(a) Two long, straight, parallel conductors carry steady currents in opposite directions. Explain the nature o

Question

Q. (a) Two long, straight, parallel conductors carry steady currents in opposite directions. Explain the nature of the force of interaction between them. Obtain an expression for the magnitude of the force between the two conductors. Hence define one ampere.

(OR)

(b) Obtain an expression for the torque

\vec{\tau}

acting on a current carrying loop in a uniform magnetic field

\vec{B}

. Draw the necessary diagram.

CBSECBSE Class XII Board 2024Subjective· 3mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Part (a)Concept understanding — Force Between Parallel Wires

Force Between Parallel Current-Carrying Wires

Imagine two long, straight wires placed side by side, each carrying an electric current. You already know that a current-carrying wire creates a magnetic field around it. And you know that a wire placed in a magnetic field experiences a magnetic force. So here, each wire sits inside the magnetic field created by the other wire. That is the whole story — each wire feels a force because of the other wire's magnetic field.

The direction of that force — attraction or repulsion — depends on whether the currents flow in the same direction or opposite directions.

The Intuition

Take two wires with currents in the same direction. Use the right-hand thumb rule: for wire 1, the magnetic field lines circle around it. At the location of wire 2, that field points in a particular direction. Now apply the right-hand rule for force on a current-carrying wire (Fleming's left-hand rule works too): the current in wire 2, crossed with the field from wire 1, gives a force toward wire 1. The same reasoning from wire 2's perspective gives a force on wire 1 toward wire 2. So they attract.

If the currents are opposite, the field directions reverse, and the forces point away from each other — they repel.

Tip

A quick memory aid: Same direction → Attract; Opposite direction → Repel. This is the opposite of what you might guess from electric charges, where like charges repel. Don't mix them up.

The Precise Statement

For two long, straight, parallel wires separated by a distance $d$ , carrying steady currents $I_1$ and $I_2$ , the magnitude of the force per unit length on either wire is:

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi d}

where $\mu_0 = 4\pi \times 10^{-7} \, \text{N/A}^2$ is the permeability of free space.

The force is attractive if the currents are in the same direction, repulsive if they are opposite.

Where Does This Formula Come From?

Wire 1 produces a magnetic field at the location of wire 2. The magnitude of that field is:

B_1 = \frac{\mu_0 I_1}{2 \pi d}

This field is perpendicular to wire 2. The magnetic force on a length $L$ of wire 2 carrying current $I_2$ in a perpendicular field $B_1$ is:

F = I_2 L B_1

Substitute $B_1$ :

F = I_2 L \cdot \frac{\mu_0 I_1}{2 \pi d}

Divide both sides by $L$ to get force per unit length:

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi d}

That is the entire derivation — two simple steps: field from one wire, then force on the other.

Important

This formula assumes the wires are infinitely long (or at least very long compared to $d$ ) and thin. It gives the force per unit length, which is constant along the wires.

The Definition of the Ampere

This effect is so fundamental that it defines the SI unit of current. One ampere is defined as the constant current which, when flowing through two infinitely long, straight, parallel wires of negligible cross-section placed one metre apart in vacuum, produces a force of exactly $2 \times 10^{-7}$ newtons per metre of length between them. …

Part (b)Concept understanding — Torque on a Current Loop

Torque on a Current Loop

The Intuition First

Imagine a compass needle in the Earth's magnetic field. The needle always turns until it points north. Why? Because the needle itself is a tiny magnet, and the field exerts a twist — a torque — that tries to align it.

A current-carrying loop behaves exactly like that tiny magnet. It has a magnetic moment $\vec{m}$ , which is like its own internal compass arrow. When you place this loop in an external magnetic field $\vec{B}$ , the field pulls on one side of the loop and pushes on the other, creating a turning effect.

The loop doesn't feel a net force (if the field is uniform), but it does feel a torque. That torque always tries to rotate the loop so that its magnetic moment points along the field — just like a compass needle.

The Key Players

The magnetic moment of a planar current loop is:

\vec{m} = I \, A \, \hat{n}

where $I$ is the current, $A$ is the area of the loop, and $\hat{n}$ is a unit vector perpendicular to the plane of the loop (direction given by the right-hand rule: curl your fingers along the current, your thumb points along $\vec{m}$ ).

The external field $\vec{B}$ is uniform — same magnitude and direction everywhere in the region of the loop.

The Torque: Two Equivalent Forms

The torque on the loop is:

\tau = m B \sin\theta

where $\theta$ is the angle between $\vec{m}$ and $\vec{B}$ . The torque is maximum when $\vec{m}$ is perpendicular to $\vec{B}$ ( $\theta = 90^\circ$ ), and zero when they are parallel or antiparallel ( $\theta = 0^\circ$ or $180^\circ$ ).

The vector form captures both magnitude and direction:

\vec{\tau} = \vec{m} \times \vec{B}

\vec{\tau} = \vec{m} \times \vec{B}

The cross product tells you: the torque is perpendicular to both $\vec{m}$ and $\vec{B}$ , and its direction is given by the right-hand rule. This torque always rotates $\vec{m}$ toward $\vec{B}$ .

Why It Happens (The Physics)

Consider a rectangular loop of sides $a$ and $b$ , carrying current $I$ , placed in a uniform field $\vec{B}$ . Let the plane of the loop make an angle $\theta$ with the field.

The two sides of length $a$ are perpendicular to $\vec{B}$ . On each of these sides, the magnetic force is $F = I a B$ , but the forces on opposite sides are in opposite directions. These two forces form a couple — equal and opposite, not along the same line — which produces a torque.

The lever arm for each force is $(b/2) \sin\theta$ , so the net torque is:

\tau = 2 \times (I a B) \times \frac{b}{2} \sin\theta = I (ab) B \sin\theta = I A B \sin\theta

Since $m = I A$ , we get $\tau = m B \sin\theta$ .

Tip

For a rectangular loop, the torque comes only from the sides perpendicular to the field. The sides parallel to the field experience forces that are either zero or along the axis — they contribute nothing to the torque.

The Stable Equilibrium

When $\vec{m}$ is aligned with $\vec{B}$ ( $\theta = 0$ ), the torque is zero. This is a stable equilibrium — if you nudge the loop slightly, the torque brings it back. …

Start your 14-day free trial to unlock the full solution →

(a) Force per unit length between parallel wires is $F/l=\dfrac{\mu_0 I_1 I_2}{2\pi d}$ — repulsive for the opposite currents given; this defines the ampere ( $2\times10^{-7}\ \text{N m}^{-1}$ at $1\ \text{m}$ ). (b) Torque on a loop is $\tau=NIAB\sin\theta$ , i.e. $\vec\tau=\vec m\times\vec B$ .

A rectangular current-carrying coil of N turns in a uniform magnetic field B, with its normal making angle theta with B, showing the forces F on the two current-carrying sides that form a couple producing the torque tau = NIAB sin(theta) on the loop.

Part (a)

When a current flows in a wire it creates a magnetic field around it. A second current-carrying wire placed in that field experiences a force. Applying this twice (each wire in the other's field) gives the mutual force.

1. Field of conductor 1 at conductor 2. For an infinitely long straight wire, Ampère's law gives, at perpendicular distance $d$ ,

B_1=\frac{\mu_0 I_1}{2\pi d}.

2. Force on conductor 2. A length $l$ of conductor 2 (current $I_2$ ) lies in this field, which is perpendicular to it, so

\vec F=I_2\vec l\times\vec B_1,\qquad F=I_2 l B_1=\frac{\mu_0 I_1 I_2 l}{2\pi d}.

3. Force per unit length.

\frac{F}{l}=\frac{\mu_0 I_1 I_2}{2\pi d}

4. Nature of the force. By the right-hand rule applied to both wires: parallel (same-direction) currents attract, antiparallel (opposite-direction) currents repel. The conductors in this question carry opposite currents, so the force is repulsive.

5. Definition of the ampere. Setting $I_1=I_2=1\ \text{A}$ , $d=1\ \text{m}$ :

$\frac{F}{l}=\frac{(4\pi\times10^{-7})(1)(1)}{2\pi(1)}=2\times10^{-7}\ \text{N m}^{-1}.$ …

Start your 14-day free trial to unlock the full solution →

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 19 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. If vector m be the magnetic moment of a magnetic dipole placed in a magnetic field of induction vector B, the torque experienced by the dipole will be$
(a) m . B
(b) |m| / |B|
(c) m x B
(d) |m| |B|
›Reveal solutionSolution

The torque on a magnetic dipole is $\vec{\tau} = \vec{m}\times\vec{B}$ , analogous to torque on an electric dipole $\vec{p}\times\vec{E}$ .

When a magnetic dipole of moment $\vec{m}$ is placed in a uniform magnetic field $\vec{B}$ , it experiences a torque that tends to align it with the field. This torque is given by the vector product:

$\vec{\tau} = \vec{m}\times\vec{B}$

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. Two long parallel wires each carrying a current of 1 A in the same direction, are placed 1 m apart. The force of attraction between them is$
(a) 2 x 10^7 N/m
(b) 2 x 10^-4 N/m
(c) 2 x 10^-7 N/m
(d) 4 x 10^-7 N/m
›Reveal solutionSolution

Two parallel current-carrying wires attract if their currents are in the same direction; the force per unit length is mu_0I1I2/(2pid).

Each current-carrying wire produces a magnetic field around it, and this field exerts a force on the other current-carrying wire (F = I*L x B). The standard result for the force per unit length between two long straight parallel wires carrying currents I1 and I2, separated by a distance d, is

F/L = mu_0 * I1 * I2 / (2 * pi * d)

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. Assertion (A): Two infinitely long straight conductors carrying current in the same direction attract each other. Reason (R): The net magnetic field at a point exactly halfway between two infinitely long straight conductors carrying current in the same direction is zero.$
(a) Both Assertion and Reason are true, and reason is the correct explanation
(b) Both Assertion and Reason are true, but the Reason is not the correct explanation
(c) Assertion is true, but Reason is false.
(d) Assertion is false, but Reason is true.
›Reveal solutionSolution

Both statements are individually correct, but the field being zero at the midpoint is not why the two wires attract each other.

Checking the Assertion: Two infinitely long straight parallel conductors carrying current in the SAME direction do attract each other. Each wire sits in the magnetic field created by the other wire, and using $\vec F = I\vec L \times \vec B$ (or the right-hand/Fleming's left-hand rule), the force on each wire due to the other's field points towards the other wire. So the Assertion is TRUE.

Checking the Reason: Take the two wires along the $y$ -axis at $x=-a$ and $x=+a$ , both carrying current $I$ in the $+y$ direction. At the midpoint (origin), using $\vec B = \dfrac{\mu_0 I}{2\pi r}\hat\phi$ with $\hat\phi = \hat I \times \hat r$ : the field due to the left wire points in $+\hat y'$ s perpendicular direction (say $+\hat z$ ), while the field due to the right wire (displacement now in $-\hat x$ from that wire) points in the opposite transverse direction ( $-\hat z$ ). Since both wires are equidistant and carry equal current, these two fields are equal in magnitude and opposite in direction — they cancel exactly. So the net field at the midpoint IS zero when the currents …

Start your 14-day free trial to see the full solution →
CBSE 2026Set SEM31 markMCQ
$Q. The ratio of the radii of two circular loops is 1 : 2. The ratio of their magnetic moments is 1 : 2. The ratio of currents flowing through them is$
(a) 1 : 1
(b) 2 : 1
(c) 4 : 1
(d) 1 : 4
›Reveal solutionSolution

Magnetic moment M = I·(πr²), so I = M/(πr²) ∝ M/r². Substituting the given ratios gives I₁ : I₂ = 2 : 1. Option (b).

Step 1 — magnetic moment of a current loop (NCERT/CBSE Class 12 Physics, Moving Charges and Magnetism): M = I·A = I·πr².

Step 2 — so current I = M/(πr²), i.e. I ∝ M/r².

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set D1 markMCQ
$Q. Dimensional formula of permeability is (A) [MLT^-2 A^-2] (B) [MLT^2 A^-2] (C) [MLT^2 A^2] (D) [MLT^-2 A]$
›Reveal solutionSolution

Using the force per unit length between two wires, μ₀ works out to dimensions [M L T⁻² A⁻²].

The force per unit length between two parallel current-carrying wires is

$\frac{F}{\ell} = \frac{\mu_0 I_1 I_2}{2\pi d}$

Solving for μ₀:

$\mu_0 = \frac{2\pi d\,(F/\ell)}{I_1 I_2}$

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. Assertion: The turns of a spring come close to each other, when current is passed through it. Reason: It is because, the turns of a spring carry current in same direction and hence attract each other.$
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not a correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
›Reveal solutionSolution

Adjacent turns of a current-carrying spring act like parallel wires carrying current in the same direction, which attract each other by the magnetic force between parallel currents — so the coils are pulled together.

Two straight parallel conductors carrying currents in the SAME direction attract each other (force per unit length $F/l = \mu_0 I_1 I_2 / 2\pi d$ , attractive for like-directed currents, repulsive for opposite). A spring is essentially a coil of many closely-spaced turns; each turn carries current in the same sense as its neighbours. Treating adjacent turns as parallel current-carrying wires, they attract each other, so the spring's turns are pulled closer together ( …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. Two circular loops having ratio of their radii 1 : 2 possess same magnetic moment. The ratio of their circulating currents will be$
(a) 4 : 1
(b) 1 : 4
(c) 2 : 1
(d) 1 : 2
›Reveal solutionSolution

Magnetic moment m = IA = Iπr²; equal m with r ratio 1:2 forces the current ratio to be 4:1 (inverse of the area ratio).

Magnetic moment of a current loop is $m = I\pi r^2$ . Let the radii be $r_1:r_2 = 1:2$ and the moments be equal, $m_1=m_2$ : …

Start your 14-day free trial to see the full solution →
CBSE 2024Set 55/1/11 markMCQ
$Q. For question 15, two statements are given - one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false and Reason (R) is also false. Assertion (A) : Two long parallel wires, freely suspended and connected in series to a battery, move apart. Reason (R) : Two wires carrying current in opposite directions repel each other.$
›Reveal solutionSolution
Connected in series, the two freely suspended parallel wires carry equal currents in opposite directions — the current goes out along one wire and returns along the other. Antiparallel currents repel, so the wires move apart. Both statements are true and the Reason is exactly why the wires separate. The correct option is (A).

The physical setup: what does "in series" mean here?

When two long parallel wires hang freely side by side and are joined in series to a battery, there is a single current path: the current leaves the battery, travels along the first wire, crosses over at the far end, and comes back along the second wire to the battery. Because the second wire carries the return current, the two adjacent wires carry equal currents in opposite directions — antiparallel currents.

Force between the wires

Each wire sits in the magnetic field created by the other. For two long parallel wires a distance $d$ apart carrying currents $I_1$ and $I_2$ , the force per unit length on either wire is
$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$
with the standard direction rule: parallel (same-direction) currents attract; antiparallel (opposite-direction) currents repel. You can check this with $\vec{F} = I\vec{L} \times \vec{B}$ : for opposite currents, the field of wire 1 at wire 2 gives a force on wire 2 pointing away from wire 1, and by Newton's third law wire 1 is pushed away from wire 2 with equal magnitude.

Evaluating the statements

Assertion (A): "Two long parallel wires, freely suspended and connected in series to a battery, move apart." As shown above, the series connection makes the currents antiparallel, the wires repel, and — being freely suspended — they move apart. True. …
Start your 14-day free trial to see the full solution →
CBSE 2024Set A11 mark
$Q.The torque on a rectangular current loop in a uniform magnetic field increases by ———————— the area of the loop. Fill in the blank choosing the appropriate answer from the bracket: (decreasing, interference, helium, greater, diffraction, increasing)$
›Reveal solutionSolution

increasing (the torque is directly proportional to the area of the loop). …

Start your 14-day free trial to see the full solution →
CBSE 2024Set A1 markMCQ
$Q. The nature of electron beams moving with uniform velocity in the same direction will be (A) converging (B) diverging (C) parallel (D) none of these$
›Reveal solutionSolution
Like charges repel electrostatically; this force exceeds the magnetic attraction at ordinary speeds, so the beams diverge.

Two parallel electron beams experience two effects:

As parallel currents in the same direction, the magnetic force is attractive.

As streams of like (negative) charges, the electrostatic force is repulsive. …
Start your 14-day free trial to see the full solution →
CBSE 2024Set A1 markMCQ
$Q. The value of torque (τ) experienced by current loop of magnetic moment (m) placed in magnetic field (B) is (A) τ = m \times B (B) τ = B \times m (C) τ = m/B (D) τ = B/m$
›Reveal solutionSolution

A current loop behaves like a magnetic dipole; the torque on it is the cross product of its magnetic moment and the field, τ = m × B.

A planar current loop carrying current I and enclosing area A has a magnetic (dipole) moment $\vec{m} = I\vec{A}$ , directed along the normal to the loop (right-hand rule).

When this dipole is placed in a uniform magnetic field $\vec{B}$ , the two sides of the loop carry equal and opposite forces that form a couple. The resulting torque is

$\vec{\tau} = \vec{m} \times \vec{B}, \qquad \tau = mB\sin\theta$

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. Two long parallel wires each carrying a current of 1 A in the same direction, are placed 1 m apart. The force of attraction between them is$
(a) 2 x 10^-7 N/m
(b) 2 x 10^-4 N/m
(c) 1 x 10^-7 N/m
(d) 4 x 10^-7 N/m
›Reveal solutionSolution

Two parallel current-carrying wires attract each other (same direction) with a force per unit length given by mu0 I1 I2 / (2pid).

The force per unit length between two long parallel wires carrying currents $I_1$ and $I_2$ , separated by distance $d$ , is

$\dfrac{F}{l} = \dfrac{\mu_0 I_1 I_2}{2\pi d}$

Substituting $\mu_0 = 4\pi\times10^{-7}\ \text{T m/A}$ , $I_1 = I_2 = 1\ \text{A}$ , $d = 1\ \text{m}$ :

$\dfrac{F}{l} = \dfrac{4\pi\times10^{-7}\times 1\times 1}{2\pi\times 1} = 2\times10^{-7}\ \text{N/m}$

…

Start your 14-day free trial to see the full solution →

Q.(a) Two long, straight, parallel conductors carry steady currents in opposite directions. Explain the nature of the force of interaction between them. Obtain an expression for the magnitude of the force between the two conductors. Hence define one ampere.

Part (a)

🎓Unlock everything free for 14 days

Unlock everything free for 14 days