Magnetic force on current-carrying conductor: Question bank

Ground rules you must have loaded

Before the traps, pin down the four facts every question below attacks. Cloze-test yourself:

The force is a cross product, so it is perpendicular to both the wire and the field.
The magnitude carries a == $sin θ$ == factor, where $θ$ is the angle between the wire and $B$ .
$θ$ is measured between the ==current direction and $B$ ==, not between the force and anything.
A closed loop in a uniform field feels zero net force (but possibly a torque).

True or false — justify

State true/false AND give the one-sentence reason. A bare "true" scores nothing.

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

False. If the wire is parallel to

B

then

θ = 0

and

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

; alignment kills the force.

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

True.

F = B I L sin θ

is linear in

I

, so

I \to 2 I

gives

F \to 2 F

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

False. The force

I L \times B

is perpendicular to

L

(the current direction), so it has no component along the wire to push carriers faster.

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

False. For a closed loop

\oint d ℓ = 0

, so

F = I (\oint d ℓ) \times B = 0

; there is no net force (there can be a torque).

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

True.

F = I L \times B

; flipping both signs multiplies by

(- 1) (- 1) = + 1

, so

F

is identical.

Reversing only the current reverses the force.

True.

F \propto L

; sending

L \to - L

flips the cross product, so

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. The magnitude is

B I L sin θ

, so it depends on each segment's orientation

θ

; different angles give different forces.

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

and

\int d ℓ

is just the end-to-end displacement, identical for both.

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

False. The force is also zero when the wire is parallel to a perfectly strong field (

sin 0^{\circ} = 0

); zero force does not imply zero

B

Spot the error

Each line contains a wrong statement. Say what's wrong and correct it.

"The force pushes the wire along the direction of

B

Wrong — the force is a cross product, so it is perpendicular to

B

(and to the wire), never along the field.

F = B I L

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

Wrong — that's only the

θ = 9 0^{\circ}

case; in general

F = B I L sin θ

, and it drops to zero at

θ = 0^{\circ}

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

Wrong — the left hand is the motor rule (Fleming's Left-Hand Rule); the right hand gives

v \times B

for a positive charge / generator direction.

θ

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

Wrong —

θ

is the angle between the current (wire) and $B$ ; the force is always

9 0^{\circ}

from

B

regardless.

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

Wrong — in a uniform field there is no stronger side; opposite sides give equal-and-opposite forces, net force zero, leaving only a torque.

"Each electron in the wire feels

q v \times B

, so the wire force is

q v_{d} B

Wrong — that's the force on one carrier; multiply by

N = n A L

carriers, which reassembles into

I L \times B

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

I L \times B

Wrong — electrons have

q < 0

and drift opposite to

I

; the two sign flips cancel, so the force agrees with

I L \times B

using conventional current.

Why questions

Explain the mechanism, not just the outcome.

Why is the force zero when the wire lies along

B

The carriers' drift velocity is then parallel to

B

, and

v ∥ B \Rightarrow v \times B = 0

; no carrier feels a magnetic force, so neither does the wire.

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

Only the projection

L sin θ

of the wire crosses field lines; the component along

B

contributes nothing to

L \times B

, which is why the

sin θ

appears.

Why can we replace the microscopic

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

with the single macroscopic

I

Because

I = n A q v_{d}

, so

n A q v_{d} = I L

; the current bundles all the microscopic detail into one measurable number.

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

The forces on opposite sides cancel in sum (

\oint d ℓ = 0

) but act along different lines, producing a couple — a torque — which is what turns a motor.

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

Both encode

v \times B

; the left-hand version is just a mnemonic tuned for "field, current, force" on a wire, giving the same physical direction — the hands don't disagree.

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

In a uniform

B

the field factors out of the sum, leaving

F = I (\int d ℓ) \times B

, and

\int d ℓ

collapses to the start-to-end vector regardless of the path.

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

The magnetic force does no work on the carriers along their motion; a cross product is perpendicular to the current, so it can only shove the wire sideways, never forward.

Edge cases

The boundary scenarios the formula quietly handles.

θ = 0^{\circ}

(wire parallel to

B

): what is the force?

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

; the maximum-alignment case gives no force, the opposite of gravity-like intuition.

θ = 9 0^{\circ}

(wire perpendicular to

B

): what is the force?

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

, the maximum force for given

B, I, L

θ = 18 0^{\circ}

(wire antiparallel to

B

): what is the force?

sin 18 0^{\circ} = 0

, so

F = 0

; antiparallel is just as force-free as parallel — sign of alignment doesn't matter, only that it's aligned.

I = 0

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

Zero — with no moving charges there is nothing for

B

to act on;

F = B I L sin θ

vanishes because

I = 0

B = 0

(current flows, no field): force?

Zero — no field means no magnetic force at all,

F = B \cdot (\dots) = 0

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

Not necessarily. The cancellation

\oint d ℓ = 0

needed a uniform

B

; if

B

varies across the loop, sides can feel unequal forces and a net force survives.

What happens to

F

θ

increases from

0^{\circ}

9 0^{\circ}

It rises smoothly from

0

to its maximum

B I L

following

sin θ

— the graph is the first quarter of a sine curve, steepest change near

0^{\circ}

A wire in the plane of the page with

B

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

No. The force is perpendicular to

B

; if the wire and field both lie in the page plane,

L \times B

points out of or into the page, never within it.

Connections

Parent: Force on a current-carrying conductor — the formula every trap here tests.
Lorentz force on a moving charge — where $F = q v \times B$ (and every sign trap) originates.
Drift velocity and current — justifies $n A q v_{d} = I L$ , the "why $I$ ?" answer.
Torque on a current loop — the resolution to the "loop is pushed sideways" trap.
Electric motor / Moving-coil galvanometer — where zero-net-force-but-torque actually does work.
Force between two parallel currents — an application where non-uniform fields matter.
Cross product (vectors) — the perpendicularity that defeats the "force along the field" intuition.