Derive an expression for the magnetic force $\vec{F}$ acting on a straight conductor of length $L$ carrying current $I$ in an external magnetic field $\vec{B}$. Is it valid when the conductor is in zig-zag form? Justify.

> [!TLDR] > The magnetic force on a straight current-carrying conductor of length $L$ is derived from the force on individual moving charges, resulting in $\vec{F} = I(\vec{L} \times \vec{B})$. This expression remains valid for a zig-zag conductor, provided $\vec{L}$ is taken as the net displacement vector from the start to the end of the conductor. When a current flows through a conductor, it means that charges are moving. We know that a moving charge experiences a force when it is in a magnetic field. Therefore, it is logical that a conductor carrying current, which is essentially a collection of moving charges, will also experience a force when placed in an external magnetic field. The total force on the conductor is the sum of the forces on all the individual moving charges within it. Let's derive the expression for this force. ### Derivation of Magnetic Force on a Straight Conductor 1. **Force on a single moving charge:** The fundamental principle is the Lorentz force law, which states that a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ experiences a force $\vec{F}_q$ given by: $$ \vec{F}_q = q(\vec{v} \times \vec{B}) $$ In a conductor, the charges responsible for current are typically electrons, which have a drift velocity $\vec{v}_d$. For conventional current $I$, we consider the flow of positive charges, so the direction of $\vec{v}_d$ is taken along the direction of current. 2. **Relating current to charge flow:** Consider a small segment of the conductor of length $dL$. Let $A$ be the cross-sectional area of the conductor, and $n$ be the number density of charge carriers (number of charge carriers per unit volume). If each charge carrier has charge $q$, then the total charge $dQ$ contained in this small segment of volume $A \cdot dL$ is: $$ dQ = (n A dL) q $$ The current $I$ flowing through the conductor is related to the drift velocity $\vec{v}_d$ by the formula: $$ I = n A q v_d $$ Here, $

Derive an expression for the magnetic force vecF acting on a straight conductor of length L carrying current I

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\vec{F}

CBSECBSE Class XII Board 2024Subjective· 2mImportance★★★★★

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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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The magnetic force on a straight current-carrying conductor of length $L$ is derived from the force on individual moving charges, resulting in $\vec{F} = I(\vec{L} \times \vec{B})$ . This expression remains valid for a zig-zag conductor, provided $\vec{L}$ is taken as the net displacement vector from the start to the end of the conductor.

When a current flows through a conductor, it means that charges are moving. We know that a moving charge experiences a force when it is in a magnetic field. Therefore, it is logical that a conductor carrying current, which is essentially a collection of moving charges, will also experience a force when placed in an external magnetic field. The total force on the conductor is the sum of the forces on all the individual moving charges within it.

Let's derive the expression for this force.

Derivation of Magnetic Force on a Straight Conductor

Force on a single moving charge: The fundamental principle is the Lorentz force law, which states that a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ experiences a force $\vec{F}_q$ given by:

\vec{F}_q = q(\vec{v} \times \vec{B})

In a conductor, the charges responsible for current are typically electrons, which have a drift velocity $\vec{v}_d$. For conventional current $I$, we consider the flow of positive charges, so the direction of $\vec{v}_d$ is taken along the direction of current.

2. Relating current to charge flow:

Consider a small segment of the conductor of length $dL$ . Let $A$ be the cross-sectional area of the conductor, and $n$ be the number density of charge carriers (number of charge carriers per unit volume). If each charge carrier has charge $q$ , then the total charge $dQ$ contained in this small segment of volume $A \cdot dL$ is:

dQ = (n A dL) q

The current $I$ flowing through the conductor is related to the drift velocity $\vec{v}_d$ by the formula:

I = n A q v_d

Here, $v_d$ is the magnitude of the drift velocity.

3. Force on a small current element:

Now, let's find the force $d\vec{F}$ acting on this small segment $dL$ . All the charge carriers $dQ$ within this segment are moving with an average drift velocity $\vec{v}_d$ . So, the force on this segment is the sum of forces on all these charges:

d\vec{F} = dQ (\vec{v}_d \times \vec{B})

Substitute the expression for $dQ$:

d\vec{F} = (n A dL q) (\vec{v}_d \times \vec{B})

We can rearrange the terms. Notice that $n A q v_d$ is the current $I$. Also, we can define a vector $d\vec{L}$ whose magnitude is $dL$ and whose direction is along the direction of the current (which is the direction of $\vec{v}_d$).
Therefore, we can write $\vec{v}_d dL$ as $v_d d\vec{L}$ (if $d\vec{L}$ is in the direction of $\vec{v}_d$) or more precisely, $d\vec{L}$ is a vector representing the infinitesimal displacement in the direction of current.

d\vec{F} = (n A q v_d) (d\vec{L} \times \vec{B})

Substituting $I = n A q v_d$:

d\vec{F} = I (d\vec{L} \times \vec{B})

This is the force on an infinitesimal current element $I d\vec{L}$.

4. Total force on a straight conductor:

To find the total force $\vec{F}$ on a straight conductor of finite length $L$ , we integrate the force $d\vec{F}$ over the entire length of the conductor. For a straight conductor, the direction of $d\vec{L}$ is constant along its length. Let $\vec{L}$ be a vector representing the length of the conductor, pointing in the direction of the current.

\vec{F} = \int_0^L I (d\vec{L} \times \vec{B})

Since $I$ and $\vec{B}$ are constant along the straight conductor, we can take them out of the integral:

\vec{F} = I \left( \int_0^L d\vec{L} \right) \times \vec{B}

The integral $\int_0^L d\vec{L}$ simply gives the total displacement vector $\vec{L}$ from the start to the end of the conductor.

\vec{F} = I (\vec{L} \times \vec{B})

\vec{F}

Validity for a Zig-zag Conductor

The expression $\vec{F} = I(\vec{L} \times \vec{B})$ is derived for a straight conductor. However, it is valid for a zig-zag conductor, provided we interpret $\vec{L}$ correctly.

Consider a zig-zag conductor made up of several small straight segments, say $\Delta \vec{l}_1, \Delta \vec{l}_2, \dots, \Delta \vec{l}_N$ . The current $I$ flows sequentially through these segments. The magnetic field $\vec{B}$ is uniform.

Force on each segment: The force on each individual straight segment $\Delta \vec{l}_i$ is given by:

\Delta \vec{F}_i = I (\Delta \vec{l}_i \times \vec{B})

Total force by superposition: …

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

CBSE 2024Set 55/1/11 markMCQ
$I$
›Reveal solutionSolution
In a uniform magnetic field a closed current loop feels zero net force, and with the loop's plane already perpendicular to $\vec{B}$ the torque is zero too — so it neither translates nor rotates. But each current element feels a radial force $d\vec{F} = I\,d\vec{l}\times\vec{B}$ , and for a clockwise current with $\vec{B}$ along $+z$ this force points radially inward on every element. The loop tends to shrink. The correct option is (C).

The key here is to look past the two "global" effects (net force and torque) — both of which vanish in this configuration — and examine the force on each individual element of the loop.

Set up the geometry. The loop lies in the $x$ – $y$ plane and the field is $\vec{B} = B\hat{k}$ (along the $z$ -axis). The current $I$ flows clockwise as seen from the $+z$ direction, so by the right-hand rule the loop's magnetic moment $\vec{m} = IA\hat{n}$ points along $-\hat{k}$ , anti-parallel to $\vec{B}$ .

No translation. For any closed loop in a uniform field the net force is

$\vec{F} = I\oint d\vec{l}\times\vec{B} = I\left(\oint d\vec{l}\right)\times\vec{B} = \vec{0},$
because $\oint d\vec{l} = 0$ around a closed path. This immediately rules out options (A) and (B) — the loop cannot move along the $x$ - or $y$ -axis.

No rotation. The torque is $\vec{\tau} = \vec{m}\times\vec{B}$ . Here $\vec{m}$ is anti-parallel to $\vec{B}$ , so $\vec{\tau} = 0$ — the loop does not turn.

Force on each element — the deciding step. Take the element of the loop at the point $(R, 0, 0)$ . For a clockwise current (seen from $+z$ ), the tangent there points along $-\hat{j}$ , so $d\vec{l} = -dl\,\hat{j}$ and

$d\vec{F} = I\,d\vec{l}\times\vec{B} = I(-dl\,\hat{j})\times(B\hat{k}) = -IB\,dl\,(\hat{j}\times\hat{k}) = -IB\,dl\,\hat{i},$
which points along $-\hat{i}$ — i.e. radially inward, toward the centre of the loop. By the symmetry of the circle the same is true at every point: each element is pushed straight toward the centre. …
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CBSE 2024Set 55/1/11 markMCQ
$10\ \text{cm}$
›Reveal solutionSolution
The magnetic force on a current-carrying wire is given by $\vec{F} = I (\vec{L} \times \vec{B})$ . Here, the wire is along the $y$ -axis, so only the $z$ -component of $\vec{B}$ contributes, producing a force of $-(0.8\ \text{mN})\hat{i}$ . The correct option is (B).

The key idea is that a magnetic field exerts a force on a moving charge, and a current-carrying wire is just a collection of moving charges. The force on a straight wire of length $\vec{L}$ (a vector pointing in the direction of current) in a uniform magnetic field $\vec{B}$ is $\vec{F} = I (\vec{L} \times \vec{B})$ . This is a cross product, so only the component of $\vec{B}$ perpendicular to the wire matters.

Let’s work through it step by step.

Identify the vector length of the wire. The wire is $10\ \text{cm} = 0.10\ \text{m}$ long, lying along the $y$ -axis, with current in the positive $y$ -direction. So the length vector is:

$\vec{L} = (0.10\ \text{m})\,\hat{j}$

Write the magnetic field in SI units. The field is given as $\vec{B} = (5\ \text{mT})\hat{j} - (8\ \text{mT})\hat{k}$ . Since $1\ \text{mT} = 10^{-3}\ \text{T}$ , we have:

$\vec{B} = (5 \times 10^{-3})\,\hat{j} - (8 \times 10^{-3})\,\hat{k}\ \text{T}$

Apply the force formula. The current is $I = 1.0\ \text{A}$ . So:

$\vec{F} = I (\vec{L} \times \vec{B}) = (1.0) \left[ (0.10\,\hat{j}) \times \left(5\times10^{-3}\,\hat{j} - 8\times10^{-3}\,\hat{k}\right) \right]$
Compute the cross product term by term. Remember:

$\hat{j} \times \hat{j} = 0$ (parallel vectors give zero cross product)

$\hat{j} \times \hat{k} = \hat{i}$ (right-hand rule: $y$ cross $z$ gives $x$ )

So:
$0.10\,\hat{j} \times (5\times10^{-3}\,\hat{j}) = 0$ $0.10\,\hat{j} \times (-8\times10^{-3}\,\hat{k}) = (0.10)(-8\times10^{-3})\,(\hat{j} \times \hat{k}) = -8\times10^{-4}\,\hat{i}$
Therefore:
$\vec{L} \times \vec{B} = -8\times10^{-4}\,\hat{i}\ \text{T·m}$
Multiplying by $I = 1.0\ \text{A}$ :

$\vec{F} = -8\times10^{-4}\,\hat{i}\ \text{N}$ …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. The dimensional formula of the magnetic field intensity is :$
(a) ML^3 T^-2 A^-2
(b) ML^0 T^-2 A^-2
(c) ML^0 T^-2 A^-1
(d) Dimensionless
›Reveal solutionSolution

The magnetic field B has dimensions M L^0 T^-2 A^-1.

From the magnetic force F = q v B, we get B = F/(q v).

Force F = [M L T^-2]; charge q = [A T]; velocity v = [L T^-1].

q v = [A T][L T^-1] = [A L].

B = [M L T^-2] / [A L] = [M T^-2 A^-1] = M L^0 T^-2 A^-1.

…

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CBSE 2022Set ANNUAL1 markMCQ
$l$
(a) $\sqrt{2}\,\beta I l$
(b) $\sqrt{\dfrac{2}{3}}\,\beta I l$
(c) $\sqrt{\dfrac{1}{2}}\,\beta I l$
(d) $\sqrt{\dfrac{1}{3}}\,\beta I l$
›Reveal solutionSolution

Computing the cross product $\vec l\times\vec B$ for the wire along $\hat j$ and the given $\vec B$ vector, then taking its magnitude, gives $F=\sqrt{2/3}\,\beta Il$ .

Working

The Lorentz force on a current-carrying wire is $\vec F=I\vec l\times\vec B$ .

Here $\vec l=l\hat j$ (current along $Y$ ) and $\vec B=\dfrac{\beta}{\sqrt3}(\hat i+\hat j+\hat k)$ .

$\vec l\times\vec B=l\hat j\times\dfrac{\beta}{\sqrt3}(\hat i+\hat j+\hat k)=\dfrac{l\beta}{\sqrt3}\left[\hat j\times\hat i+\hat j\times\hat j+\hat j\times\hat k\right]$

Using $\hat j\times\hat i=-\hat k$ , $\hat j\times\hat j=0$ , $\hat j\times\hat k=\hat i$ :

…

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CBSE 2020Set 55/3/11 markMCQ
$\vec{B}$
›Reveal solutionSolution
Figure — CBSE 2020 55/3/1 Q10

The magnetic force on a current-carrying wire in a uniform field depends only on the vector from start to end of the wire, not its shape. For the isosceles right triangle, the force on QR equals the negative of the force on PQ, so the answer is $-F$ .

The key insight here is a beautiful simplification: in a uniform magnetic field, the net magnetic force on any current-carrying wire segment depends only on the vector displacement between its endpoints, not on the path the wire takes between them. This is because the force on a small element $d\vec{l}$ is $I\,d\vec{l} \times \vec{B}$ , and when $\vec{B}$ is constant, the integral $\int d\vec{l}$ over the wire is just the straight-line vector from start to end.

Let's apply this to the triangular loop.

Set up the geometry. The loop PQR is an isosceles right triangle with the right angle at P. So PR and PQ are the perpendicular legs, and QR is the hypotenuse. The uniform magnetic field $\vec{B}$ points along PR. Let's assign directions: take PR along the $+y$ axis, and PQ along the $+x$ axis. Then the current direction matters — we need to be consistent. The loop is closed, so current flows P → Q → R → P (or the reverse; the magnitude of force is unaffected by sign, but direction matters for comparing forces).

Force on arm PQ. Arm PQ is a straight wire of length $L$ (say) along the $x$ -axis. The current in PQ flows from P to Q, so $d\vec{l}$ is along $+\hat{x}$ . The magnetic field is $\vec{B} = B\,\hat{y}$ . The force on a straight wire is $\vec{F}_{PQ} = I\,(\vec{L}_{PQ} \times \vec{B})$ , where $\vec{L}_{PQ}$ is the vector from P to Q (length $L$ , direction $+\hat{x}$ ).

Compute: $\vec{L}_{PQ} \times \vec{B} = (L\hat{x}) \times (B\hat{y}) = LB\,(\hat{x} \times \hat{y}) = LB\,\hat{z}$ .

So $\vec{F}_{PQ} = I L B \,\hat{z}$ . The magnitude is $F = I L B$ , and it points out of the plane (say upward). The problem states this force is $F$ , so $F = I L B$ .

Force on arm QR. Arm QR is the hypotenuse. Its vector from Q to R: Q is at $(L,0)$ , R is at $(0,L)$ (since PR = PQ = L for an isosceles right triangle). So $\vec{L}_{QR} = \vec{R} - \vec{Q} = (0 - L)\hat{x} + (L - 0)\hat{y} = -L\hat{x} + L\hat{y}$ .

The force: $\vec{F}_{QR} = I\,(\vec{L}_{QR} \times \vec{B}) = I\,[(-L\hat{x} + L\hat{y}) \times (B\hat{y})]$ .

Compute the cross product term by term:

$(-L\hat{x}) \times (B\hat{y}) = -L B\,(\hat{x} \times \hat{y}) = -L B\,\hat{z}$ .

$(L\hat{y}) \times (B\hat{y}) = L B\,(\hat{y} \times \hat{y}) = 0$ . So $\vec{F}_{QR} = -I L B \,\hat{z} = -F\,\hat{z}$ . …
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CBSE 2019Set ANNUAL1 mark
$Q. Write formula for force on a current carrying conductor in a magnetic field.$
›Reveal solutionSolution

A conductor of length L carrying current I in a magnetic field B feels a force F = I(L x B), of magnitude BIL sin(theta).

Each charge carrier drifting in the wire experiences a magnetic force. Summing over all carriers in a length L gives the force on the whole conductor:

F = I (L x B)

where L points along the direction of conventional current and has magnitude equal to the length of the conductor.

Magnitude: F = B I L sin(theta), theta = angle between L and B. …

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Q.Derive an expression for the magnetic force F⃗\vec{F}F acting on a straight conductor of length LLL carrying current III in an external magnetic field B⃗\vec{B}B. Is it valid when the conductor is in zig-zag form? Justify.

Derivation of Magnetic Force on a Straight Conductor

Validity for a Zig-zag Conductor

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Q.Derive an expression for the magnetic force $\vec{F}$ acting on a straight conductor of length $L$ carrying current $I$ in an external magnetic field $\vec{B}$ . Is it valid when the conductor is in zig-zag form? Justify.

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