Do magnetic forces obey Newton's third law. Verify for two current elements $d\vec{l}_1 = dl\,\hat{i}$ located at the origin and $d\vec{l}_2 = dl\,\hat{j}$ located at $(0, R, 0)$. Both carry current $I$.

Do magnetic forces obey Newton's third law. Verify for two current elements dvecl₁ = dlhati located at the ori

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d\vec{l}_1 = dl\,\hat{i}

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Concept understanding — Magnetic Force on Current

Magnetic Force on Current

Imagine a garden hose spraying pure water — bring a magnet near the stream and nothing happens, because water is electrically neutral. But if that stream carried electric charge (a current), the magnet would push the whole stream sideways. That is the essence of this concept: a current-carrying wire placed in a magnetic field experiences a sideways force, because every moving charge inside the wire feels the Lorentz force, and since the charges cannot leave the wire, they drag the wire along with them.

From a single charge to a wire

A single charge $q$ moving with velocity $\vec v$ in a field $\vec B$ feels $\vec F = q(\vec v \times \vec B)$ . A current is just many such charges drifting together, so summing their individual forces over the whole wire gives a net force on it.

Note

The metal lattice itself is neutral and stationary — only the free electrons drift. The magnetic force acts on those drifting electrons, which then collide with the lattice and transfer the push to the entire wire.

The formula

For a straight wire of length $L$ carrying current $I$ in a uniform field $\vec B$ :

\vec F = I(\vec L \times \vec B), \qquad F = ILB\sin\theta

where $\vec L$ points along the current and $\theta$ is the angle between the wire and $\vec B$ .

Watch out

The force is zero when the wire runs parallel to the field ( $\theta=0^\circ$ or $180^\circ$ ) and maximum when perpendicular ( $\theta=90^\circ$ ) — the magnetic force only responds to the component of current motion that is perpendicular to $\vec B$ .

Direction: the right-hand rule

Point your index finger along the current ( $\vec L$ ), your middle finger along the field ( $\vec B$ ); your thumb then gives the force direction — this is just the cross product $\vec L \times \vec B$ read off by hand. Because it is a cross product, swapping the two vectors reverses the force.

Worked example

A $0.5\ \text{m}$ wire carries $3\ \text{A}$ from east to west, in a uniform field of $0.2\ \text{T}$ pointing north.

$\theta = 90^\circ$ (the wire and the field are perpendicular), so $F = ILB\sin\theta = (3)(0.5)(0.2)(1) = 0.3\ \text{N}$ . …

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Magnetic forces between isolated current elements do not, in general, obey Newton's third law. For the given perpendicular elements, the force on element 1 due to element 2 is zero, while the force on element 2 due to element 1 is nonzero - clearly not equal and opposite.

Why Newton's third law can fail here

Newton's third law requires that the force of A on B be equal and opposite to the force of B on A. This holds for complete, closed circuits (verified experimentally), but it is not guaranteed for two isolated current elements $d\vec{l}_1$ , $d\vec{l}_2$ considered on their own - because the magnetic force each element feels depends on the relative orientation of the two elements through a cross product, and that geometric relationship is not symmetric in general. (The "missing" momentum is actually carried by the electromagnetic field itself - but that is beyond what we need to verify here.)

Setting up the geometry

Element 1: $d\vec{l}_1 = dl\,\hat{i}$ , located at the origin $(0,0,0)$ .
Element 2: $d\vec{l}_2 = dl\,\hat{j}$ , located at $(0,R,0)$ .

Both carry current $I$ . The unit vector from 1 to 2 is $\hat{r}_{12}=\hat{j}$ (distance $R$ ); the unit vector from 2 to 1 is $\hat{r}_{21}=-\hat{j}$ (same distance $R$ ).

Force on element 2 due to element 1

Field at element 2's location, produced by element 1 (Biot-Savart):

d\vec{B}_1 = \frac{\mu_0 I}{4\pi}\frac{d\vec{l}_1\times\hat{r}_{12}}{R^2} = \frac{\mu_0 I}{4\pi R^2}\, dl\,(\hat{i}\times\hat{j}) = \frac{\mu_0 I\,dl}{4\pi R^2}\hat{k}.

Force on element 2 in this field:

d\vec{F}_{2\text{ due to }1} = I\,d\vec{l}_2\times d\vec{B}_1 = I(dl\,\hat{j})\times\left(\frac{\mu_0 I\,dl}{4\pi R^2}\hat{k}\right) = \frac{\mu_0 I^2 dl^2}{4\pi R^2}(\hat{j}\times\hat{k}) = \frac{\mu_0 I^2 dl^2}{4\pi R^2}\hat{i}.

This is nonzero, pointing along $+\hat{i}$ .

Force on element 1 due to element 2

Field at element 1's location, produced by element 2:

d\vec{B}_2 = \frac{\mu_0 I}{4\pi}\frac{d\vec{l}_2\times\hat{r}_{21}}{R^2} = \frac{\mu_0 I\,dl}{4\pi R^2}\big(\hat{j}\times(-\hat{j})\big) = 0,

since the cross product of any vector with itself (or its negative) is zero.

So the force on element 1:

d\vec{F}_{1\text{ due to }2} = I\,d\vec{l}_1\times d\vec{B}_2 = I\,d\vec{l}_1\times 0 = 0.

Comparing the two …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 20 on this concept.

AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
$Q. If a current is passing in a spring, it (A) gets compressed (B) gets expanded (C) oscillates (D) remains unchanged$
›Reveal solutionSolution
This tests the force between parallel current-carrying conductors, applied to the loops of a helical spring. Answer: it gets compressed.

Concept and Intuition

A spring carrying current is essentially a stack of closely-spaced current loops (like a solenoid). Two parallel wires carrying current in the same direction attract one another (this is the basic result behind the definition of the ampere). Since adjacent turns of the spring carry current flowing the same rotational way, neighbouring turns effectively attract each other, pulling the coils closer together.

Step-by-Step Solution

Model the spring as a helix of closely-spaced circular current loops — locally, adjacent turns look like short parallel segments carrying current in the same direction.

The magnetic force between two parallel currents flowing in the same direction is attractive (Ampère's force law).

This attraction pulls each turn of the spring toward its neighbours. …
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AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$Q. A current carrying wire AB is placed near an another long, straight current carrying wire CD as shown in figure. If AB is free to move, then AB will have [FIGURE: CD is a long straight vertical wire (dashed, current direction upward) and AB is a horizontal wire (current flowing from A to B, left to right) placed to the right of CD, oriented perpendicular to CD, with its left end A level with the midpoint of CD] (A) Translational motion only (B) Rotational motion only (C) Both translational and rotational motion (D) Neither translational nor rotational motion$
›Reveal solutionSolution
The field from CD is non-uniform along AB, so the forces on different parts of AB are unequal in magnitude but parallel in direction — this produces both a net translation and a net torque; the answer is: both.

Concept and Intuition

Treating CD as an infinite straight wire, the magnetic field it produces at any external point in the plane containing the two wires is perpendicular to that plane, with magnitude $B=\dfrac{\mu_0 I}{2\pi r}$ falling off with distance $r$ from CD. Because AB is laid out along the direction of increasing $r$ (A is closer to CD than B), the field is strongest near A and weakest near B. The force on each element of AB, $d\vec F=I\,d\vec l\times\vec B$ , therefore points the same way everywhere along AB (perpendicular to AB, parallel to CD) — but its magnitude steadily decreases from A to B.

Step-by-Step Solution

Because the field magnitude is unequal along AB (strong near A, weak near B) but always points the same way, integrating $d\vec F$ along the length of AB gives a non-zero net force — so AB translates. …
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AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
$\log_{10}3 = 0.4771$
›Reveal solutionSolution
The attractive force per unit length between parallel same-direction currents is $\mu_0I_1I_2/2\pi d$ ; integrating it from 10 cm to 30 cm gives the work done per unit length, about $17.6\times10^{-5}$ J/m.

Concept and Intuition

Currents flowing in the same direction in two parallel wires attract each other. To pull them apart (increase their separation), an external agent must do positive work against this attractive force. Since the force is not constant with separation ( $f\propto 1/d$ ), the work must be found by integrating force over distance, not by a simple $F\times\Delta d$ .

Step-by-Step Solution

Force per unit length at separation $d$ : $f(d)=\dfrac{\mu_0 I_1 I_2}{2\pi d}$ .

Work done per unit length moving from $d_1=0.1$ m to $d_2=0.3$ m:

$\dfrac{W}{L}=\int_{0.1}^{0.3} f(d)\,dd=\dfrac{\mu_0 I_1I_2}{2\pi}\ln\left(\dfrac{d_2}{d_1}\right)=\dfrac{\mu_0 I_1I_2}{2\pi}\ln 3$

$\dfrac{\mu_0}{2\pi}=2\times10^{-7}$ ; $I_1I_2=20\times40=800$ .

$\ln 3 = 2.303\log_{10}3=2.303\times0.4771=1.0989$ . …
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AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$30°$
›Reveal solutionSolution
Force per unit length on a wire in a magnetic field is $f=BI\sin\theta$ ; plugging in the given values gives $0.6\ \mathrm{N\,m^{-1}}$ .

Concept and Intuition

A straight current-carrying wire of length $L$ placed in a uniform magnetic field $B$ experiences a force $F = BIL\sin\theta$ , where $\theta$ is the angle between the current direction and the field. This comes directly from the Lorentz force acting on the moving charge carriers, summed over the wire. Dividing both sides by $L$ gives the force per unit length, $f = BI\sin\theta$ — a useful quantity because it doesn't depend on how long the wire actually is.

Step-by-Step Solution

Write the force-per-length formula: $f = BI\sin\theta$ .

Substitute $B = 0.15$ T, $I = 8$ A, $\theta = 30°$ .

$\sin 30° = 0.5$ .

$f = 0.15 \times 8 \times 0.5 = 1.2\times0.5 = 0.6\ \mathrm{N\,m^{-1}}$ .

Common Mistakes …
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AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
$4\sqrt{3}$
›Reveal solutionSolution
The force per unit length on a current-carrying wire in a magnetic field is $F/L = BI\sin\theta$ ; plugging in the given values yields $1.2\,\mathrm{N/m}$ .

Concept and Intuition

A current-carrying conductor in an external magnetic field experiences a force $\vec F = I\vec L \times \vec B$ , whose magnitude per unit length is $F/L = BI\sin\theta$ , where $\theta$ is the angle between the current direction and the field.

Step-by-Step Solution

Given $I = 4\sqrt3\,\mathrm{A}$ , $B = 200\,\mathrm{mT} = 0.2\,\mathrm{T}$ , $\theta = 60^\circ$ .

$\sin60^\circ = \dfrac{\sqrt3}{2}$ .

$F/L = BI\sin\theta = 0.2 \times 4\sqrt3 \times \dfrac{\sqrt3}{2}$ . …
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AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
$\dfrac{mv}{Bl}$
›Reveal solutionSolution
Equating the magnetic impulse ( $Bl$ times the total charge passed) to the rod's momentum change gives $q=mv/(Bl)$ .

Concept and Intuition

A current-carrying rod in a magnetic field experiences a force $F=BIl$ . If the current is a brief, arbitrary pulse rather than constant, we cannot use $F=ma$ directly with a single value, but the impulse (time-integral of force) is straightforward: $J=\int F\,dt = Bl\int I\,dt = Bl\,q$ , since $\int I\,dt$ is exactly the total charge $q$ that flowed.

Step-by-Step Solution

Instantaneous force on the rod: $F(t) = B\,I(t)\,l$ .

Impulse (total change in momentum): $J = \int_0^t F\,dt = Bl\int_0^t I\,dt = Bl\,q$ .

Since the rod starts at rest and ends with speed $v$ : $J = mv - 0 = mv$ . …
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AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
$Q. A wire of 60 cm length and mass 10 g is suspended by a pair of flexible leads in a magnetic field of 0.60 T as shown in the figure. The magnitude of the current required to remove the tension in the supporting leads is [FIGURE] (a horizontal wire of length 60 cm suspended by two vertical flexible leads inside a uniform magnetic field directed into the page, shown by rows of \times symbols) (A) 0.47 A (B) 0.17 A (C) 0.27 A (D) 0.32 A$
›Reveal solutionSolution
The key idea is that the magnetic force on the current-carrying wire must exactly balance its weight to remove tension in the leads. Using the formula $F = I L B$ and setting it equal to $mg$ , the required current is $I = \frac{mg}{LB} \approx 0.27\ \text{A}$ , so the correct option is (C).

Concept and Intuition: Magnetic Force on a Current-Carrying Wire

When a wire carries an electric current and sits in a magnetic field, the field exerts a force on the moving charges inside the wire. This force is given by the simple but powerful equation:
$\vec{F} = I \vec{L} \times \vec{B}$
Here, $I$ is the current, $\vec{L}$ is a vector pointing along the wire in the direction of the current (with magnitude equal to the wire's length), and $\vec{B}$ is the magnetic field. The magnitude of the force is $F = I L B \sin\theta$ , where $\theta$ is the angle between the wire and the field.

In this problem, the wire is horizontal, and the magnetic field points directly into the page (shown by the "×" symbols). The wire is perpendicular to the field, so $\theta = 90^\circ$ and $\sin\theta = 1$ . The force is therefore $F = I L B$ , and its direction is given by the right-hand rule: point your fingers in the direction of the current, curl them toward the field (into the page), and your thumb gives the force direction. For a current flowing left-to-right, the force points upward. That upward magnetic force can lift the wire, reducing the tension in the flexible leads. To remove the tension entirely, the magnetic force must exactly balance the wire's weight.

Step-by-Step Solution

Identify the forces acting on the wire. The wire has weight $mg$ pulling it downward. The flexible leads can only provide tension; if the magnetic force exactly cancels the weight, the leads go slack (tension = 0). So we require:

$F_{\text{magnetic}} = mg$

Write the magnetic force expression. Since the wire is perpendicular to the field, the magnitude is:

$F_{\text{magnetic}} = I L B$
where $L = 60\ \text{cm} = 0.60\ \text{m}$ , $B = 0.60\ \text{T}$ , and $I$ is the unknown current.

Write the weight expression. Mass $m = 10\ \text{g} = 0.010\ \text{kg}$ , and $g = 9.8\ \text{m/s}^2$ . So:
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AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
$3.2 \times 10^{-5}$
›Reveal solutionSolution
The current flows exactly parallel to Earth's horizontal magnetic field (both south-to-north), so the magnetic force on it is zero.

Concept and Intuition

The force per unit length on a current-carrying wire in a magnetic field is $\dfrac{F}{L}=BI\sin\theta$ , where $\theta$ is the angle between the current direction and the field direction. Earth's horizontal magnetic field component at most places points from geographic south to geographic north (toward magnetic north, ignoring declination). If the wire's current also flows south to north, the current and field vectors are parallel — and a magnetic force requires a component of current perpendicular to the field, which is absent here.

Step-by-Step Solution

Direction of current: south to north.

Direction of Earth's horizontal field component: also south to north (that's the defining direction of the horizontal component).

Angle between current and field: $\theta=0°$ . …
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AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
$6.25\times10^{-5}$
›Reveal solutionSolution
Only the two sides of the loop parallel to the wire feel a net (unbalanced) force since they're at different distances; the perpendicular sides' forces cancel by symmetry, leaving a clean two-term calculation.

Concept and Intuition

Two parallel current-carrying wires attract or repel with force per unit length $\dfrac{\mu_0 I_1I_2}{2\pi d}$ . For a rectangular loop near a long straight wire, the near and far sides parallel to the wire experience forces in opposite senses (because loop current flows in opposite directions along those two sides) but of different magnitude (since they're at different distances), so they don't cancel — their difference is the net force. The two sides perpendicular to the wire experience forces that are equal and opposite by symmetry (same current, mirror-image geometry along the wire's length) and cancel exactly.

Step-by-Step Solution

Near side: 10 cm from wire, force per length $= \dfrac{\mu_0 I_1I_2}{2\pi d_1}$ .

Far side: 10+10=20 cm from wire, force per length $= \dfrac{\mu_0 I_1I_2}{2\pi d_2}$ . …
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AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
$Q. A magnetic field of 1T is acting on a wire carrying current of 1A so that the wire is at rest in air. If the mass of the wire is 100 g then the length of the wire is (A) 1 m (B) 0.5 m (C) 2 m (D) 0.25 m$
›Reveal solutionSolution
The magnetic force $BIL$ on the current-carrying wire must equal its weight $mg$ for it to float; solving for $L$ gives about 1 m.

Concept and Intuition

A current-carrying wire in a magnetic field experiences a force $F=BIL$ (for field perpendicular to the wire). If this force is directed upward and equals the wire's weight, the wire can remain suspended in equilibrium — a classic magnetic levitation setup.

Step-by-Step Solution

Equilibrium condition: $BIL=mg$ .

$m=100\ \mathrm g=0.1\ \mathrm{kg}$ , $B=1$ T, $I=1$ A, $g\approx9.8\ \mathrm{ms^{-2}}$ . …
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AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
$3\times10^{-5}$
›Reveal solutionSolution
This tests the force on a current-carrying conductor in a magnetic field, $F=BIL\sin\theta$ , when the current and field are parallel. Answer: the force is zero.

Concept and Intuition

The magnetic force per unit length on a straight current-carrying wire is $f=BI\sin\theta$ , where $\theta$ is the angle between the current's direction and the magnetic field. This force vanishes whenever the current flows parallel (or antiparallel) to the field, because $\vec F = I\vec L\times\vec B$ and the cross product of two parallel vectors is zero.

Step-by-Step Solution

The conductor carries current from south to north.

The Earth's horizontal magnetic field at the location also points from geographic south to geographic north. …
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AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
$Q. A wire loop of irregular shape carrying current is placed in an external magnetic field. If the wire is flexible, the shape of the loop changes to (A) helical (B) circular (C) straight line (D) parabolic$
›Reveal solutionSolution
A flexible current loop free to move under the magnetic forces on it settles into a circular shape, since that shape balances the force per unit length uniformly all around.

Concept and Intuition

Each element of a current loop in a magnetic field feels a force $d\vec F = I\, d\vec l \times \vec B$ . If the wire is rigid, its shape stays fixed and net force/torque may act on it as a whole. If instead the wire is flexible, each small segment can move independently in response to the local force on it, so the loop reshapes itself. The shape that gives a uniform, self-consistent outward force per unit length everywhere along the loop (mechanical equilibrium, like a stretched elastic band forced outward uniformly) is the circle — any irregular shape has regions of higher curvature that would keep deforming until the whole loop becomes circular.

Step-by-Step Solution

Recognize that a flexible current loop is like an elastic ring being pushed outward (or inward) by magnetic forces at every point.

The equilibrium shape under such uniformly-distributed radial forces is the one with constant curvature — a circle. …
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Q.Do magnetic forces obey Newton's third law. Verify for two current elements $d\vec{l}_1 = dl\,\hat{i}$ located at the origin and $d\vec{l}_2 = dl\,\hat{j}$ located at $(0, R, 0)$ . Both carry current $I$ .

Concept understanding — Magnetic Force on Current

Magnetic Force on Current

From a single charge to a wire

The formula

Direction: the right-hand rule

Worked example

Why Newton's third law can fail here

Setting up the geometry

Force on element 2 due to element 1

Force on element 1 due to element 2

Comparing the two …

Method: Testing Newton's Third Law Between Two Current Elements

Steps

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.Do magnetic forces obey Newton's third law. Verify for two current elements dl⃗1=dl i^d\vec{l}_1 = dl\,\hat{i}dl1​=dli^ located at the origin and dl⃗2=dl j^d\vec{l}_2 = dl\,\hat{j}dl2​=dlj^​ located at (0,R,0)(0, R, 0)(0,R,0). Both carry current III.

Why Newton's third law can fail here

Setting up the geometry

Force on element 2 due to element 1

Force on element 1 due to element 2

Comparing the two …

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Q.Do magnetic forces obey Newton's third law. Verify for two current elements $d\vec{l}_1 = dl\,\hat{i}$ located at the origin and $d\vec{l}_2 = dl\,\hat{j}$ located at $(0, R, 0)$ . Both carry current $I$ .

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