Exercises — Magnetic field of straight wire, circular loop, solenoid, toroid

WHAT tool: infinite straight wire → $B=\frac{{\mu }_{0}I}{2\pi r}$ (Ampère's law ka result, choose kiya kyunki wire mein circular symmetry hai). WHY: hume sirf current aur distance di gayi hai — yeh textbook straight-wire case hai. $r=5\text{cm}=0.05\text{m}$ convert karo. $B=\frac{(4\pi \times 10^{-7})(10)}{2\pi (0.05)}=\frac{4\pi \times 10^{-6}}{0.1\pi }=4\times 10^{-5}\text{T}$ Direction: grip rule — field wire ke around circle karta hai. Answer: $B=4\times 10^{-5}\text{T}=40\mu \text{T}$.

WHAT tool: $B=\frac{{\mu }_{0}I}{2a}$ — centre-of-loop special case ($x=0$). WHY: field exact centre par puchha gaya hai, isliye hume full axial formula ki zarurat nahi. $a=0.04\text{m}$. $B=\frac{(4\pi \times 10^{-7})(3)}{2(0.04)}=\frac{12\pi \times 10^{-7}}{0.08}=1.5\pi \times 10^{-5}\approx 4.71\times 10^{-5}\text{T}$ Direction: right-hand curl rule — thumb axis ke saath $\vec{B}$ ko loop plane ke perpendicular deta hai. Answer: $B\approx 4.71\times 10^{-5}\text{T}$.

WHAT tool: $B={\mu }_{0}nI$ — "bare" formula, koi radius nahi chahiye. WHY: "deep inside a long solenoid" exactly wahi jagah hai jahan ideal formula apply hota hai. $B=(4\pi \times 10^{-7})(500)(2)=4\pi \times 10^{-4}\approx 1.257\times 10^{-3}\text{T}$ Direction: axis ke saath, sense current ke saath fingers curl karke milta hai. Answer: $B\approx 1.26\times 10^{-3}\text{T}=1.26\text{mT}$.

WHAT tool: $B=\frac{{\mu }_{0}I{a}^2}{2(a^2+x^2{)}^{3/2}}$ — full axial formula (Biot–Savart). WHY this tool: field point centre se off hai ($x=0.10\text{m}\ne 0$), isliye simple ${\mu }_{0}I/2a$ centre formula ab apply nahi hoti. Aur Ampère's law shortcut kyun nahi chalega? Ampère's law tabhi $B$ deta hai jab tum ek closed path dhundh sako jis par $B$ ki *har jagah same magnitude ho aur path ke saath point kare — tab $\oint \vec{B}\cdot d\vec{l}=B\times (\text{length})$. Ek single loop ke axis par, field magnitude point-to-point change karta hai aur field lines kisi bhi circle ya line se curve kar jaati hain jo tum draw karo, isliye koi aisa "nice" path exist nahi karta. $\oint \vec{B}\cdot d\vec{l}={\mu }_0{I}_{\text{enc}}$ equation still true hai, bas $B$ ke liye solve nahi ho sakti. Isliye hum Biot–Savart ko element by element integrate karne par vaapis aate hain. Neeche figure dekho: har element ke field ke sirf axial (green) components survive karte hain; sideways parts pairs mein cancel ho jaate hain.

Yahan $a=x=0.10$, isliye $a^2+x^2=0.02{\text{m}}^2$ aur $(a^2+x^2{)}^{3/2}=(0.02{)}^{1.5}$. $B=\frac{(4\pi \times 10^{-7})(5)(0.01)}{2(0.02{)}^{3/2}}$ Ab $(0.02{)}^{3/2}=0.02\sqrt{0.02}=0.02\times 0.14142=2.828\times 10^{-3}$. $B=\frac{(4\pi \times 10^{-7})(5)(0.01)}{2(2.828\times 10^{-3})}=\frac{2\pi \times 10^{-8}}{5.657\times 10^{-3}}\approx 1.111\times 10^{-5}\text{T}$ Centre value: $B_0=\frac{{\mu }_{0}I}{2a}=\frac{(4\pi \times 10^{-7})(5)}{0.2}\approx 3.14\times 10^{-5}\text{T}$. Ratio: $B/B_0=a^3/(a^2+x^2{)}^{3/2}=(0.10{)}^3/(0.02{)}^{1.5}=10^{-3}/2.828\times 10^{-3}\approx 0.354$. Direction: surviving field (Figure s01 mein green axial arrow) axis ke saath point karta hai, iska sense right-hand curl rule se milta hai — loop ke current ke saath fingers curl karo, aur tumhara thumb axis ke saath $\vec{B}$ ki direction deta hai (centre par wahi direction, bas weak). Answer: $B\approx 1.11\times 10^{-5}\text{T}$ axis ke saath directed, centre value ka lagbhag 35%.

WHAT tool: $B=\frac{{\mu }_{0}NI}{2\pi r}$ (Ampère's law, doughnut ke andar circular symmetry). WHY: toroid ke andar field $r$ par depend karta hai; mean circle ke saath mean radius use karte hain. $B=\frac{(4\pi \times 10^{-7})(1200)(4)}{2\pi (0.15)}=\frac{4\pi \times 10^{-7}\times 4800}{0.3\pi }=\frac{1.92\times 10^{-3}\pi }{0.3\pi }$ $\pi$ cancel kyun karein: same factor $\pi$ fraction ke upar aur neeche dono mein hai, isliye divide hokar $1$ ho jaata hai — upar-neeche shared factor kabhi value nahi badalta, bas arithmetic clean karta hai. Jo bachta hai woh plain division hai: $B=\frac{1.92\times 10^{-3}}{0.3}=6.4\times 10^{-3}\text{T}$ Direction: field core ke andar circularly chalta hai, sense winding current ke saath fingers curl karke milta hai. Answer: $B=6.4\times 10^{-3}\text{T}=6.4\text{mT}$.

Pehle, idealization ki warning. Clean result $B=\frac{1}{2}{\mu }_{0}nI$ sirf ideal semi-infinite solenoid ke liye exact hai — ek taraf infinitely lamba, thin walls, tightly wound. Ek real finite solenoid ka end field ${\mu }_{0}nI$ ke roughly half ke close to but not exactly hota hai: ek lamba-lekin-finite coil ke middle ke paas field $\approx {\mu }_{0}nI$ hota hai, aur har mouth par roughly one-half tak taper karta hai, phir bahar aur decrease karta rehta hai. Isliye "end par half" ko ek excellent estimate ki tarah use karo uss coil ke liye jo apni width se bahut zyada lambi ho, aur stubby coils ke liye chhote corrections ki umeed rakho. WHAT tool: semi-infinite solenoid ke end par, $B=\frac{1}{2}{\mu }_{0}nI$ — deep-inside value ka bilkul half. WHY exactly half (short derivation): solenoid ko current loops ke stack ki tarah socho. Deep inside, loops field point ke dono taraf extend karte hain, aur unke axial contributions milke ${\mu }_{0}nI$ dete hain. Ise loop-fields ka angles par integral model karo: ek infinite solenoid $\theta =0$ se $\theta =180°$ tak contributions collect karta hai (dono taraf ke loops), poora ${\mu }_{0}nI$ deta hai. Semi-infinite solenoid ke open mouth par, winding field point ke sirf ek taraf exist karti hai, isliye integral sirf half angular range ($\theta =90°$ se $180°$) par chalta hai. Us integral ki symmetry se, har half equally contribute karta hai, isliye end value bilkul $\frac{1}{2}{\mu }_{0}nI$ hoti hai. $B=\frac{1}{2}(4\pi \times 10^{-7})(500)(2)=\frac{1}{2}(1.257\times 10^{-3})\approx 6.28\times 10^{-4}\text{T}$ Answer: $B\approx 6.28\times 10^{-4}\text{T}=0.628\text{mT}$.

WHAT tool: finite-wire Biot–Savart result $B=\frac{{\mu }_{0}I}{4\pi r}(\sin {\theta }_1+\sin {\theta }_2)$, jahan ${\theta }_1,{\theta }_2$ woh angles hain jo har wire end perpendicular foot se measure karke subtend karta hai. WHY this tool (${\mu }_{0}I/2\pi r$ shortcut nahi): ${\mu }_{0}I/2\pi r$ formula ek infinite wire assume karta hai taaki Ampère's law $B$ ko circular loop se bahar nikal sake. Ek finite wire us symmetry ko tod deta hai (dono ends finite angles par hain), isliye hume Biot–Savart par vaapis jaana padega aur har element ka contribution integrate karna padega — integral $(\sin {\theta }_1+\sin {\theta }_2)$ factor deta hai, jo parent note mein puri tarah derive kiya gaya hai. Dono angles ko $90°$ set karne par infinite-wire result vapas milta hai, confirm karta hai ki dono limit mein agree karte hain. Geometry figure dekho:

Algebra ko figure se jodna: Figure s02 mein, $r$ length ki dashed line P se wire ka perpendicular hai; far end us foot se wire ke saath $\ell$ distance par baitha hai. Woh do legs ($r$ across, $\ell$ along) ek right triangle banate hain jiska hypotenuse $\sqrt{{\ell }^2+r^2}$ hai, P se wire ke end tak ki straight-line distance. Formula mein angle $\theta$ woh angle hai P par perpendicular aur us hypotenuse ke beech, isliye iska sine = opposite / hypotenuse = $\ell /\sqrt{{\ell }^2+r^2}$ — yahi $\sin \theta$ hai jo Biot–Savart integral maangta hai. Kyunki P midpoint ke opposite baitha hai, left taraf ka triangle right taraf ke jaisa mirror hai, isliye ${\theta }_1={\theta }_2=\theta$. $\sin \theta =\frac{\ell }{\sqrt{{\ell }^2+r^2}}=\frac{0.12}{\sqrt{0.12^2+0.05^2}}=\frac{0.12}{\sqrt{0.0144+0.0025}}=\frac{0.12}{0.13}\approx 0.9231$ $B=\frac{{\mu }_{0}I}{4\pi r}(2\sin \theta )=\frac{(4\pi \times 10^{-7})(6)}{4\pi (0.05)}(2\times 0.9231)$ $=\frac{6\times 10^{-7}}{0.05}(1.8462)=(1.2\times 10^{-5})(1.8462)\approx 2.215\times 10^{-5}\text{T}$ Answer: $B\approx 2.22\times 10^{-5}\text{T}$. Notice karo yeh infinite-wire value $\frac{{\mu }_{0}I}{2\pi r}=2.4\times 10^{-5}\text{T}$ se kam hai, kyunki ek finite wire mein fewer current elements contribute karte hain. ✓

WHAT tool: superposition — har wire ke field ko vectorially add karo. Midpoint par, har wire $r=0.10\text{m}$ door hai. WHY vectors matter: same direction ke currents ke liye, midpoint par dono fields opposite directions mein point karte hain (grip rule), isliye woh subtract hote hain. $B_A=\frac{{\mu }_0{I}_A}{2\pi r}=\frac{(4\pi \times 10^{-7})(8)}{2\pi (0.10)}=1.6\times 10^{-5}\text{T}$ $B_B=\frac{{\mu }_0{I}_B}{2\pi r}=\frac{(4\pi \times 10^{-7})(12)}{2\pi (0.10)}=2.4\times 10^{-5}\text{T}$ Opposite directions → net magnitude: $B_{\text{net}}=|B_B-B_A|=(2.4-1.6)\times 10^{-5}=0.8\times 10^{-5}\text{T}=8\times 10^{-6}\text{T}$ Answer: $B_{\text{net}}=8\times 10^{-6}\text{T}$, stronger wire (B) ki direction mein directed. (Dekhte hain Magnetic force on a current-carrying conductor mein yeh net field phir ek teesri wire ko kya karta hai.)

Step 1 — WHAT/WHY $N$ nikalo: har turn rod ki ek circumference hai, length $2\pi \rho$. Isliye turns ki number = total wire length ÷ circumference: $N=\frac{L_{\text{wire}}}{2\pi \rho }=\frac{20}{2\pi (0.01)}=\frac{20}{0.0628}\approx 318.3\text{turns}$ Step 2 — WHAT/WHY turns per metre: solenoid formula ko $n$ chahiye = length ke har metre mein packed turns, isliye turn count ko coil length ${\ell }_s$ se divide karo: $n=\frac{N}{{\ell }_s}=\frac{318.3}{0.25}\approx 1273\text{turns/m}$ Step 3 — field: ab $B={\mu }_{0}nI$ apply karo: $B={\mu }_{0}nI=(4\pi \times 10^{-7})(1273)(3)\approx 4.80\times 10^{-3}\text{T}$ Answer: $B\approx 4.80\times 10^{-3}\text{T}=4.80\text{mT}$. Insight: rod radius $\rho$ sirf is through enter hua ki kitne turns fit hote hain — yeh final ${\mu }_{0}nI$ mein appear nahi karta. Radius $N$ ko affect karta hai, field-per-turn ko nahi.

Equation set up karo (yahi synthesis hai: do alag formulas equal set kiye): $\frac{{\mu }_{0}I{a}^2}{2(a^2+x^2{)}^{3/2}}=\frac{{\mu }_{0}I}{2\pi a}$ Dono sides se ${\mu }_{0}I/2$ cancel karo: $\frac{a^2}{(a^2+x^2{)}^{3/2}}=\frac{1}{\pi a}$ Cross-multiply karo: $\pi a\cdot a^2=(a^2+x^2{)}^{3/2}$, yaani $\pi a^3=(a^2+x^2{)}^{3/2}$. Dono sides ko $2/3$ power tak raise karo: $(\pi a^3{)}^{2/3}=a^2+x^2\Rightarrow {\pi }^{2/3}{a}^2=a^2+x^2$ $x^2=a^2({\pi }^{2/3}-1)\Rightarrow x=a\sqrt{{\pi }^{2/3}-1}$ Numerically ${\pi }^{2/3}=2.145$, isliye $x=a\sqrt{1.145}\approx 1.070a$. Answer: $x\approx 1.07a$ — axis ke saath sirf ek radius se thoda aage.

(a) WHAT tool: $B=\frac{{\mu }_{0}NI}{2\pi r}$, har radius par evaluate kiya (field uniform nahi — yeh $1/r$ ke saath girta hai). Pehle common numerator simplify karo. ${\mu }_{0}NI=(4\pi \times 10^{-7})(2000)(5)=4\pi \times 10^{-3}$. $2\pi$ se divide karne par $\frac{4\pi \times 10^{-3}}{2\pi }=2\times 10^{-3}$ milta hai. Isliye $B(r)=\frac{2\times 10^{-3}}{r}$. $B_{\text{inner}}=\frac{2\times 10^{-3}}{0.10}=2.0\times 10^{-2}\text{T}=20\text{mT}$ $B_{\text{outer}}=\frac{2\times 10^{-3}}{0.14}\approx 1.43\times 10^{-2}\text{T}=14.3\text{mT}$ (b) Percentage variation (inner, badi value ke relative): $\frac{B_{\text{inner}}-B_{\text{outer}}}{B_{\text{inner}}}=\frac{20-14.29}{20}\approx 0.2857=28.6\%$ Answer: inner $=20\text{mT}$, outer $\approx 14.3\text{mT}$; field core ke across lagbhag 28.6% drop karta hai.

WHY Ampère's law: cylindrical symmetry $B$ ko integral se bahar nikalne deta hai, isliye $\oint \vec{B}\cdot d\vec{l}=B(2\pi r)={\mu }_0{I}_{\text{enc}}$. Twist yeh hai ki wire ke andar sirf part of current enclosed hota hai.

Figure dekho — enclosed current Amperian circle ki area ke saath badhta hai:

(a) Andar ($r<R$): current density uniform hai, isliye enclosed current area ratio ke saath scale karta hai: $I_{\text{enc}}=I\frac{\pi r^2}{\pi R^2}=I\frac{r^2}{R^2}$ $B(2\pi r)={\mu }_{0}I\frac{r^2}{R^2}\Rightarrow B=\frac{{\mu }_{0}Ir}{2\pi R^2}$ $r=0.01$, $R=0.02$ par: $B=\frac{(4\pi \times 10^{-7})(30)(0.01)}{2\pi (0.02{)}^2}=\frac{(4\pi \times 10^{-7})(0.3)}{2\pi (4\times 10^{-4})}=\frac{1.2\pi \times 10^{-7}}{8\pi \times 10^{-4}}=1.5\times 10^{-4}\text{T}$

(b) Bahar ($r>R$): ab poora current loop ko thread karta hai, $I_{\text{enc}}=I$, isliye hum ordinary straight-wire formula par vapas aate hain: $B(2\pi r)={\mu }_{0}I\Rightarrow B=\frac{{\mu }_{0}I}{2\pi r}$ $r=0.04$ par: $B=\frac{(4\pi \times 10^{-7})(30)}{2\pi (0.04)}=\frac{1.2\pi \times 10^{-5}}{0.08\pi }=1.5\times 10^{-4}\text{T}$

Answer: dono $1.5\times 10^{-4}\text{T}$ dete hain. Yeh numeric coincidence isliye hota hai kyunki $r_{\text{inside}}=0.01$ aur $r_{\text{outside}}=0.04$ same value par land karne ke liye choose kiye gaye the. Generally, andar $B\propto r$ (centre se linearly zero se utha karta hai), aur bahar $B\propto 1/r$ (girta hai), maximum surface $r=R$ par: wahan $B_{\max }=\frac{{\mu }_{0}I}{2\pi R}=\frac{(4\pi \times 10^{-7})(30)}{2\pi (0.02)}=3.0\times 10^{-4}\text{T}$.