The Timpson Lyceum · University Physics II (2026) – Electricity & Magnetism

Chapter 28 — Sources of Magnetic Field

Problems with worked solutions · 13 points · 3 extra credit

Chapter 28 — Sources of Magnetic Field

Core: 13 points · Extra credit: 3 points. Show your reasoning; partial credit is given.

1. 1. (2 pts ••) A point charge $q=+2.0\,\text{nC}$ moves with speed $v=4.0\times10^{6}\,\text{m/s}$. Consider a field point a distance $r=3.0\,\text{mm}$ from the charge.

   (a) Find the magnitude of the magnetic field this charge produces at that point when the point lies in the direction perpendicular to the velocity.

   (b) Find the field magnitude at the same distance for a point whose position vector makes a $30^\circ$ angle with the velocity.

   Solution

   1.

   (a) For a moving point charge, $B=\dfrac{\mu_0}{4\pi}\dfrac{q\,v\sin\phi}{r^2}$, where $\phi$ is the angle between $\vec v$ and the line to the field point. Perpendicular means $\phi=90^\circ$, so $\sin\phi=1$:

   $$B=\frac{\mu_0}{4\pi}\frac{qv}{r^2}=(1.0\times10^{-7})\frac{(2.0\times10^{-9})(4.0\times10^{6})}{(3.0\times10^{-3})^2}\approx 8.9\times10^{-5}\,\text{T}.$$

   (b) The only change is the factor $\sin\phi=\sin30^\circ=0.5$:

   $$B=(8.9\times10^{-5})(0.5)\approx 4.4\times10^{-5}\,\text{T}.$$

2. 2. (2 pts ••) A long, straight wire carries a steady current $I=8.0\,\text{A}$.

   (a) Find the magnetic field magnitude at a perpendicular distance $r=5.0\,\text{cm}$ from the wire.

   (b) At what perpendicular distance does the field drop to $20\,\mu\text{T}$?

   Solution

   2.

   (a) For a long straight wire, $B=\dfrac{\mu_0 I}{2\pi r}=\dfrac{(4\pi\times10^{-7})(8.0)}{2\pi(0.050)}\approx 3.2\times10^{-5}\,\text{T}$.

   (b) Solve $B=\dfrac{\mu_0 I}{2\pi r}$ for $r$: $r=\dfrac{\mu_0 I}{2\pi B}=\dfrac{(4\pi\times10^{-7})(8.0)}{2\pi(20\times10^{-6})}\approx 0.080\,\text{m}=8.0\,\text{cm}$.

3. 3. (2 pts ••) Two long, straight, parallel wires are separated by $d=10\,\text{cm}$. Each carries a current of $6.0\,\text{A}$, but in opposite directions. Find the magnitude of the net magnetic field at the point midway between the wires, and state the reasoning that lets you add the two contributions.

   Solution

   3. Each wire is $r=5.0\,\text{cm}$ from the midpoint, so each produces $B_1=\dfrac{\mu_0 I}{2\pi r}=\dfrac{(4\pi\times10^{-7})(6.0)}{2\pi(0.050)}=2.4\times10^{-5}\,\text{T}$.

   Because the currents run in opposite directions, at the midpoint the two field vectors (each circling its own wire by the right-hand rule) point the same way, so they add:

   $$B=2(2.4\times10^{-5})\approx 4.8\times10^{-5}\,\text{T}.$$

4. 4. (2 pts ••) A flat circular coil has $N=50$ turns, radius $R=4.0\,\text{cm}$, and carries a current $I=2.5\,\text{A}$.

   (a) Find the magnetic field magnitude at the center of the coil.

   (b) Find the field magnitude on the coil's axis at a distance $x=3.0\,\text{cm}$ from the center.

   Solution

   4.

   (a) On-axis field of $N$ turns at the center ($x=0$): $B=\dfrac{\mu_0 N I}{2R}=\dfrac{(4\pi\times10^{-7})(50)(2.5)}{2(0.040)}\approx 2.0\times10^{-3}\,\text{T}$.

   (b) $B=\dfrac{\mu_0 N I R^2}{2(R^2+x^2)^{3/2}}$. With $R^2+x^2=(0.040)^2+(0.030)^2=2.5\times10^{-3}\,\text{m}^2$ and $(R^2+x^2)^{3/2}=1.25\times10^{-4}$:

   $$B=\frac{(4\pi\times10^{-7})(50)(2.5)(0.040)^2}{2(1.25\times10^{-4})}\approx 1.0\times10^{-3}\,\text{T}.$$

5. 5. (2 pts ••) Two long, straight, parallel wires are $2.0\,\text{cm}$ apart. One carries $I_1=10\,\text{A}$ and the other $I_2=15\,\text{A}$ in the same direction.

   (a) Find the force per unit length each wire exerts on the other.

   (b) State whether the wires attract or repel, and why.

   Solution

   5.

   (a) $\dfrac{F}{L}=\dfrac{\mu_0 I_1 I_2}{2\pi d}=\dfrac{(4\pi\times10^{-7})(10)(15)}{2\pi(0.020)}\approx 1.5\times10^{-3}\,\text{N/m}$.

   (b) The wires attract. Each wire sits in the magnetic field of the other; for currents in the same direction the force $I\vec L\times\vec B$ on each points toward the other wire (parallel currents attract, antiparallel repel).

6. 6. (1 pt •) A solenoid is $40\,\text{cm}$ long and is wound with $1200$ turns of wire carrying a current of $3.0\,\text{A}$. Treating it as an ideal (long) solenoid, find the magnetic field magnitude inside it.

   Solution

   6. Turns per unit length $n=\dfrac{1200}{0.40}=3000\,\text{m}^{-1}$. Inside an ideal solenoid $B=\mu_0 n I=(4\pi\times10^{-7})(3000)(3.0)\approx 1.1\times10^{-2}\,\text{T}$.

7. 7. (2 pts ••) A toroid has $N=500$ turns and carries a current $I=4.0\,\text{A}$. A point inside the toroid's core lies a distance $r=10\,\text{cm}$ from the central axis.

   (a) Use Ampère's law to find the magnetic field magnitude at that point.

   (b) Explain, using the same Ampèrian-loop argument, why the field is zero in the hole at the very center of the toroid (on the central axis).

   Solution

   7.

   (a) Apply Ampère's law to a circular Ampèrian loop of radius $r$ inside the core, concentric with the axis. By symmetry $B$ is constant along it and tangent to it, and the loop encloses all $N$ turns once: $B(2\pi r)=\mu_0 N I$, so

   $$B=\frac{\mu_0 N I}{2\pi r}=\frac{(4\pi\times10^{-7})(500)(4.0)}{2\pi(0.10)}\approx 4.0\times10^{-3}\,\text{T}.$$

   (b) Take an Ampèrian loop lying in the central hole (radius smaller than the toroid's inner radius). It encloses no current, so $\oint\vec B\cdot d\vec\ell=\mu_0 I_\text{enc}=0$. Combined with the toroid's symmetry (any field would have to be uniform and tangential around the loop), this forces $B=0$ there — the ideal toroid confines its field entirely to the core.

8. Extra Credit (3 pts •••)

   A single circular loop of radius $R$ carries a steady current $I$. Using the Biot–Savart law, derive the magnetic field on the axis of the loop at a distance $x$ from the center.

   (a) Set up the Biot–Savart integral for a current element $I\,d\vec{\ell}$, and explain carefully why only the axial component of $d\vec B$ survives when you integrate around the full loop.

   (b) Carry out the integration to obtain

   $$B_x=\frac{\mu_0 I R^2}{2\,(R^2+x^2)^{3/2}} .$$

   (c) Show that your result reduces to the correct field at the center of the loop, and find its numerical value for $R=4.0\,\text{cm}$, $I=2.5\,\text{A}$, at the point $x=R$.

   Solution

   Extra Credit

   (a) Place the loop in the $yz$-plane centered at the origin, with the axis along $x$. A current element $I\,d\vec\ell$ on the ring is a distance $s=\sqrt{R^2+x^2}$ from the axial field point $P$, and $d\vec\ell$ is perpendicular to the vector $\hat r$ from the element to $P$ (the element is tangent to the ring; $\hat r$ lies in the plane containing the axis and the element). Biot–Savart gives

   $$dB=\frac{\mu_0}{4\pi}\frac{I\,|d\vec\ell\times\hat r|}{s^2}=\frac{\mu_0}{4\pi}\frac{I\,d\ell}{R^2+x^2}.$$

   Each $d\vec B$ is perpendicular to $\hat r$ and so makes an angle with the axis. As we go around the loop, the components of $d\vec B$ perpendicular to the axis point in all directions and cancel in pairs (the element diametrically opposite contributes an equal, opposite transverse component); only the axial components survive. The axial component carries a factor $\cos\theta=\dfrac{R}{\sqrt{R^2+x^2}}$, where $\theta$ is the angle between $d\vec B$ and the axis.

   (b) Keeping only the axial part and integrating $d\ell$ around the full circumference ($\oint d\ell=2\pi R$):

   $$B_x=\int \frac{\mu_0}{4\pi}\frac{I\,d\ell}{R^2+x^2}\cdot\frac{R}{\sqrt{R^2+x^2}} =\frac{\mu_0 I R}{4\pi (R^2+x^2)^{3/2}}\oint d\ell =\frac{\mu_0 I R}{4\pi (R^2+x^2)^{3/2}}(2\pi R),$$ $$\boxed{\,B_x=\frac{\mu_0 I R^2}{2\,(R^2+x^2)^{3/2}}\,}.$$

   For $N$ turns, multiply by $N$.

   (c) At the center, $x=0$, so $(R^2+x^2)^{3/2}=R^3$ and

   $$B_x(0)=\frac{\mu_0 I R^2}{2R^3}=\frac{\mu_0 I}{2R},$$

   the standard center-of-loop result. Numerically at $x=R$, using $R=0.040\,\text{m}$, $I=2.5\,\text{A}$: $(R^2+R^2)^{3/2}=(2R^2)^{3/2}=2^{3/2}R^3$, so

   $$B_x=\frac{\mu_0 I R^2}{2\cdot 2^{3/2}R^3}=\frac{\mu_0 I}{2^{5/2}R}=\frac{(4\pi\times10^{-7})(2.5)}{2^{5/2}(0.040)}\approx 1.4\times10^{-5}\,\text{T}.$$