Physics 1C: Magnetic Forces, Induction, and Circuits – Visual Study Guide

Q1. An electron at point A has a speed $v_0 = 1.60 \times 10^6\ \mathrm{m/s}$. Find the magnitude of the magnetic field that will cause the electron to follow the semicircular path from A to B (radius $R = 5\ \mathrm{cm}$).

Background

Topic: Motion of charged particles in magnetic fields

This question tests your understanding of how a magnetic field can exert a force on a moving charge, causing it to move in a circular (or semicircular) path. The force responsible is the magnetic Lorentz force, which acts as a centripetal force for the electron's motion.

Key Terms and Formulas

• Magnetic force: $F_B = qvB$ (when $\vec{v} \perp \vec{B}$)

• Centripetal force: $F_C = \dfrac{mv^2}{R}$

• Equating forces: $qvB = \dfrac{mv^2}{R}$

• Solving for $B$: $B = \dfrac{mv}{qR}$

• Electron mass: $m = 9.11 \times 10^{-31}\ \mathrm{kg}$

• Electron charge: $q = 1.6 \times 10^{-19}\ \mathrm{C}$

• Radius: $R = 0.05\ \mathrm{m}$

Step-by-Step Guidance

1. Identify the forces acting on the electron: The magnetic force provides the centripetal force needed for circular motion.

2. Write the equation for the magnetic force and set it equal to the centripetal force: $qvB = \dfrac{mv^2}{R}$

3. Rearrange the equation to solve for $B$: $B = \dfrac{mv}{qR}$

4. Substitute the known values for $m$, $v_0$, $q$, and $R$ into the equation. Be sure to use SI units for all quantities.

Try solving on your own before revealing the answer!

Final Answer: $B = 1.82 \times 10^{-4}\ \mathrm{T}$

Plugging in the values: $B = \dfrac{9.11 \times 10^{-31} \times 1.60 \times 10^6}{1.6 \times 10^{-19} \times 0.05}$ gives the required magnetic field.

This field will cause the electron to follow the specified semicircular path.

Q2. Find the magnitude of the drift velocity of electrons in the x-direction for a silver ribbon carrying a current of 130 A, with $n = 5.7 \times 10^{28}\ \mathrm{m}^{-3}$, $y_1 = 0.25\ \mathrm{mm}$, $z_1 = 14\ \mathrm{mm}$.

Background

Topic: Current, drift velocity, and the Hall effect

This question examines your understanding of how current relates to the motion of charge carriers (electrons) in a conductor, and how to calculate the drift velocity given the current and physical dimensions of the conductor.

Key Terms and Formulas

• Current: $I = n e A v_d$

• Drift velocity: $v_d = \dfrac{I}{n e A}$

• Area: $A = y_1 z_1$ (convert mm to m)

• Elementary charge: $e = 1.6 \times 10^{-19}\ \mathrm{C}$

Step-by-Step Guidance

1. Convert $y_1$ and $z_1$ to meters: $y_1 = 0.25 \times 10^{-3}\ \mathrm{m}$, $z_1 = 14 \times 10^{-3}\ \mathrm{m}$.

2. Calculate the cross-sectional area $A = y_1 z_1$.

3. Plug the values for $I$, $n$, $e$, and $A$ into the drift velocity formula: $v_d = \dfrac{I}{n e A}$

4. Check that all units are in SI before calculating.

Try solving on your own before revealing the answer!

Final Answer: $v_d = 4.1 \times 10^{-3}\ \mathrm{m/s}$

Using the formula and plugging in the values gives the drift velocity of electrons in the ribbon.

Q3. Calculate the magnitude of the magnetic force on a voice coil in a loudspeaker with 60 turns, diameter 1.40 cm, current 0.960 A, and a magnetic field of 0.200 T at $60.0^\circ$ to the normal.

Background

Topic: Magnetic force on a current-carrying loop

This question tests your ability to calculate the force on a coil in a non-uniform magnetic field, considering the geometry and orientation of the coil.

Key Terms and Formulas

• Magnetic force on a coil: $F = N I B d \cos(\theta)$ (where $d$ is the diameter, $N$ is the number of turns, $\theta$ is the angle between $\vec{B}$ and the normal to the coil)

• For a circular coil, integrate over the circumference if needed

• Convert diameter to meters: $d = 0.014\ \mathrm{m}$

Step-by-Step Guidance

1. Convert the diameter to meters if necessary.

2. Calculate $\cos(30^\circ)$ since the field is $60^\circ$ from the normal (complementary angle).

3. Plug the values into the formula: $F = \pi N I B d \cos(30^\circ)$

4. Be careful with the units and the trigonometric calculation.

Try solving on your own before revealing the answer!

Final Answer: $F = 0.439\ \mathrm{N}$

Substituting all values gives the net force on the coil in the specified direction.

Q4. For a square loop of wire in the $xy$-plane with current $I$ and a non-uniform magnetic field $\vec{B} = \frac{B_0 z}{L} \hat{j} + \frac{B_0 y}{L} \hat{k}$, find the magnitude and direction of the magnetic force on each side of the loop.

Background

Topic: Magnetic force on current-carrying loops in non-uniform fields

This question tests your ability to use the differential form of the magnetic force and integrate along each segment of the loop, considering the spatial variation of the field.

Key Terms and Formulas

• Differential force: $d\vec{F} = I d\vec{l} \times \vec{B}$

• Integrate along each side, using the appropriate limits and directions for $d\vec{l}$ and $\vec{B}$

Step-by-Step Guidance

1. For each side, write $d\vec{l}$ and $\vec{B}$ at that segment.

2. Compute the cross product $d\vec{l} \times \vec{B}$ for each side.

3. Integrate over the length of each side to find the total force.

4. Determine the direction of each force using the right-hand rule and vector analysis.

Try solving on your own before revealing the answer!

Final Answer: Side 1: $F_1 = \frac{1}{2} I L B_0$ in $+x$; Side 2: $F_2 = I B_0 L$ in $-y$; Side 3: $F_3 = \frac{1}{2} I L B_0$ in $-x$; Side 4: $F_4 = 0$

The net force is $I L B_0$ in the $-y$ direction, as only side 2's force survives after cancellation.