Worked examples — Magnetic field of straight wire, circular loop, solenoid, toroid

1.8.24 · D3 · Physics › Electromagnetism › Magnetic field of straight wire, circular loop, solenoid, to

Yeh page parent topic ko tab tak drill karta hai jab tak koi bhi case tumhe surprise na kar sake. Pehle hum har tarah ke questions lay out karte hain jo chaar formulas produce kar sakte hain, phir har cell ke liye ek example work karte hain — zero cases, limiting cases, ek real-world word problem, aur ek exam twist bhi shamil hai.

Neeche sab kuch sirf chaar boxed results aur do master laws (Biot-Savart Law, Ampere's Law) se hi kaam karta hai. Agar koi symbol aaye, toh woh parent note mein define kiya gaya hai; jahan naya symbol aata hai, main usse pehle yahan define karta hoon.

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The scenario matrix

Isse ek checklist ki tarah padho. Har row ek class of problem hai; last column us worked example ka naam batata hai jo ise cover karta hai. Dhyan raho ki Ex 4 do alag cells cover karta hai (D aur E) — usi coil ka use karke on-axis/far-field case aur degenerate centre case dono illustrate kiye gaye hain.

#	Geometry	Case class jo yeh test karta hai	Covered by
A	Straight wire	Basic magnitude + direction (wire ke around field ka sign)	Ex 1
B	Straight wire	Do wires ki Superposition — same vs opposite current sense	Ex 2
C	Straight wire	Finite wire, aur degenerate "wire ki apni line par point" ($\theta =0$)	Ex 3
D	Circular loop	Centre vs on-axis; limiting case $x\gg a$ (far field)	Ex 4 (part b, c)
E	Circular loop	Zero/degenerate point ($x=0$ recovered) + $N$-turn coil	Ex 4 (part a)
F	Solenoid	Interior field, end ($\frac{1}{2}$), radius-independence	Ex 5
G	Toroid	Inner edge vs outer edge (woh $1/r$ variation), outside = 0	Ex 6
H	Real-world	Word problem: ek target field hit karne ke liye coil size karna	Ex 7
I	Exam twist	Loop + straight wire ko combine karna; jab fields cancel ho sakti hain aur kab nahi	Ex 8

Poore note mein, ${\mu }_0=4\pi \times 10^{-7}\text{T⋅m/A}$, toh handy combo $\frac{{\mu }_0}{2\pi }=2\times 10^{-7}\text{T⋅m/A}$ hai.

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Ex 1 — Straight wire, magnitude + direction (Cell A)

Forecast: Order of magnitude guess karo — micro-tesla? milli-tesla? Aur apna right thumb upar point karo: tumhari east side par tumhari ungliyan kis taraf sweep karti hain — tumhari taraf ya door?

1. Formula choose karo. Infinite straight wire → $B=\frac{{\mu }_{0}I}{2\pi r}$. Yeh step kyun? Ek wire, poori circular symmetry ⇒ Ampère's law ne parent note mein pehle se integral kar diya.
2. Units convert karo. $r=5\text{cm}=0.05\text{m}$. Yeh step kyun? ${\mu }_0$ metres mein hai; cm mix karne se $100\times$ error aata hai.
3. Plug in karo. $B=\frac{(4\pi \times 10^{-7})(10)}{2\pi (0.05)}=4\times 10^{-5}\text{T}=40\mu \text{T}.$ Yeh step kyun? $4\pi /2\pi =2$ cleanly cancel ho jaata hai.
4. Direction (right-hand grip). Thumb upar point karta hai (current up), ungliyan upar se dekhe jaane par counter-clockwise curl karti hain. Tumhari east side par ungliyan north ki taraf point karti hain. Yeh step kyun? Direction optional nahi hai — ek field answer ka aadha hissa direction hi hoti hai.

Verify: Units: $\frac{(\text{T⋅m/A})(\text{A})}{\text{m}}=\text{T}$ ✓. Magnitude $40\mu \text{T}$ Earth ke field ($\sim 50\mu \text{T}$) se comparable hai — 10 A par haath ki chaudai ki doori ke liye sensible hai.

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Ex 2 — Do parallel wires, superposition (Cell B)

Forecast: Kis case mein midpoint field zero ho jaata hai? Padhne se pehle apna right hand trust karo.

1. Midpoint par har wire ka field. Dono $r=d/2=0.10\text{m}$ door hain: $B_1=B_2=\frac{{\mu }_{0}I}{2\pi r}=\frac{(2\times 10^{-7})(6)}{0.10}=1.2\times 10^{-5}\text{T}.$ Yeh step kyun? Superposition: total field = vector sum har wire ke field ka. Pehle har piece lo.
2. Same direction (a) — har field ki direction explicitly track karo. Maano dono currents upar, page-plane se bahar point karti hain. Grip rule apply karo har wire par midpoint par:
   • Left wire ka field wahan circulate karta hai aur upar point karta hai (ise $+y$ kaho): ${\vec{B}}_1=+B_1\hat{y}$.
   • Right wire ka field wahan neeche point karta hai ($-y$): ${\vec{B}}_2=-B_2\hat{y}$. Kyunki midpoint do wires ki opposite sides par baitha hai, wohi counter-clockwise circulation opposite local directions deti hai. Vectors add karne par: ${\vec{B}}_{\text{net}}=B_1\hat{y}-B_2\hat{y}=0.$ Yeh step kyun? Yeh subtraction arbitrary nahi hai — yeh is wajah se aata hai ki do field vectors literally $+\hat{y}$ aur $-\hat{y}$ along point karte hain. Same magnitudes, opposite directions ⇒ genuine zero (matrix cell "net zero").
3. Opposite direction (b). Right wire ki current neeche flip karo. Isse midpoint par uska field $-\hat{y}$ se $+\hat{y}$ ho jaata hai, toh ab ${\vec{B}}_2=+B_2\hat{y}$ aur dono align karte hain: ${\vec{B}}_{\text{net}}=B_1\hat{y}+B_2\hat{y}=2.4\times 10^{-5}\text{T}(\text{along }+\hat{y})=24\mu \text{T}.$ Yeh step kyun? Ek current reverse karne se uski field direction reverse ho jaati hai, cancelling subtraction ko reinforcing addition mein badal deta hai.

Verify: Sanity: same-direction case mein do wires ke fields symmetry se cancel hone chahiye (midpoint equidistant baitha hai opposite local field directions ke saath) — aur hote hain. Magnetic force on a current-carrying conductor dekho yeh samajhne ke liye ki same-direction currents phir bhi ek doosre ko attract kyun karte hain jabki midpoint field zero hai.

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Ex 3 — Finite wire aur degenerate on-line point (Cell C)

Forecast: Kya ek finite wire us same $r$ par infinite wire se zyada ya kam field degi? Inequality ki direction guess karo.

Naya tool — Biot-Savart Law se finite-wire result: $B=\frac{{\mu }_{0}I}{4\pi r}(\sin {\theta }_1+\sin {\theta }_2),$ jahan ${\theta }_1,{\theta }_2$ woh angles hain jo do ends banate hain, point par daale gaye perpendicular se measure kiye jaate hain. Yeh angles kyun? Yeh record karte hain ki wire kitna "wrap around" karti hai point ko; ek infinite wire dono ko $90^{\circ }$ tak fill karti hai. Yahan constant $\frac{{\mu }_0}{4\pi }=10^{-7}$ hai (nahi $\frac{{\mu }_0}{2\pi }$) — hum $\frac{{\mu }_0}{4\pi }$ ko poore example mein consistently use karenge.

1. Angles find karo. Symmetric point ⇒ ${\theta }_1={\theta }_2=\theta$ jahan $\sin \theta =\frac{L}{\sqrt{L^2+r^2}}=\frac{4}{\sqrt{4^2+3^2}}=\frac{4}{5}=\mathrm{0.8.}$ Yeh step kyun? $\sin \theta$ = (half-length)/(end tak ki doori); right triangle end–foot–point ki pure geometry.
2. Plug in karo — ek consistent constant. $\frac{{\mu }_0}{4\pi }=10^{-7}$ poore mein use karte hue: $\frac{{\mu }_{0}I}{4\pi r}=\frac{(10^{-7})(8)}{0.03}=2.667\times 10^{-5}\text{T}.$ Phir $(\sin {\theta }_1+\sin {\theta }_2)=0.8+0.8=1.6$ se multiply karo: $B=(2.667\times 10^{-5})(1.6)=4.27\times 10^{-5}\text{T}=42.7\mu \text{T}.$ Yeh step kyun? Biot–Savart ka finite-wire formula $\frac{{\mu }_0}{4\pi }$ constant carry karta hai; hum ise factors of 2 mix karne se bachne ke liye fixed rakhte hain.
3. Infinite wire se compare karo. Infinite-wire formula doosra combo use karta hai $\frac{{\mu }_0}{2\pi }=2\times 10^{-7}$: $B_{\infty }=\frac{{\mu }_{0}I}{2\pi r}=\frac{(2\times 10^{-7})(8)}{0.03}=5.33\times 10^{-5}\text{T}.$ Finite wire kam deta hai ($42.7<53.3\mu \text{T}$), kyunki $\sin \theta <1$ hai. Yeh step kyun? Forecast confirm karta hai: ek choti wire mein kam "wrap" hai, isliye weaker field. (Do formulas alag constants use karte hain sirf isliye kyunki ek mein $4\pi$ aur doosre mein $2\pi$ baked in hai; limit mein agree karte hain — Verify dekho.)
4. Line par degenerate point. Agar field point wire ke extension par lie karta hai, toh $d\vec{l}$ aur $\hat{r}$ parallel hain, isliye $d\vec{l}\times \hat{r}=0$ ⇒ $B=0$. Yeh step kyun? Cross product un contributions ko kill kar deta hai jo wire ke seedhe neeche stare karne wale point se aate hain — yeh ek aisa case hai jo students bhool jaate hain.

Verify: Limit check: $L\to \infty$ hone do ⇒ $\sin \theta \to 1$ ⇒ $B\to \frac{{\mu }_{0}I}{4\pi r}(1+1)=\frac{{\mu }_{0}I}{2\pi r}$, infinite wire recover ho jaati hai — do constants exactly reconcile hote hain. ✓

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Ex 4 — Loop: centre, on-axis, far-field limit, $N$ turns (Cells D & E)

Forecast: Centre ya $x=12\text{cm}$ wala point — kaun stronger hoga? Roughly kis factor se?

1. Centre ($x=0$), $N$ turns — yeh Cell E hai. $B=\frac{{\mu }_{0}NI}{2a}=\frac{(4\pi \times 10^{-7})(100)(2)}{2(0.05)}=\frac{2.513\times 10^{-4}}{0.1}=2.51\times 10^{-3}\text{T}.$ Yeh step kyun? $x=0$ axis formula ko ${\mu }_{0}NI/2a$ par collapse karta hai (degenerate on-axis point). Turns ke liye $N$ se multiply karo.
2. On-axis at $x=12\text{cm}$ — yeh Cell D hai. $B=\frac{{\mu }_{0}NI{a}^2}{2(a^2+x^2{)}^{3/2}}.$ Yahan $a^2+x^2=0.05^2+0.12^2=0.0025+0.0144=0.0169$, toh $(a^2+x^2{)}^{3/2}=0.0169^{1.5}=2.197\times 10^{-3}$. $B=\frac{(4\pi \times 10^{-7})(100)(2)(0.0025)}{2(2.197\times 10^{-3})}=\frac{6.283\times 10^{-7}}{4.394\times 10^{-3}}=1.43\times 10^{-4}\text{T}.$ Yeh step kyun? Centre se hat kar Biot–Savart ka axial result use karna zaroor hai — yahan koi Amperian shortcut nahi hai.
3. Far-field limit — yeh bhi Cell D hai. Jab $x\gg a$ ho, root ke neeche $a^2$ drop karo: $(a^2+x^2{)}^{3/2}\to x^3$, jo deta hai $B\approx \frac{{\mu }_{0}NI{a}^2}{2x^3}\propto \frac{1}{x^3}.$ Yeh step kyun? Yeh $1/x^3$ ek magnetic dipole ki fingerprint hai — loop door se ek hi hota hai.

Verify: Ratio check: $\frac{B_{\text{centre}}}{B_{x=12}}=\frac{(a^2+x^2{)}^{3/2}}{a^3}=\frac{2.197\times 10^{-3}}{1.25\times 10^{-4}}=17.6$, aur directly $2.51\times 10^{-3}/1.43\times 10^{-4}\approx 17.6$ ✓.

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Ex 5 — Solenoid: interior, end, radius-independence (Cell F)

Forecast: Kaun bada hai — deep-inside field ya end field? Exactly kis factor se?

1. Turns per metre. $n=N/L=1000/0.5=2000{\text{m}}^{-1}$. Yeh step kyun? Solenoid formula $B={\mu }_{0}nI$ ko $n$ chahiye, $N$ nahi.
2. Interior. $B={\mu }_{0}nI=(4\pi \times 10^{-7})(2000)(3)=7.54\times 10^{-3}\text{T}.$ Yeh step kyun? Ampère ke rectangular loop ne andar position-independent field diya — yeh ek single number poore bulk mein kahin bhi hold karta hai.
3. End of a (semi-infinite) solenoid. $B_{\text{end}}=\frac{1}{2}{\mu }_{0}nI=3.77\times 10^{-3}\text{T}.$ Yeh step kyun? Mouth par, solenoid ka sirf "half" hi har taraf lie karta hai, isliye exactly half field milta hai — ek limiting/degenerate boundary case.
4. Radius double karo. $B$ unchanged rahega: ${\mu }_{0}nI$ mein koi $r$ nahi hai. Yeh step kyun? Formula mein koi radius nahi hai, isliye coil ko wide karne se (fixed $n$ aur $I$ par) bulk field change nahi hoti — field sirf is baat par depend karti hai ki turns kitne tightly packed hain, tube ki width par nahi.

Verify: Ratio $B_{\text{interior}}/B_{\text{end}}=2$ exactly ✓. Units: $(\text{T⋅m/A})({\text{m}}^{-1})(\text{A})=\text{T}$ ✓.

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Ex 6 — Toroid: inner vs outer edge, outside zero (Cell G)

Forecast: Field kaahan strongest hai — inner ya outer edge? Yaad karo $B\propto 1/r$.

1. Inner edge. $B_{\text{in}}=\frac{{\mu }_{0}NI}{2\pi r_{\text{in}}}=\frac{(2\times 10^{-7})(500)(4)}{0.08}=5.0\times 10^{-3}\text{T}.$ Yeh step kyun? Core ke andar Ampère ka circular loop saare $N$ turns enclose karta hai; chota $r$ ⇒ bada $B$.
2. Outer edge. $B_{\text{out}}=\frac{{\mu }_{0}NI}{2\pi r_{\text{out}}}=\frac{(2\times 10^{-7})(500)(4)}{0.10}=4.0\times 10^{-3}\text{T}.$ Yeh step kyun? Same enclosed current, bada $r$ ⇒ weaker field — $1/r$ non-uniformity in action.
3. Toroid ke bilkul bahar. Core ke bahar ek Amperian circle zero net current enclose karta hai (har turn ise do baar cross karta hai, andar aur bahar) ⇒ $B=0$. Yeh step kyun? Yeh zero toroid ki signature hai — field andar trapped hai.

Verify: Ratio $B_{\text{in}}/B_{\text{out}}=r_{\text{out}}/r_{\text{in}}=0.10/0.08=1.25$, aur $5.0/4.0=1.25$ ✓. Inner indeed stronger hai, forecast confirm karta hai.

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Ex 7 — Real-world word problem (Cell H)

Forecast: Required current "a few amps" hoga ya "hundreds of amps"? Compute karne se pehle guess karo.

1. $B={\mu }_{0}nI$ ko $I$ ke liye solve karo. $I=\frac{B}{{\mu }_{0}n}=\frac{0.01}{(4\pi \times 10^{-7})(4000)}=\frac{0.01}{5.027\times 10^{-3}}=1.99\text{A}.$ Yeh step kyun? Demo $B$ fix karta hai; hum solenoid formula invert karke unknown current nikalte hain.
2. Supply-limited version. $I=2\text{A}$ capped ke saath, $n$ ke liye solve karo: $n=\frac{B}{{\mu }_{0}I}=\frac{0.01}{(4\pi \times 10^{-7})(2)}=3.98\times 10^3{\text{m}}^{-1}\approx 4000\text{turns/m}.$ Yeh step kyun? Ab $I$ fixed hai aur $n$ design knob hai — same equation, alag unknown.

Verify: Dono answers essentially same operating point par land karte hain ($\approx 2\text{A}$, $\approx 4000\text{turns/m}$), yeh ek self-consistency check hai. ~2 A demo current bahut reasonable hai — forecast "a few amps" se match karta hai. Produce kiya gaya uniform field phir ek pickup coil ke through flux se demo drive kar sakta hai (Faraday's law of induction).

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Ex 8 — Exam twist: loop + wire, jab fields cancel hoti hain (Cell I)

Forecast: Wire ki field axis ke around circle karti hai; loop ki centre field uske along point karti hai. Kya yeh do fields parallel bhi hain? Yeh ek single fact decide karta hai ki cancellation possible hai ya nahi — ab guess karo.

1. Loop centre field (magnitude aur direction). $B_L=\frac{{\mu }_0{I}_L}{2a}=\frac{(4\pi \times 10^{-7})(3)}{2(0.04)}=4.71\times 10^{-5}\text{T},\text{axis along directed (page se bahar).}$ Yeh step kyun? Loop ka contribution establish karo aur, crucially, uski direction — yeh axial hai.
2. Wire current jo us magnitude se match kare. Straight-wire field ko $B_L$ ke equal karo: $\frac{{\mu }_0{I}_W}{2\pi r}=B_L\Rightarrow I_W=\frac{B_{L}2\pi r}{{\mu }_0}=\frac{B_{L}r}{2\times 10^{-7}}=\frac{(4.71\times 10^{-5})(0.02)}{2\times 10^{-7}}=4.71\text{A}.$ Yeh step kyun? Target magnitude reproduce karne wala current nikalne ke liye straight-wire law invert karo. (Numerically yahan $I_L$ ke saath same digits aana chosen numbers ka coincidence hai.)
3. Kya cancel ho sakta hai? — trap, geometry se resolved. Wire ki field perpendicular hai wire ke: yeh axis ke around circle karti hai, isliye yeh loop ke plane mein lie karti hai. Loop ki centre field axis along point karti hai. $90^{\circ }$ par do vectors kabhi anti-parallel nahi ho sakte, isliye yeh cancel nahi ho sakte. Net magnitude hai $B_{\text{net}}=\sqrt{B_L^2+B_W^2}=\sqrt{2}{B}_L=6.66\times 10^{-5}\text{T},$ axis se $45^{\circ }$ par directed — kabhi zero nahi. Yeh step kyun? Exam ka bait hai "unhe cancel karo." Cancellation ke liye anti-parallel vectors chahiye; yahan woh orthogonal hain, isliye answer firmly "no matter what $I_W$ is, net non-zero hai."

Verify: $I_W=4.71\text{A}$ ke saath do magnitudes equal hain, toh net $\sqrt{2}{B}_L=6.66\times 10^{-5}\text{T}$ hai axis se $45^{\circ }$ par — bilkul non-zero, step 3 confirm karta hai.

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Recall Rapid self-test (right side cover karo)

Infinite wire ka field, 10 A at 5 cm ::: $40\mu \text{T}$ Do 6 A wires ka midpoint, same direction, 20 cm apart ::: $0$ (fields opposite point karti hain, cancel) Same, opposite direction ::: $24\mu \text{T}$ Wire ki apni line par lie karne wala point ::: $B=0$ (kyunki $d\vec{l}\parallel \hat{r}$) Loop ka $1/x^3$ far-field matlab loop behave karta hai ek jaisa ::: magnetic dipole Toroid ke bilkul bahar field ::: $0$ (zero enclosed current) Kya wire-on-axis field loop ki centre field cancel kar sakti hai? ::: Nahi — woh perpendicular hain