Magnetic force on charge — F = qv × B: Question bank

1.8.20 · D5 · Physics › Electromagnetism › Magnetic force on charge — F = qv × B

Neeche ke har jawab mein do ideas hi saara kaam karte hain, toh inhe pehle pin kar lo:

Recall Wo do facts jinpe har jawab tikta hai

Fact A — sirf sideways. $F = q v \times B$ hamesha $v$ ke perpendicular hota hai. Do vectors ka Cross product dono se perpendicular hota hai. Fact B — perpendicular motion chahiye. Magnitude hai $∣ F ∣ = ∣ q ∣ v B sin θ$ , jahan $θ$ , $v$ aur $B$ ke beech ka angle hai. Sirf $v$ ka woh hissa jo $B$ ke across hai, wahi matter karta hai; $B$ ke along waala hissa kuch contribute nahi karta.

True or false — justify

A magnetic field can hold a stationary electron in place against gravity.

False. Jab

v = 0

ho, toh

F = q v \times B = 0

: ek magnetic field ek aisi charge ko ignore karta hai jo move nahi kar rahi, isliye woh kisi cheez ko rest mein support nahi kar sakta.

A magnetic force can change the kinetic energy of a charged particle.

False. Power hoti hai

F \cdot v

, aur kyunki

F ⊥ v

hai (Fact A), yeh dot product hamesha

0

hota hai; kinetic energy

\frac{1}{2} m v^{2}

constant rehti hai.

If you double the speed of a proton in a fixed field, its circular period doubles.

False.

T = \frac{2 π m}{q B}

mein

v

hai hi nahi;

v

double karne se radius double hoti hai (

r = m v / q B

) lekin period bilkul same rehti hai.

A magnetic force can change a particle's velocity.

True. Velocity ek vector hai: force uski direction badal deta hai, chahe woh kabhi uski magnitude (speed) nahi badalta. "Speed nahi badal sakta" ka matlab yeh nahi ki "velocity nahi badal sakta."

A proton and an electron fired with the same

v

into the same

B

curve the same way.

False.

q

ki sign flip karne se

F = q v \times B

flip ho jaata hai, isliye dono opposite directions mein curve karte hain (haalaanki shape, ek circle, same rehta hai).

v

is exactly parallel to

B

, the particle travels in a straight line.

True.

sin θ = sin 0 = 0

, isliye

F = 0

aur path ko bend karne ke liye kuch bhi nahi hai.

Reversing the direction of

B

reverses the direction of the force.

True.

v \times (- B) = - (v \times B)

; cross product linear hota hai, isliye

B

flip karne se

F

flip ho jaata hai.

A faster particle in the same field always feels a stronger force.

False in general.

∣ F ∣ = ∣ q ∣ v B sin θ

v

ke saath tabhi badhta hai jab

sin θ \neq = 0

ho. Agar

v ∥ B

ho, toh force zero hi rehta hai chahe kitni bhi speed ho.

Spot the error

F = q v \cdot B

, so the force is a number pointing along

v

Law mein cross product use hota hai, dot nahi. Dot product ek scalar deta hai; force ek vector hai, aur woh

v

ke perpendicular point karta hai, uske along nahi.

"The force is largest when

v

is parallel to

B

Yeh ulta hai:

sin θ

maximum hota hai

θ = 9 0^{\circ}

pe (perpendicular) aur zero hota hai

θ = 0^{\circ}

pe (parallel). Maximum force ke liye

v ⊥ B

chahiye.

"For an electron, point fingers along

v

, curl to

B

, thumb gives the force."

Thumb

v \times B

deta hai. Negative charge ke liye tumhe ise flip karna hoga: force thumb ke opposite point karta hai.

"Since force is

q v B sin θ

, an electron feels a negative force."

Magnitude

∣ q ∣ v B sin θ

kabhi negative nahi hoti —

∣ q ∣

ek absolute value hai.

q

ki sign direction decide karti hai, jo vector form mein handle hoti hai, magnitude mein nahi.

"The period is

T = 2 π m / q B

, so a heavier particle laps faster."

Ulta hai.

T \propto m

: ek heavier particle same field mein har lap mein zyada time leta hai.

"A magnetic field speeds a particle up as it enters, then it circles."

Koi speed-up phase nahi hota. Jis instant charge move karta hai, force purely sideways hota hai (Fact A); speed pehle moment se hi constant rehti hai.

Why questions

Why must a magnetic force be perpendicular to the velocity?

Kyunki woh

q v \times B

hai, aur cross product by definition dono input vectors ke perpendicular hota hai —

v

bhi shamil hai (Fact A).

Why does the cyclotron period not depend on speed?

Zyada speed, radius

r = m v / q B

ko exactly usi proportion mein bada kar deti hai, isliye longer path ko higher speed pe same time mein cover kiya jaata hai:

T = 2 π r / v = 2 π m / q B

, aur

v

cancel ho jaata hai.

Why does a charge moving along

B

feel no force?

Sirf

v

ka woh component jo

B

ke across hai wahi matter karta hai, aur jab dono parallel hote hain toh woh component zero hota hai:

sin 0 = 0

, jo

∣ F ∣ = 0

deta hai (Fact B).

Why can only an electric field, never a magnetic one, do work on a charge?

Electric force

q E

v

ke along ek component rakh sakti hai, isliye

F \cdot v \neq = 0

ho sakta hai; magnetic force hamesha

⊥ v

hoti hai, isliye uski power hamesha

0

hoti hai.

Why does a velocity with both parallel and perpendicular parts trace a helix?

Perpendicular part circle banata hai (constant sideways force), parallel part freely drift karta hai (koi force nahi); circle + steady drift =

B

ke along ek helix.

Why does flipping the sign of the charge flip the curving direction but not the radius?

Sign

F

ki direction reverse kar deti hai (doosri taraf curve), lekin

r = m v / (∣ q ∣ B)

mein sirf

q

ki magnitude use hoti hai, isliye circle same size ka rehta hai.

Edge cases

What happens to the radius

r = m v / q B

as the field

B \to 0

r \to \infty

: circle infinitely bada ho jaata hai, yani path seedhi line mein badal jaati hai — koi field nahi, koi bending nahi.

What is the force if the charge is neutral (

q = 0

Zero.

F = 0 \cdot (v \times B) = 0

; ek magnetic field sirf moving charge pe act karta hai, aur yahan koi charge hai hi nahi.

A charge moves so that

v

and

B

are antiparallel (

θ = 18 0^{\circ}

). What's the force?

Zero.

sin 18 0^{\circ} = 0

, isliye

∣ F ∣ = 0

— antiparallel bilkul utna hi "force-free" hai jitna parallel.

As the angle

θ

between

v

and

B

goes from

0^{\circ}

9 0^{\circ}

, how does the force change?

Woh

sin θ

ke saath smoothly badhta hai:

0

se (parallel) lekar apne maximum

∣ q ∣ v B

tak (perpendicular), ek sine curve ki shape trace karta hua.

In the limit of very strong

B

(with

v

q

m

fixed), what happens to the circle?

r = m v / q B \to

bahut chota aur

T = 2 π m / q B \to

bahut chota: particle ek tight, fast loop mein spiral karta hai — field ke zariye tightly "trap" ho jaata hai.

A particle enters a field region moving exactly perpendicular to

B

and exits it. Has its speed changed?

Nahi. Puri time

F ⊥ v

rehta hai isliye koi work nahi hota; woh same speed se nikalta hai jis speed se daakhil hua tha, sirf ek naye direction mein aim karta hua.

Connections

Magnetic force on charge — F = qv × B (index 1.8.20) — woh parent law jinhe ye traps probe karte hain.
Cross product — kyun force dono inputs ke perpendicular hoti hai (Fact A).
Lorentz force law — electric part $q E$ add karo yeh dekhne ke liye ki actually work kaun karta hai.
Centripetal force and circular motion — "sideways force ⇒ circle" ki machinery.
Cyclotron — speed-independent period $T = 2 π m / q B$ ka exploit karta hai.
Mass spectrometer — masses sort karne ke liye $r = m v / q B$ use karta hai.
Velocity selector — electric aur magnetic forces ko balance karta hai (sirf $E$ hi work karti hai).
Magnetic force on a current-carrying wire — wahi law bahut saari moving charges pe sum kiya gaya.