Biot-Savart law — magnetic field from current element: Question bank

1.8.22 · D5 · Physics › Electromagnetism › Biot-Savart law — magnetic field from current element

Shuru karne se pehle, un vocabulary terms ko pin down karo jo yeh questions baar baar use karte hain (sab neeche wale figure se anchored hain).

Definition Har question mein kaam aane wale players

Field point $P$ — space mein woh specific location jahan hum magnetic field jaanna chahte hain. Yeh wire par nahi hota; kuch doori par hota hai.
Perpendicular distance $a$ — wire se $P$ tak ki sabse chhoti (seedhi) doori. Wire par woh point jo $P$ ke sabse kareeb hai, woh perpendicular foot $F$ hai.
Current element $d l$ — wire ka ek tiny segment, jis direction mein current flow karta hai usi direction mein point karta hua.
$\overset{r}{^}$ — unit vector jo us current element se $P$ ki taraf point karta hai; aur $θ$ woh angle hai jo $d l$ aur $\overset{r}{^}$ ke beech hota hai.
End-angles $θ_{1}, θ_{2}$ — ek finite straight wire ke liye, $P$ par khade ho, perpendicular foot $F$ ko dekho, phir apni line of sight ko wire ke har end ki taraf swing karo. Jo do angles tum sweep karte ho (perpendicular $P F$ se measure kiye hue) woh $θ_{1}$ aur $θ_{2}$ hain.
Axial distance $x$ — ek circular loop ke liye, woh straight-line distance jo axis ke along (loop ke centre se guzarne wali line, loop ke plane ke perpendicular) centre se field point tak measure ki jaati hai. Centre par $x = 0$ hota hai.

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

Teen anchors parent se, inhi naamon ke saath restate kiye gaye:

Biot–Savart magnitude hai $d B = \frac{μ _{0}}{4 π} \frac{I d l sin θ}{r ^{2}}$ , jahan $θ$ segment $d l$ aur $\overset{r}{^}$ ke beech ka angle hai.
$d B$ ki direction ek cross product $d l \times \overset{r}{^}$ se aati hai — dono ke perpendicular.
Final fields ( $μ_{0} I /2 π a$ , $μ_{0} I /2 R$ ) integrating ka result hain; yeh law itself nahi hain.

Poore note mein refer hone wala finite-wire formula hai (figure dekho): $B = \frac{μ _{0} I}{4 π a} (sin θ_{1} + sin θ_{2})$ .

Definition End-angles

$θ_{1}, θ_{2}$ ke liye sign convention Har end-angle ko $P$ par, perpendicular line $P F$ se shuru karke wire ke end ki taraf swing karte hue measure karo. Isko sign do:

Agar end foot $F$ ke ek side par hai (maan lo, upar), uska angle positive hai.
Agar end $F$ ke opposite side par hai (neeche), tab bhi usse positive enter kiya jaata hai jab $F$ dono ends ke beech ho — opposite sides ke do contributions add hote hain, isliye formula $sin θ_{1} + sin θ_{2}$ likhta hai.
Agar $P$ ek end se aage lie karta hai taaki foot $F$ wire ke bahar pade (dono ends $F$ ke same side par), toh us nearer end ka angle same side par measure hota hai jaise doosra end, aur use negative enter karna padta hai (ya equivalently, $sin θ_{2} - sin θ_{1}$ use karo). Dono contributions tab partly cancel ho jaate hain, correctly ek weaker field dete hue.

Safe rule: ek angle positive hai jab uska end $F$ ke reference se opposite side par ho, negative jab same side par ho. Wire ke bagal mein points ke liye angles kabhi $P F$ se $9 0^{\circ}$ se zyada nahi hote; jab $P$ ek end ke level par ho, woh angle exactly $0$ hota hai.

True or false — justify

Doubling the current doubles the magnetic field everywhere.

True.

B

linearly

I

par depend karta hai (law mein

d B \propto I

, aur integration linear hai), isliye

I

ko scale karne se

B

bhi har point par same factor se scale hota hai.

The formula

B = \frac{μ _{0} I}{2 π a}

applies to a wire of any length.

False. Woh result ek infinitely long straight wire assume karta hai (dono end-angles

θ_{1} = θ_{2} = 9 0^{\circ}

). Ek finite wire ko

B = \frac{μ _{0} I}{4 π a} (sin θ_{1} + sin θ_{2})

chahiye, jo chhota hota hai.

A single current element

I d l

produces a field that points radially outward, like a point charge's field.

False. Point charge ka field radial hota hai, lekin

d B \propto d l \times \overset{r}{^}

\overset{r}{^}

ke perpendicular hota hai — field element ke around circles mein wrap karta hai, kabhi

\overset{r}{^}

ke along point nahi karta.

Directly ahead of a current element (in the direction it points), the field it makes is zero.

True. Wahan

θ = 0

hai, isliye

sin θ = 0

aur

d l \times \overset{r}{^} = 0

. Ek segment field sideways spray karta hai, kabhi apni khud ki direction ke along nahi.

The magnetic field of an infinite straight wire falls off as

1/ r^{2}

False. Har element

1/ r^{2}

contribute karta hai, lekin wire ke along infinitely many elements ko sum karne se net

1/ a

dependence milti hai:

B = μ_{0} I /2 π a

On the axis of a circular loop, far away, the field falls as

1/ x^{2}

like Coulomb's law.

False. On-axis

B = \frac{μ _{0} I R ^{2}}{2 ( R ^{2} + x ^{2} ) ^{3/2}}

hai (

x

centre se axial distance ke saath), jo

x ≫ R

ke liye

1/ x^{3}

ki tarah jaata hai — yeh ek magnetic dipole ki signature hai, monopole ki nahi.

At the perpendicular foot's field point

P

, all the

d B

contributions from a straight wire point the same way.

True. Har element ke liye, direction

d l \times \overset{r}{^}

(Right-Hand Rule se)

P

par identical nikalta hai — wire aur

P

wale plane ke perpendicular — isliye woh magnitudes ki tarah add hote hain bina kisi cancellation ke. Figure ki orientation mein (wire vertical,

P

daayein, current upar) woh common direction seedha page se bahar tumhari taraf hai.

Biot–Savart and Ampère's Law can give different answers for the same infinite wire.

False. Unhe agree karna hi padega; dono

μ_{0} I /2 π a

dete hain. Ampère's law sirf symmetric geometries ke liye ek faster route hai.

At the exact centre of a current-carrying circular loop,

sin θ = 1

for every element.

True. Tangent

d l

ek circle par radius

\overset{r}{^}

ke hamesha perpendicular hota hai, isliye

θ = 9 0^{\circ}

aur

sin θ = 1

throughout.

The field at

P

on the perpendicular bisector of a finite wire is stronger than

μ_{0} I /2 π a

at the same distance

a

False. Ek finite wire mein fewer contributing elements hote hain, isliye

sin θ_{1} + sin θ_{2} < 2

, jo infinite case se weaker field deta hai.

Spot the error

A student writes

d B = \frac{μ _{0}}{4 π} \frac{I d l}{r ^{2}}

and integrates for a straight wire.

sin θ

factor missing hai. Magnetism ek cross product se aata hai, isliye directional

sin θ

essential hai; iske bina tum overcount karte ho aur wrong constant milta hai.

Someone measures

r

from the coordinate origin to the field point

P

for every element.

Biot–Savart mein

r

current element se

P

tak run karta hai, aur yeh har element ke liye alag hota hai. Ek fixed origin use karne se wrong

r

aur wrong

θ

milte hain.

To find the loop's centre field, a student adds all

d B

as scalars without checking their directions.

Mathematical error yeh hai ki ek vector sum ko scalar sum ki tarah treat kiya gaya hai. Yeh yahaan sirf isliye valid hai kyunki centre par har

d B

parallel hai (sab axis ke along), isliye

∣ \sum d B ∣ = \sum ∣ d B ∣

. Centre se door — jaise loop ki axis par kisi distance par —

d B

alag alag directions mein point karte hain aur yeh step wrong hai.

A solution claims the field of a current element points along

d l

d B ⊥ d l

hamesha, kyunki

d l \times \overset{r}{^}

d l

ke perpendicular hota hai. Field ka koi bhi part segment ke along nahi hota.

For a finite wire a student uses angles measured from the wire, not from the perpendicular

P F

B = \frac{μ _{0} I}{4 π a} (sin θ_{1} + sin θ_{2})

mein, angles

θ_{1}, θ_{2}

P

par perpendicular

P F

se har end tak measure kiye jaate hain. Wire-axis convention use karne se

sin \leftrightarrow cos

flip ho jaata hai aur wrong answer milta hai.

A student concludes

μ_{0}

has units of tesla.

μ_{0} = 4 π \times 1 0^{- 7} T\cdotm/A

— iske units mein metres aur amps specifically isliye hain taaki poora formula tesla mein aaye. Unhe drop karne se dimensional checks fail ho jaate hain.

Someone treats

\overset{r}{^}

as if it always points from

P

back to the element.

\overset{r}{^}

element se $P$ ki taraf point karta hai. Use reverse karne se cross product ka sign flip ho jaata hai aur field direction ulta ho jaata hai — tumhe field wrong direction mein circulate karta milega.

A student applies

B = \frac{μ _{0} I}{4 π a} (sin θ_{1} + sin θ_{2})

with both angles positive even though

P

is beyond one end of the wire.

Jab foot

F

wire ke bahar pade (dono ends

F

ke same side par), ek angle ko negative sign ke saath enter karna padta hai, taaki terms partly cancel ho jayein:

B = \frac{μ _{0} I}{4 π a} (sin θ_{2} - sin θ_{1})

. Dono positive rakhne se field overestimate ho jaata hai.

Why questions

Why does the field circle around the wire instead of pointing toward or away from it?

Cross product

d l \times \overset{r}{^}

force karta hai

d B

ko current aur point ki line dono ke perpendicular hone ke liye. Yeh "curl" nature reflect karta hai ki koi magnetic monopoles nahi hote — sirf loops hote hain.

Why does a segment make maximum field to its side and none straight ahead?

sin θ

ki wajah se: sideways matlab

θ = 9 0^{\circ}

(

sin = 1

, maximum), seedha aage matlab

θ = 0^{\circ}

(

sin = 0

, zero). Moving charge field apni motion ke perpendicular throw karta hai.

Why does the infinite wire's

1/ r^{2}

per element become an overall

1/ a

Pehle geometry samjho. Foot

F

ko origin par rakho aur ek segment wire par height

y

par maano; angle

θ

us segment par

d l

(wire ke upar) aur

\overset{r}{^}

(

P

ki taraf) ke beech measure hota hai. Right triangle

F

–segment–

P

mein, side

a = P F

θ

ke opposite hai aur

r

segment se

P

tak ka hypotenuse hai, isliye

sin θ = a / r

, yaani

r = a / sin θ

. Wire ke along height bhi

y = a cot θ

hai (

F

se measure ki gayi, aur

tan (angle at P) = y / a

; equivalently segment-angle

θ

use karke

y = a cot θ

y = a cot θ

differentiate karne par

d y = - a d θ / sin^{2} θ

milta hai, isliye

d l = ∣ d y ∣ = a d θ / sin^{2} θ

r = a / sin θ

aur

d l = a d θ / sin^{2} θ

d B = \frac{μ _{0} I}{4 π} \frac{s i n θ d l}{r ^{2}}

mein substitute karo:

a

's aur

sin

's collapse ho kar

d B = \frac{μ _{0} I}{4 π a} sin θ d θ

de dete hain.

θ

0

π

tak integrate karne par

\int_{0}^{π} sin θ d θ = 2

milta hai, isliye

B = \frac{μ _{0} I}{2 π a}

—

a

ki sirf ek power bachti hai kyunki

d l

a

ka ek factor

1/ r^{2}

ke do mein se ek cancel kar deta hai.

Why can Ampère's law replace this messy integral for a straight wire but not for a finite one?

Ampère's law ko high symmetry chahiye taaki

B

ek chosen loop ke along constant ho. Ek infinite wire mein woh symmetry hai; ek finite wire mein nahi, isliye direct Biot–Savart integration par fall back karna padta hai.

Why do only the axial components of

d B

survive on a loop's axis?

Symmetry ki wajah se, har element ke liye ek diametrically opposite element hota hai jiska

d B

equal-and-opposite radial (off-axis) component rakhta hai; woh pairs mein cancel ho jaate hain, sirf axis ke along point karne wale components bacha kar, jo sab add hote hain.

Why is Biot–Savart called the magnetic analogue of Coulomb's Law, yet fundamentally different?

Dono

1/ r^{2}

falloff aur ek source element share karte hain. Lekin Coulomb's field

\overset{r}{^}

ke along point karta hai (ek source point), jabki Biot–Savart ka cross product field ko circulate karta hai — magnetism mein koi isolated poles nahi hote.

Edge cases

What is the field at a point lying on the wire's own line (the longitudinal line through the wire, beyond its end)?

Har element ke liye,

\overset{r}{^}

d l

ke parallel ya antiparallel hota hai, isliye

θ = 0

18 0^{\circ}

aur

sin θ = 0

. Wire ki apni line par net field zero hota hai.

What happens to

B = μ_{0} I /2 π a

a \to 0

(approaching the wire)?

Yeh infinity tak diverge karta hai. Yeh ek idealization hai: ek real wire ki finite thickness hoti hai, isliye current pheli hoti hai aur field andar finite rehta hai.

What is the field at the centre of a half-loop (semicircle) of radius

R

Full-loop value ka aadha:

B = \frac{μ _{0} I}{4 R}

. Sirf aadha arc-length

π R

contribute karta hai, aur centre par har element ka

sin θ = 1

hi rehta hai.

Do the two straight lead-in wires that connect to a loop at its centre contribute to the centre field?

Nahi — agar woh radially centre ki taraf run karti hain, toh

d l

wahan

\overset{r}{^}

ke parallel hota hai (

θ = 0

), isliye woh straight sections centre par kuch add nahi karti.

What does the on-axis loop formula give at the very centre (

x = 0

B = \frac{μ _{0} I R ^{2}}{2 R ^{3}} = \frac{μ _{0} I}{2 R}

, jo direct centre result se exactly match karta hai — ek built-in consistency check.

In the finite-wire formula, what does one angle equalling

0

mean physically?

Iska matlab hai ki point

P

wire ke ek end ke level par hai (perpendicular foot

F

us end ke saath coincide karta hai). Tab sirf doosre end ka

sin θ

contribute karta hai.

What if

P

lies beyond one end of the wire, so the foot

F

falls off the wire entirely?

Dono ends ab

F

ke same side par hain. Nearer end ke angle ko negative enter karo, jo

B = \frac{μ _{0} I}{4 π a} (sin θ_{2} - sin θ_{1})

deta hai — contributions partly cancel ho jaate hain aur field symmetric placement se weaker hota hai.

If a wire carries zero current, what field does Biot–Savart predict?

Har jagah exactly zero, kyunki

d B \propto I

. Koi current nahi matlab koi source element nahi aur koi field nahi — law cleanly degenerate ho jaata hai.

Connections

Parent topic — full derivations jinhe yeh traps test karti hain.
Ampère's Law — symmetry shortcut jo Biot–Savart se agree karna chahiye.
Magnetic Field of a Solenoid — integrated loops; apna loop reasoning wahan test karo.
Magnetic Dipole Moment — kyun far-axis loop field $1/ x^{3}$ ki tarah jaata hai.
Coulomb's Law — $1/ r^{2}$ analogue jisse contrast karo.
Lorentz Force — resulting $B$ kya karta hai.
Right-Hand Rule — upar ki saari directions fix karta hai.