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

WHAT: "Bahut lamba straight wire" pehchano ⇒ $B=\frac{{\mu }_{0}I}{2\pi a}$ use karo. WHY: Wire infinite bataya gaya hai aur hum perpendicular distance par field chahte hain — yeh exactly wahi closed form hai jo humne derive kiya, to koi integration nahi chahiye. $B=\frac{{\mu }_{0}I}{2\pi a}=\frac{(4\pi \times 10^{-7})(5)}{2\pi (0.10)}=\frac{2\times 10^{-7}\times 5}{0.10}=1.0\times 10^{-5}\text{T}.$ (Handy trick: $\frac{{\mu }_0}{2\pi }=2\times 10^{-7}$, to $B=2\times 10^{-7}\cdot \frac{I}{a}$.) Answer: $B=1.0\times 10^{-5}\text{T}=10\mu \text{T}$.

WHAT: Loop ka centre ⇒ $B=\frac{{\mu }_{0}I}{2R}$. WHY: Centre par har element same distance $R$ par hota hai aur saare $d\vec{B}$ axis ke saath same direction mein point karte hain, jisse clean ${\mu }_{0}I/2R$ milta hai. $B=\frac{(4\pi \times 10^{-7})(2)}{2(0.05)}=\frac{4\pi \times 10^{-7}\times 2}{0.10}=2.51\times 10^{-5}\text{T}.$ 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: Wire infinite nahi hai, isliye ${\mu }_{0}I/2\pi a$ use nahi kar sakte. Kyunki $P$ midpoint ke saamne hai, dono ends symmetric hain: ${\theta }_1={\theta }_2$. Geometry (Figure s02 exactly yahi angles dikhata hai): har end foot se $L/2=0.10\text{m}$ horizontal distance par hai, perpendicular distance $a=0.05$ ke saath. Ek end ka angle, foot se measured: $\sin \theta =\frac{L/2}{\sqrt{a^2+(L/2{)}^2}}=\frac{0.10}{\sqrt{0.05^2+0.10^2}}=\frac{0.10}{\sqrt{0.0125}}=\frac{0.10}{0.1118}=\mathrm{0.894.}$ To $\sin {\theta }_1+\sin {\theta }_2=2(0.894)=1.789$. $B=\frac{{\mu }_{0}I}{4\pi a}(1.789)=10^{-7}\cdot \frac{3}{0.05}\cdot 1.789=10^{-7}\cdot 60\cdot 1.789=1.07\times 10^{-5}\text{T}.$ 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: Axis par centre se door, $d\vec{B}$ vectors tilt karte hain; sirf unke axial components bachte hain (radial wale symmetry se cancel ho jaate hain), aur yahi projection ${\mu }_{0}I/2R$ ko is $R^2/(R^2+x^2{)}^{3/2}$ form mein badal deta hai. Yahan $R^2=0.01$, $R^2+x^2=0.02$, $(0.02{)}^{3/2}=0.02\sqrt{0.02}=0.02\times 0.1414=2.83\times 10^{-3}$. $B=\frac{(4\pi \times 10^{-7})(4)(0.01)}{2(2.83\times 10^{-3})}=\frac{(4\pi \times 10^{-7})(0.04)}{5.66\times 10^{-3}}.$ Numerator $=4\pi \times 10^{-7}\times 0.04=5.027\times 10^{-8}$. Divide karo: $B=8.89\times 10^{-6}\text{T}$. Answer: $B\approx 8.89\times 10^{-6}\text{T}$.

WHAT: Midpoint par do infinite-wire fields superpose karo, har ek $a=d/2=0.05\text{m}$ par. WHY & direction: Har wire par midpoint ke liye right-hand rule lagao (upar restated). Dono currents ko upar point karte draw karo: midpoint wire 1 ke right mein hai (to uska field out of the page, $+$) aur wire 2 ke left mein (to uska field into the page, $-$). Out-of-page ko positive maanein to, do signed fields $+B_1$ aur $-B_2$ hain — yeh subtract karte hain. Har 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 ⇒ midpoint par perfect cancellation.)

WHAT: Same do magnitudes, lekin ek current reverse karo. WHY: $I_2$ reverse karne se midpoint par uske field ka sign flip ho jaata hai (right-hand rule): pehle $-B_2$ tha, ab $+B_2$ ho gaya. To dono signed fields ab out of the page ($+$) hain — yeh add karte hain. $B_{\text{net}}=+B_1+B_2=4.0\times 10^{-5}+4.0\times 10^{-5}=8.0\times 10^{-5}\text{T (out of page)}.$ Answer: $B_{\text{net}}=8.0\times 10^{-5}\text{T}$.

WHAT: Poora loop ${\mu }_{0}I/2R$ deta hai; semicircle exactly half arc hai, to half field. WHY: Centre par har arc element distance $R$ par hota hai $\theta =90^{\circ }$ ke saath aur saare $d\vec{B}$ parallel hain (right-hand rule se saare out of the page). Field arc length ke proportional hai; semicircle ki arc length poore loop ki half hoti hai. (Straight radial leads kuch contribute nahi karte: unke liye $d\vec{l}\parallel \hat{r}$, to $\sin \theta =0$.) $B=\frac{1}{2}\cdot \frac{{\mu }_{0}I}{2R}=\frac{{\mu }_{0}I}{4R}=\frac{(4\pi \times 10^{-7})(6)}{4(0.08)}=\frac{7.54\times 10^{-6}}{0.32}=2.36\times 10^{-5}\text{T}.$ Answer: $B\approx 2.36\times 10^{-5}\text{T}$.

WHAT: Har side ko ek finite wire maano aur $\frac{{\mu }_{0}I}{4\pi a}(\sin {\theta }_1+\sin {\theta }_2)$ use karo; charon contributions equal aur same-sign hain, isliye ek ko 4 se multiply karo. WHY: Koi single stock formula square ke liye fit nahi hota, lekin iski sides sirf char finite wires hain — decompose karo, phir superpose karo. Right-hand rule se, current ko square ke around chalane par charon $d\vec{B}$ centre par same taraf point karte hain (out of the page), isliye charon signs agree karte hain. Ek side ki geometry: centre har side se perpendicular distance $a=s/2=0.10\text{m}$ par hai. Perpendicular foot side ke midpoint par padta hai, isliye har end wire ke along $s/2$ door hai: $\sin {\theta }_1=\sin {\theta }_2=\frac{s/2}{\sqrt{(s/2{)}^2+(s/2{)}^2}}=\frac{1}{\sqrt{2}}=\mathrm{0.707.}$ Ek side: $B_1=\frac{{\mu }_{0}I}{4\pi a}(2\cdot 0.707)=10^{-7}\cdot \frac{5}{0.10}\cdot 1.414=7.07\times 10^{-6}\text{T}.$ Char sides: $B=4B_1=2.83\times 10^{-5}\text{T}.$ Answer: $B\approx 2.83\times 10^{-5}\text{T}$. Sanity check: radius $R=s/2=0.10$ ka ek circle ${\mu }_{0}I/2R=3.14\times 10^{-5}$ deta — square se thoda bada, as expected (square ke far corners zyaada door hain). ✔

WHAT: Do on-axis loop fields, har ek $x=0.10$ par, dono same taraf point karte hain ⇒ add karo. WHY: Same-sense currents midpoint par same-direction (same-sign) axial fields dete hain, isliye hum ek on-axis result double kar dete hain. Ek 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, lekin $P$ middle ke saamne nahi hai — do angles alag hain. WHY: Hum general $\frac{{\mu }_{0}I}{4\pi a}(\sin {\theta }_1+\sin {\theta }_2)$ rakhte hain, lekin har $\theta$ apne right triangle se compute karna padega. Yeh fully general case hai jismein ek zero angle bhi shamil hai. Lower end ($y=0$): perpendicular ka foot is end par hi hai, isliye us tak jaane wali line wire ke perpendicular hai ⇒ foot se angle $=0$ ⇒ $\sin {\theta }_1=0$. Upper end ($y=0.30$): horizontal offset $0.30$, isliye $\sin {\theta }_2=\frac{0.30}{\sqrt{0.10^2+0.30^2}}=\frac{0.30}{\sqrt{0.10}}=\frac{0.30}{0.3162}=\mathrm{0.9487.}$ $B=\frac{{\mu }_{0}I}{4\pi a}(0+0.9487)=10^{-7}\cdot \frac{8}{0.10}\cdot 0.9487=10^{-7}\cdot 80\cdot 0.9487=7.59\times 10^{-6}\text{T}.$ Answer: $B\approx 7.59\times 10^{-6}\text{T}$. Case note: agar $P$ middle ke saamne hota, to dono angles equal aur non-zero hote — L2 case. Agar wire $y=+\infty$ tak extend hoti, to $\sin {\theta }_2\to 1$ aur $\sin {\theta }_1=0$ ke saath, humein exactly infinite-wire field ka half milta, $\frac{{\mu }_{0}I}{4\pi a}$ — "semi-infinite wire ka apne end par field." ✔

WHAT: $\frac{B_{\text{axis}}(x)}{B_{\text{centre}}}=\frac{1}{2}$ set karo aur $x$ ke liye solve karo. WHY: Yeh test karta hai ki tum on-axis form ko symbolically manipulate kar sakte ho ya sirf numbers plug karte ho. $\frac{{\mu }_{0}I{R}^2/[2(R^2+x^2{)}^{3/2}]}{{\mu }_{0}I/(2R)}=\frac{R^3}{(R^2+x^2{)}^{3/2}}=\frac{1}{2}.$ Dono sides ko $2/3$ power par raise karo: $\frac{R^2}{R^2+x^2}={\left(\frac{1}{2}\right)}^{2/3}=2^{-2/3}.$ To $R^2+x^2=R^2{2}^{2/3}$, jisse $x^2=R^2(2^{2/3}-1)$ milta hai aur $\boxed{x=R\sqrt{2^{2/3}-1}\approx 0.766R.}$ Answer: $x\approx 0.766R$. ($R=0.10\text{m}$ ke liye, $x\approx 0.0766\text{m}$.)

WHAT: $B=\frac{{\mu }_{0}I{R}^2}{2(R^2+x^2{)}^{3/2}}$ mein $x\gg R$ ka limit lo. WHY: Limiting behaviour loop ki far-field identity reveal karta hai — yahan Biot–Savart Magnetic Dipole Moment se milta hai. Step 1 — small term drop karo. Jab $x\gg R$, bracket ke andar $R^2$ negligible hai $x^2$ ke next to, isliye $(R^2+x^2{)}^{3/2}\approx (x^2{)}^{3/2}=x^3$. Substitute karo: $B\approx \frac{{\mu }_{0}I{R}^2}{2x^3}.$ Yeh already promised $B\propto 1/x^3$ falloff dikhata hai — dur jaane par, field faster die karti hai straight wire ke $1/a$ ya point charge ke $1/r^2$ se. Step 2 — dipole form mein rewrite karo. ${\mu }_0/4\pi$ aur loop ki area $\pi R^2$ expose karne ke liye upar aur neeche multiply karo: $B\approx \frac{{\mu }_{0}I{R}^2}{2x^3}=\frac{{\mu }_0}{4\pi }\cdot \frac{2(I\pi R^2)}{x^3}=\frac{{\mu }_0}{4\pi }\cdot \frac{2m}{x^3},\boxed{m=I\pi R^2.}$ Step 3 — result padho. Quantity $m=I\pi R^2$ magnetic dipole moment hai: current times enclosed area. Far-field hai $B=\frac{{\mu }_0}{2\pi }\frac{m}{x^3}$, textbook axial dipole field. To ek chhota current loop, door se dekha jaaye, ek bar-magnet-jaise dipole se indistinguishable hota hai. ✔ Answer: $B\to \frac{{\mu }_0}{2\pi }\frac{m}{x^3}$ jahan $m=I\pi R^2$.