(a) Derive the expression for the force acting per unit length between two long straight parallel current carr

Question

Q. (a) Derive the expression for the force acting per unit length between two long straight parallel current carrying conductors. Hence define one ampere.

(b) Two long parallel straight conductors are placed 12 cm apart in air. They carry equal currents of 3 A each. Find the magnitude and direction of the magnetic field at a point midway between them (drawing a figure) when the currents in them flow in opposite directions.

(OR)

(a) Draw the schematic sketch of a cyclotron. Explain the shape of the path on which charged particle moves when the particle is accelerated by it.

(b) To convert a given galvanometer into a voltmeter of ranges 2 V, V and

\dfrac{V}{2}

volt, resistances

R_1

R_2

and

R_3

ohm respectively, are required to be connected in series with the galvanometer. Obtain the relationship between

R_1

R_2

and

R_3

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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 — Cyclotron

The Cyclotron: Why a Constant Frequency Can Accelerate a Particle to High Speeds

Imagine you want to throw a ball faster and faster, but you can only give it a small push each time. You could set up two paddles that slap the ball back and forth, each time adding a little speed. But the ball would just go in a straight line and fly away. To keep it contained, you need something to bend its path back toward you after each push.

That is the core idea of a cyclotron. It uses a magnetic field to bend the path of a charged particle into a circle, and an electric field (applied across two hollow D-shaped electrodes called "dees") to give it a kick of energy each time it crosses the gap between them. The trick is that the electric field must reverse direction at exactly the right moment — once per half-circle — so that it always pushes the particle forward, never backward.

The Surprising Fact: Frequency Does Not Depend on Speed

Here is the key insight that makes the cyclotron work. When a charged particle moves in a uniform magnetic field $B$ , it experiences a centripetal force:

F = qvB = \frac{mv^2}{r}

From this, the radius of its circular path is:

r = \frac{mv}{qB}

The time it takes to complete one full circle (the period $T$ ) is:

T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}

Notice: $v$ cancels out. The period — and therefore the frequency $f = 1/T$ — depends only on the charge $q$ , the mass $m$ , and the magnetic field $B$ . It does not depend on how fast the particle is moving.

f = \frac{qB}{2\pi m}

This is the cyclotron frequency. As the particle gains energy and its speed increases, its orbit radius grows (since $r = mv/qB$ ), but the time per revolution stays exactly the same. So you can set the alternating voltage across the dees to this fixed frequency, and it will always be in sync with the particle's motion — no matter how fast the particle gets.

How It Actually Works

A charged particle (say, a proton) is released near the centre, between the two dees.
A magnetic field perpendicular to the dees bends its path into a half-circle inside one dee.
When it reaches the gap, the electric field is oriented to accelerate it forward. The particle gains kinetic energy.
It enters the other dee with a slightly higher speed, so its next half-circle has a slightly larger radius.
By the time it returns to the gap, the electric field has reversed polarity — because exactly one half-period has passed — so it again accelerates the particle forward.
This repeats. Each crossing of the gap adds energy. The spiral path grows outward until the particle reaches the edge and is extracted.

Watch out

A common mistake is to think the particle speeds up inside the dees. It does not — the electric field is zero inside the hollow dees (they are conductors). Acceleration happens only in the gap between them. The magnetic field inside the dees merely bends the path.

The Limitation: Relativity …

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Part (a): $\dfrac{F}{l}=\dfrac{\mu_0 I_1 I_2}{2\pi d}$ (defines the ampere); for 3 A antiparallel currents 12 cm apart the field midway is $2\times10^{-5}$ T into the page.

Part (b): a cyclotron drives the ion along a spiral of growing semicircles; for voltmeter ranges $2V,V,V/2$ the series resistances satisfy $R_1-R_2=2(R_2-R_3)$ , i.e. $R_1+2R_3=3R_2$ .

Schematic top view of a cyclotron showing the two D-shaped dees D1 and D2 separated by a narrow gap with an ion source S at the centre, the high-frequency oscillator connected across the dees to accelerate the ion at every gap crossing, the resulting path of ever-increasing-radius semicircles (a spiral) traced inside the dees, and crosses marking the magnetic field directed into the plane of the page.

Part (a): Force Between Parallel Wires; Field at the Midpoint

Derivation. Wire 1 carrying $I_1$ produces, at perpendicular distance $d$ , a field $B_1=\dfrac{\mu_0 I_1}{2\pi d}$ . A parallel wire carrying $I_2$ lies in this field, so a length $l$ of it feels

F=I_2 l B_1=\frac{\mu_0 I_1 I_2\,l}{2\pi d}\ \Rightarrow\ \frac{F}{l}=\frac{\mu_0 I_1 I_2}{2\pi d}

The force is attractive when the currents are parallel, repulsive when antiparallel (Newton's third law: equal and opposite on the two wires).

Definition of one ampere. One ampere is that steady current which, flowing in two infinitely long, straight, parallel conductors of negligible cross-section placed 1 m apart in vacuum, produces a force of $2\times10^{-7}$ N per metre of length. (Check: $\dfrac{\mu_0(1)(1)}{2\pi(1)}=\dfrac{4\pi\times10^{-7}}{2\pi}=2\times10^{-7}$ N/m.)

Numerical (opposite currents). $d=12$ cm $\Rightarrow$ midpoint $r=6$ cm $=0.06$ m; $I=3$ A each.

B=\frac{\mu_0 I}{2\pi r}=\frac{4\pi\times10^{-7}\times3}{2\pi\times0.06}=\frac{6\times10^{-7}}{0.06}=1\times10^{-5}\ \text{T}

Directions (right-hand rule). Take wire 1 (left) with current up and wire 2 (right) with current down. At the midpoint the field of wire 1 points into the page, and the field of wire 2 (opposite current, on the other side) also points into the page. For antiparallel currents the two fields at the midpoint are in the same direction and therefore add:

B_{\text{net}}=B_1+B_2=2\times10^{-5}\ \text{T (into the page)}

Figure: two vertical wires 12 cm apart, currents opposite; at the midpoint mark two " $\otimes$ " (into page) arrows adding to $2\times10^{-5}$ T. …

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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 16 on this concept.

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)

…

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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 …

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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:

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

…

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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 ( …

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

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

…

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CBSE 2023Set F1 markMCQ
$Q. If T is time period and V is maximum speed of a charged particle in cyclotron, then (A) T \propto V (B) T \propto V^2 (C) T \propto 1/V (D) T \propto 1/V^2$
›Reveal solutionSolution

The cyclotron period T = 2πm/(qB) is independent of the speed V, so none of the offered proportionalities holds.

The cyclotron frequency and period come from equating the magnetic force to the centripetal force:

$qvB = \dfrac{mv^2}{r} \Rightarrow r = \dfrac{mv}{qB}$

The period is

$T = \dfrac{2\pi r}{v} = \dfrac{2\pi m}{qB}$

…

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CBSE 2023Set ANNUAL1 mark
$Q. Fill in the blank: The force between two parallel current carrying conductors (flowing in the same direction) is __________.$
›Reveal solutionSolution

Two parallel current-carrying conductors carrying current in the same direction attract each other.

Each current-carrying conductor sets up a magnetic field around itself (by the Biot-Savart/Ampere law), and the other conductor, carrying current in that field, experiences a force $F = BIL$ (via $\vec F = I\vec L\times\vec B$ ). Working out the directions with the right-hand rule shows that when the currents flow in the same direction, the force on each conductor points toward the other - i.e. the conductors …

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CBSE 2022Set ANNUAL1 mark
$Q. The name of machine that accelerates charged particles or ions to high energies is ______ (fill in the blank).$
›Reveal solutionSolution

The device that accelerates charged particles or ions to high kinetic energies using crossed static magnetic and oscillating electric fields is the cyclotron.

A cyclotron consists of two hollow D-shaped electrodes ("dees") placed in a strong uniform magnetic field, with a high-frequency alternating voltage applied between them. A charged particle injected near the centre is accelerated each time it crosses the gap between the dees, and the magnetic field bends it into a circular path of increasing radius as its speed grows. Because the time for one half-revolution …

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CBSE 2020Set ANNUAL1 markMCQ
$Q. Two long parallel wires each carrying a current of 1A in the same direction, are placed 1m apart. The force of attraction between them is$
(a) 2 x 10^-7 Nm^-1
(b) 2 x 10^-4 Nm^-1
(c) 1 x 10^-7 Nm^-1
(d) 4 x 10^-7 Nm^-1
›Reveal solutionSolution

Two parallel current-carrying wires exert a magnetic force on each other; the force per unit length is F/l = (mu0 I1 I2)/(2 pi d), and it is attractive when the currents flow in the same direction.

The magnetic field produced by wire 1 at the location of wire 2 (distance d away) is:

B1 = (mu0 I1)/(2 pi d)

This field exerts a force per unit length on wire 2 (carrying current I2):

F/l = B1 * I2 = (mu0 I1 I2)/(2 pi d)

…

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CBSE 2020Set ANNUAL1 mark
$Q. How does cyclotron increase the energy of charged particles?$
›Reveal solutionSolution

The magnetic field only bends the path (does no work); the oscillating electric field across the dee-gap does the actual accelerating, once every half-cycle, in resonance with the constant cyclotron frequency.

A cyclotron has two hollow, semicircular metal electrodes called dees ( $D_1$ , $D_2$ ), placed in a strong, uniform magnetic field $B$ perpendicular to their plane, with a narrow gap between them connected to a high-frequency oscillating voltage source.

Role of the magnetic field: Inside a dee (a field-free, hollow region electrically), the particle experiences only the magnetic force, which makes it move in a semicircular arc of radius

$r = \frac{mv}{qB}$

Since $\vec F = q\vec v\times\vec B$ is always perpendicular to $\vec v$ , the magnetic field changes only the direction of motion, doing no work and not changing the speed.

Role of the electric field: Each time the particle crosses the narrow gap between the two dees, it passes through the oscillating electric field. If the field's direction is correctly synchronised, the particle is given a push and gains kinetic energy $qV$ (where $V$ is the gap's instantaneous potential difference) every single crossing.

Why this works repeatedly (resonance condition): The time for the particle to complete a semicircle in a dee is

$t = \frac{\pi m}{qB}$

which is independent of the particle's speed and radius (since larger $v$ exactly gives a proportionally larger $r$ , keeping the transit time constant). So if the oscillator frequency is fixed at the cyclotron frequency

$u_c = \frac{qB}{2\pi m}$ …

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Q.(a) Derive the expression for the force acting per unit length between two long straight parallel current carrying conductors. Hence define one ampere.

Part (a): Force Between Parallel Wires; Field at the Midpoint

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