Exercises — Biot-Savart law — magnetic field from current element

WHAT: Recognise "very long straight wire" ⇒ use $B=\frac{{\mu }_{0}I}{2\pi a}$. WHY: The wire is stated infinite and we want field at a perpendicular distance — that is exactly the closed form we derived, so no integration is needed. $B=\frac{{\mu }_{0}I}{2\pi a}=\frac{(4\pi \times 10^{-7})(5)}{2\pi (0.10)}=\frac{2\times 10^{-7}\times 5}{0.10}=1.0\times 10^{-5}\text{T}.$ (Tidy trick: $\frac{{\mu }_0}{2\pi }=2\times 10^{-7}$, so $B=2\times 10^{-7}\cdot \frac{I}{a}$.) Answer: $B=1.0\times 10^{-5}\text{T}=10\mu \text{T}$.

WHAT: Centre of a loop ⇒ $B=\frac{{\mu }_{0}I}{2R}$. WHY: At the centre every element is the same distance $R$ away and all $d\vec{B}$ point along the axis in the same direction, giving the clean ${\mu }_{0}I/2R$. $B=\frac{(4\pi \times 10^{-7})(2)}{2(0.05)}=\frac{4\pi \times 10^{-7}\times 2}{0.10}=2.51\times 10^{-5}\text{T}.$ Answer: $B\approx 2.51\times 10^{-5}\text{T}=25.1\mu \text{T}$.

WHAT: Finite wire ⇒ $B=\frac{{\mu }_{0}I}{4\pi a}(\sin {\theta }_1+\sin {\theta }_2)$. WHY: The wire is not infinite, so we cannot use ${\mu }_{0}I/2\pi a$. Because $P$ faces the midpoint, the two ends are symmetric: ${\theta }_1={\theta }_2$. Geometry (Figure s02 shows exactly these angles): each end is a horizontal distance $L/2=0.10\text{m}$ from the foot, at perpendicular distance $a=0.05$. The angle to an end, measured from the foot, has $\sin \theta =\frac{L/2}{\sqrt{a^2+(L/2{)}^2}}=\frac{0.10}{\sqrt{0.05^2+0.10^2}}=\frac{0.10}{\sqrt{0.0125}}=\frac{0.10}{0.1118}=\mathrm{0.894.}$ So $\sin {\theta }_1+\sin {\theta }_2=2(0.894)=1.789$. $B=\frac{{\mu }_{0}I}{4\pi a}(1.789)=10^{-7}\cdot \frac{3}{0.05}\cdot 1.789=10^{-7}\cdot 60\cdot 1.789=1.07\times 10^{-5}\text{T}.$ Answer: $B\approx 1.07\times 10^{-5}\text{T}$.

WHAT: On-axis ⇒ $B=\frac{{\mu }_{0}I{R}^2}{2(R^2+x^2{)}^{3/2}}$. WHY: Off-centre on the axis, the $d\vec{B}$ vectors tilt; only their axial components survive (the radial ones cancel by symmetry), and that projection is what turns ${\mu }_{0}I/2R$ into this $R^2/(R^2+x^2{)}^{3/2}$ form. Here $R^2=0.01$, $R^2+x^2=0.02$, $(0.02{)}^{3/2}=0.02\sqrt{0.02}=0.02\times 0.1414=2.83\times 10^{-3}$. $B=\frac{(4\pi \times 10^{-7})(4)(0.01)}{2(2.83\times 10^{-3})}=\frac{(4\pi \times 10^{-7})(0.04)}{5.66\times 10^{-3}}.$ Numerator $=4\pi \times 10^{-7}\times 0.04=5.027\times 10^{-8}$. Divide: $B=8.89\times 10^{-6}\text{T}$. Answer: $B\approx 8.89\times 10^{-6}\text{T}$.

WHAT: Superpose two infinite-wire fields at the midpoint, each at distance $a=d/2=0.05\text{m}$. WHY & direction: Apply the right-hand rule (restated at the top) to each wire at the midpoint. Draw both currents pointing up: the midpoint is to the right of wire 1 (so its field is out of the page, $+$) and to the left of wire 2 (so its field is into the page, $-$). With out-of-page as positive, the two signed fields are $+B_1$ and $-B_2$ — they subtract. Each magnitude: $B_1=B_2=\frac{{\mu }_{0}I}{2\pi a}=2\times 10^{-7}\cdot \frac{10}{0.05}=4.0\times 10^{-5}\text{T}.$ Net $=+B_1-B_2=0$. Answer: $B_{\text{net}}=0\text{T}$. (Equal, opposite ⇒ perfect cancellation at the midpoint.)

WHAT: Same two magnitudes, but reverse one current. WHY: Reversing $I_2$ flips its field's sign at the midpoint (right-hand rule): it was $-B_2$, now it is $+B_2$. So both signed fields are now out of the page ($+$) — they add. $B_{\text{net}}=+B_1+B_2=4.0\times 10^{-5}+4.0\times 10^{-5}=8.0\times 10^{-5}\text{T (out of page)}.$ Answer: $B_{\text{net}}=8.0\times 10^{-5}\text{T}$.

WHAT: A full loop gives ${\mu }_{0}I/2R$; a semicircle is exactly half the arc, so half the field. WHY: At the centre every arc element sits at distance $R$ with $\theta =90^{\circ }$ and all $d\vec{B}$ are parallel (all out of the page, by the right-hand rule). The field is proportional to arc length; a semicircle has half the arc length of the full loop. (The straight radial leads contribute nothing: for them $d\vec{l}\parallel \hat{r}$, so $\sin \theta =0$.) $B=\frac{1}{2}\cdot \frac{{\mu }_{0}I}{2R}=\frac{{\mu }_{0}I}{4R}=\frac{(4\pi \times 10^{-7})(6)}{4(0.08)}=\frac{7.54\times 10^{-6}}{0.32}=2.36\times 10^{-5}\text{T}.$ Answer: $B\approx 2.36\times 10^{-5}\text{T}$.

WHAT: Treat each side as a finite wire and use $\frac{{\mu }_{0}I}{4\pi a}(\sin {\theta }_1+\sin {\theta }_2)$; all four contributions are equal and same-signed, so multiply one by 4. WHY: No single stock formula fits a square, but its sides are just four finite wires — decompose, then superpose. By the right-hand rule, walking the current around the square makes all four $d\vec{B}$ point the same way (out of the page) at the centre, so all four signs agree. Geometry for one side: the centre is at perpendicular distance $a=s/2=0.10\text{m}$ from each side. The perpendicular foot lands at the side's midpoint, so each end is $s/2$ away along the wire: $\sin {\theta }_1=\sin {\theta }_2=\frac{s/2}{\sqrt{(s/2{)}^2+(s/2{)}^2}}=\frac{1}{\sqrt{2}}=\mathrm{0.707.}$ One side: $B_1=\frac{{\mu }_{0}I}{4\pi a}(2\cdot 0.707)=10^{-7}\cdot \frac{5}{0.10}\cdot 1.414=7.07\times 10^{-6}\text{T}.$ Four sides: $B=4B_1=2.83\times 10^{-5}\text{T}.$ Answer: $B\approx 2.83\times 10^{-5}\text{T}$. Sanity check: a circle of radius $R=s/2=0.10$ would give ${\mu }_{0}I/2R=3.14\times 10^{-5}$ — a bit larger than the square, as expected (the square's far corners are further away). ✔

WHAT: Two on-axis loop fields, each at $x=0.10$, both pointing the same way ⇒ add. WHY: Same-sense currents give same-direction (same-sign) axial fields at the midpoint, so we double one on-axis result. One loop: $R^2=0.01$, $R^2+x^2=0.02$, $(0.02{)}^{3/2}=2.828\times 10^{-3}$. $B_1=\frac{{\mu }_{0}I{R}^2}{2(R^2+x^2{)}^{3/2}}=\frac{(4\pi \times 10^{-7})(3)(0.01)}{2(2.828\times 10^{-3})}=\frac{3.77\times 10^{-8}}{5.657\times 10^{-3}}=6.66\times 10^{-6}\text{T}.$ Total $B=2B_1=1.33\times 10^{-5}\text{T}.$ Answer: $B\approx 1.33\times 10^{-5}\text{T}$.

WHAT: Finite wire, but $P$ is not opposite the middle — the two angles differ. WHY: We keep the general $\frac{{\mu }_{0}I}{4\pi a}(\sin {\theta }_1+\sin {\theta }_2)$, but must compute each $\theta$ from its own right triangle. This is the fully general case that includes a zero angle. Lower end ($y=0$): the foot of the perpendicular is at this end, so the line to it is perpendicular to the wire ⇒ angle from foot $=0$ ⇒ $\sin {\theta }_1=0$. Upper end ($y=0.30$): horizontal offset $0.30$, so $\sin {\theta }_2=\frac{0.30}{\sqrt{0.10^2+0.30^2}}=\frac{0.30}{\sqrt{0.10}}=\frac{0.30}{0.3162}=\mathrm{0.9487.}$ $B=\frac{{\mu }_{0}I}{4\pi a}(0+0.9487)=10^{-7}\cdot \frac{8}{0.10}\cdot 0.9487=10^{-7}\cdot 80\cdot 0.9487=7.59\times 10^{-6}\text{T}.$ Answer: $B\approx 7.59\times 10^{-6}\text{T}$. Case note: if $P$ were opposite the middle, both angles would be equal and non-zero — the L2 case. If the wire extended to $y=+\infty$, $\sin {\theta }_2\to 1$ and, with $\sin {\theta }_1=0$, we'd get exactly half the infinite-wire field, $\frac{{\mu }_{0}I}{4\pi a}$ — the "field from a semi-infinite wire at its end." ✔

WHAT: Set $\frac{B_{\text{axis}}(x)}{B_{\text{centre}}}=\frac{1}{2}$ and solve for $x$. WHY: This tests whether you can manipulate the on-axis form symbolically rather than plug numbers. $\frac{{\mu }_{0}I{R}^2/[2(R^2+x^2{)}^{3/2}]}{{\mu }_{0}I/(2R)}=\frac{R^3}{(R^2+x^2{)}^{3/2}}=\frac{1}{2}.$ Raise both sides to the $2/3$ power: $\frac{R^2}{R^2+x^2}={\left(\frac{1}{2}\right)}^{2/3}=2^{-2/3}.$ So $R^2+x^2=R^2{2}^{2/3}$, giving $x^2=R^2(2^{2/3}-1)$ and $\boxed{x=R\sqrt{2^{2/3}-1}\approx 0.766R.}$ Answer: $x\approx 0.766R$. (For $R=0.10\text{m}$, $x\approx 0.0766\text{m}$.)

WHAT: Take the limit $x\gg R$ in $B=\frac{{\mu }_{0}I{R}^2}{2(R^2+x^2{)}^{3/2}}$. WHY: Limiting behaviour reveals the loop's far-field identity — this is where Biot–Savart meets the Magnetic Dipole Moment. Step 1 — drop the small term. When $x\gg R$, the $R^2$ inside the bracket is negligible next to $x^2$, so $(R^2+x^2{)}^{3/2}\approx (x^2{)}^{3/2}=x^3$. Substituting: $B\approx \frac{{\mu }_{0}I{R}^2}{2x^3}.$ This already shows the promised $B\propto 1/x^3$ falloff — far away, the field dies faster than a straight wire's $1/a$ or a point charge's $1/r^2$. Step 2 — rewrite in dipole form. Multiply top and bottom to expose ${\mu }_0/4\pi$ and the loop's area $\pi R^2$: $B\approx \frac{{\mu }_{0}I{R}^2}{2x^3}=\frac{{\mu }_0}{4\pi }\cdot \frac{2(I\pi R^2)}{x^3}=\frac{{\mu }_0}{4\pi }\cdot \frac{2m}{x^3},\boxed{m=I\pi R^2.}$ Step 3 — read the result. The quantity $m=I\pi R^2$ is the magnetic dipole moment: current times enclosed area. The far-field is $B=\frac{{\mu }_0}{2\pi }\frac{m}{x^3}$, the textbook axial dipole field. So a small current loop, viewed from far away, is indistinguishable from a bar-magnet-like dipole. ✔ Answer: $B\to \frac{{\mu }_0}{2\pi }\frac{m}{x^3}$ with $m=I\pi R^2$.