total internal reflection.
