Problem Statement
Solve the magnetic field/force problem: Problem 3.189 — Electromagnetic induction: Faraday’s law See problem statement for all given quantities. This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematicall
Given Information
- Current $I$ or charge $q$ and velocity $v$ as given
- Geometry (straight wire, loop, solenoid) as specified
- Permeability of free space $\mu_0 = 4\pi\times10^{-7}\,\text{T m A}^{-1}$
Physical Concepts & Formulas
Magnetic fields arise from moving charges (currents). The Biot-Savart Law gives the fundamental relation: each current element $Id\vec{l}$ contributes $d\vec{B} = (\mu_0/4\pi)\,Id\vec{l}\times\hat{r}/r^2$ to the field at a field point. For high symmetry situations (infinite straight wire, solenoid, toroid), Ampere’s Law $\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}$ is far more efficient. The magnetic force on a moving charge $\vec{F} = q\vec{v}\times\vec{B}$ is always perpendicular to the velocity — it changes direction but never does work, so it cannot change the kinetic energy of the charge.
- $d\vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec{l}\times\hat{r}}{r^2}$ — Biot-Savart Law
- $B = \dfrac{\mu_0 I}{2\pi r}$ — infinite straight wire at distance $r$
- $B = \dfrac{\mu_0 I}{2R}$ — centre of circular loop of radius $R$
- $B = \mu_0 n I$ — inside a solenoid ($n$ turns per metre)
- $\vec{F} = q\vec{v}\times\vec{B}$ — Lorentz force on moving charge
- $\vec{F} = I\vec{L}\times\vec{B}$ — force on current-carrying wire
Step-by-Step Solution
Step 1 — Choose method: Biot-Savart for arbitrary geometry; Ampere’s Law for high symmetry.
Step 2 — Set up Amperian loop: (For Ampere’s method) Choose a loop where $B$ is constant along the path.
$$\oint \vec{B}\cdot d\vec{l} = B\cdot l = \mu_0 I_{\text{enc}}$$
Step 3 — Solve for $B$:
$$B = \frac{\mu_0 I_{\text{enc}}}{l}$$
Step 4 — Direction: Use right-hand rule — curl fingers in direction of current; thumb points along $\vec{B}$.
Worked Calculation
Substituting all values with units:
Infinite straight wire, $I = 10\,\text{A}$, $r = 0.05\,\text{m}$:
$$B = \frac{\mu_0 I}{2\pi r} = \frac{4\pi\times10^{-7}\times10}{2\pi\times0.05} = \frac{4\pi\times10^{-6}}{0.1\pi} = 4\times10^{-5}\,\text{T} = 40\,\mu\text{T}$$
Answer
$$\boxed{B = \dfrac{\mu_0 I}{2\pi r}}$$
Physical Interpretation
40 μT is about the same order as Earth’s magnetic field (~50 μT). A current of 10 A in a household wire at 5 cm distance produces a significant field — this is why high-voltage power lines can affect sensitive electronic instruments. The $1/r$ dependence means the field halves every time you double the distance.
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