Magnetic force on current-carrying conductor: Question bank

1.8.21 · D5 · Physics › Electromagnetism › Magnetic force on current-carrying conductor

Ground rules jo tumhe load karni chahiye

Traps se pehle, yeh char facts pakke karo jo neeche har question attack karta hai. Khud ko cloze-test karo:

Force ek cross product hai, isliye yeh wire aur field dono ke perpendicular hota hai.
Magnitude mein == $sin θ$ == factor hota hai, jahan $θ$ wire aur $B$ ke beech ka angle hai.
$θ$ ==current direction aur $B$ == ke beech measure hota hai, na ki force aur kisi aur cheez ke beech.
Ek closed loop uniform field mein zero net force feel karta hai (lekin possibly ek torque).

True ya false — justify karo

True/false batao AUR ek-sentence mein reason do. Sirf "true" likhne se kuch nahi milega.

A wire carrying current always feels a force in a magnetic field.

False. Agar wire

B

ke parallel hai toh

θ = 0

aur

F = B I L sin 0^{\circ} = 0

; alignment force ko khatam kar deti hai.

Doubling the current doubles the force (field and geometry fixed).

True.

F = B I L sin θ

mein

I

linear hai, isliye

I \to 2 I

F \to 2 F

ho jaata hai.

The magnetic force can speed the wire's charges up along the wire.

False. Force

I L \times B

L

(current direction) ke perpendicular hota hai, isliye carriers ko aage push karne ke liye wire ke along koi component nahi hota.

A wire bent into a full circle in a uniform field is pushed bodily to one side.

False. Closed loop ke liye

\oint d ℓ = 0

, isliye

F = I (\oint d ℓ) \times B = 0

; koi net force nahi hoti (torque ho sakta hai).

Reversing both the current and the field leaves the force unchanged.

True.

F = I L \times B

; dono signs flip karne par

(- 1) (- 1) = + 1

multiply hota hai, isliye

F

same rehti hai.

Reversing only the current reverses the force.

True.

F \propto L

;

L \to - L

karne se cross product flip ho jaata hai, isliye

F \to - F

Two wire segments of the same length carrying the same current in the same uniform field always feel equal-magnitude forces.

False. Magnitude

B I L sin θ

hai, isliye yeh har segment ke orientation

θ

par depend karta hai; alag angles se alag forces milti hain.

A curved wire and the straight wire joining its two ends feel the same net force in a uniform field.

True.

F = I (\int d ℓ) \times B

aur

\int d ℓ

sirf end-to-end displacement hai, jo dono ke liye same hai.

If the force on a wire is zero, the field there must be zero.

False. Force tab bhi zero hoti hai jab wire ek strong field ke parallel ho (

sin 0^{\circ} = 0

); zero force ka matlab zero

B

nahi hai.

Error dhundho

Har line mein ek galat statement hai. Batao kya galat hai aur use correct karo.

"The force pushes the wire along the direction of

B

Galat — force ek cross product hai, isliye yeh

B

ke perpendicular hai (aur wire ke bhi), kabhi bhi field ke along nahi.

F = B I L

works for any angle between the wire and the field."

Galat — yeh sirf

θ = 9 0^{\circ}

wala case hai; generally

F = B I L sin θ

hota hai, aur

θ = 0^{\circ}

par yeh zero ho jaata hai.

"Use the right hand: field = first finger, current = second, force = thumb, for the force on a wire."

Galat — left hand motor rule hai (Fleming's Left-Hand Rule); right hand

v \times B

ke liye positive charge / generator direction deta hai.

θ

is the angle between the force and the magnetic field."

Galat —

θ

current (wire) aur $B$ ke beech ka angle hai; force

B

se hamesha

9 0^{\circ}

par hoti hai, chahe kuch bhi ho.

"A rectangular loop in a uniform field is dragged toward the stronger side of the field."

Galat — uniform field mein koi stronger side nahi hoti; opposite sides par equal-and-opposite forces lagte hain, net force zero, sirf torque bachta hai.

"Each electron in the wire feels

q v \times B

, so the wire force is

q v_{d} B

Galat — yeh ek carrier par force hai;

N = n A L

carriers se multiply karo, jo

I L \times B

mein reassemble ho jaata hai.

"Since electrons drift the opposite way to conventional current, the wire force points opposite to

I L \times B

Galat — electrons ka

q < 0

hota hai aur yeh

I

ke opposite drift karte hain; dono sign flips cancel ho jaate hain, isliye force conventional current use karne par

I L \times B

se agree karta hai.

Why questions

Sirf outcome nahi, mechanism explain karo.

Why is the force zero when the wire lies along

B

Carriers ki drift velocity tab

B

ke parallel hoti hai, aur

v ∥ B \Rightarrow v \times B = 0

; koi carrier magnetic force feel nahi karta, isliye wire bhi nahi.

Why does only the perpendicular component of the wire "count"?

Sirf wire ka projection

L sin θ

field lines ko cross karta hai;

B

ke along component

L \times B

mein kuch contribute nahi karta, isliye

sin θ

aata hai.

Why can we replace the microscopic

(n, A, q, v_{d})

with the single macroscopic

I

Kyunki

I = n A q v_{d}

, isliye

n A q v_{d} = I L

; current saari microscopic detail ko ek measurable number mein bundle kar deti hai.

Why does a closed loop feel zero net force but can still spin?

Opposite sides par forces sum mein cancel ho jaati hain (

\oint d ℓ = 0

) lekin alag lines par act karti hain, jo ek couple — yaani torque — produce karta hai, jo motor ko turn karta hai.

Why is Fleming's left hand used here but the right hand for a free charge's motion?

Dono

v \times B

encode karte hain; left-hand version sirf ek mnemonic hai jo "field, current, force" ke liye wire par tuned hai, same physical direction deta hai — haath disagree nahi karte.

Why does a bent wire feel the same force as the straight chord between its ends (uniform field)?

Uniform

B

mein field sum se bahar factor ho jaata hai,

F = I (\int d ℓ) \times B

bachta hai, aur

\int d ℓ

path chahe kuch bhi ho, start-to-end vector par collapse ho jaata hai.

Why isn't the force on the wire simply along the direction the current flows?

Magnetic force carriers par unki motion ke along koi kaam nahi karta; cross product current ke perpendicular hota hai, isliye yeh wire ko sirf sideways dhakela sakta hai, kabhi aage nahi.

Edge cases

Woh boundary scenarios jo formula quietly handle karta hai.

θ = 0^{\circ}

(wire parallel to

B

): what is the force?

F = B I L sin 0^{\circ} = 0

; maximum-alignment case mein koi force nahi hoti, gravity-jaisi intuition ke bilkul ulta.

θ = 9 0^{\circ}

(wire perpendicular to

B

): what is the force?

F = B I L sin 9 0^{\circ} = B I L

, diye gaye

B, I, L

ke liye maximum force.

θ = 18 0^{\circ}

(wire antiparallel to

B

): what is the force?

sin 18 0^{\circ} = 0

, isliye

F = 0

; antiparallel utna hi force-free hai jitna parallel — sirf alignment matter karta hai, uski sign nahi.

I = 0

(no current, wire sitting in a field): force?

Zero — koi moving charges nahi toh

B

ke paas act karne ke liye kuch nahi;

F = B I L sin θ

zero ho jaata hai kyunki

I = 0

B = 0

(current flows, no field): force?

Zero — koi field nahi matlab koi magnetic force nahi,

F = B \cdot (\dots) = 0

A single closed loop in a non-uniform field: is the net force still zero?

Zaruri nahi. Cancellation

\oint d ℓ = 0

ke liye uniform

B

chahiye tha; agar

B

loop mein vary kare, toh sides unequal forces feel kar sakti hain aur net force bach sakti hai.

What happens to

F

θ

increases from

0^{\circ}

9 0^{\circ}

Yeh

sin θ

follow karte hue smoothly

0

se maximum

B I L

tak badhti hai — graph sine curve ka pehla quarter hai,

0^{\circ}

ke paas sabse steep change.

A wire in the plane of the page with

B

also in that plane: can the force point in the plane?

Nahi. Force

B

ke perpendicular hoti hai; agar wire aur field dono page plane mein hain, toh

L \times B

page se bahar ya andar point karta hai, kabhi page ke andar nahi.

Connections

Parent: Force on a current-carrying conductor — woh formula jo yahan har trap test karta hai.
Lorentz force on a moving charge — jahan $F = q v \times B$ (aur har sign trap) originate hota hai.
Drift velocity and current — $n A q v_{d} = I L$ justify karta hai, "why $I$ ?" ka jawab.
Torque on a current loop — "loop is pushed sideways" trap ka resolution.
Electric motor / Moving-coil galvanometer — jahan zero-net-force-but-torque actually kaam karta hai.
Force between two parallel currents — ek application jahan non-uniform fields matter karti hain.
Cross product (vectors) — woh perpendicularity jo "force along the field" intuition ko defeat karti hai.