A charged particle would continue to move with a constant velocity in a region wherein, (a) E = 0,B ≠ 0. (b) E

NCERT Exemplar · Q10

Q. A charged particle would continue to move with a constant velocity in a region wherein,

(a)

E = 0,\ B \neq 0

(b)

E \neq 0,\ B \neq 0

(c)

E \neq 0,\ B = 0

(d)

E = 0,\ B = 0

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Concept understanding — Magnetic Force Balance

Magnetic Force Balance

When a current-carrying wire or coil sits in a magnetic field, it feels a force $F = BIL\sin\theta$ (or, for a point charge, $F=qvB\sin\theta$ ). On its own that force just pushes the conductor - but in many real situations the push is deliberately set up to CANCEL another force, so the whole system sits in equilibrium. That equilibrium condition - magnetic force balanced against weight, against another wire's magnetic force, or against a mechanical counterweight - is what "magnetic force balance" means, and it is also historically how the ampere itself was defined.

The balance condition

Whenever a conductor is in equilibrium under a magnetic force and one other force, the two must be equal and opposite:

BIL\sin\theta = F_{\text{other}}

Solving this equation for whichever quantity is unknown ( $B$ , $I$ , $L$ , or the other force) is the entire skill in this class of problem - the only new step, beyond the force law itself, is correctly identifying what the magnetic force is opposing.

Case 1: a wire suspended against gravity

A straight horizontal wire of mass $m$ and length $l$ , carrying current $I$ , can be held up ("floated") in mid-air by a horizontal magnetic field perpendicular to it. The upward magnetic force must equal the downward weight:

BIl = mg \quad\Longrightarrow\quad B=\frac{mg}{Il}

For example, a $200\,\text{g}$ , $1.5\,\text{m}$ wire carrying $2\,\text{A}$ needs $B=\dfrac{(0.2)(9.8)}{(2)(1.5)}\approx0.65\,\text{T}$ to stay suspended.

Case 2: two wires balancing each other

Two long parallel wires carrying currents $I_1,I_2$ exert a force per unit length on each other of $\dfrac{\mu_0 I_1I_2}{2\pi d}$ (attractive if the currents run the same way, repulsive if opposite). If one wire is free to move, this magnetic force can itself balance that wire's weight:

\frac{\mu_0 I^2 L}{2\pi h}=mg \quad\Longrightarrow\quad h=\frac{\mu_0 I^2 L}{2\pi mg}

This is exactly how a "current balance" apparatus works, and historically it is how the ampere was defined: the current that, flowing in two infinite parallel wires one metre apart, produces a force of exactly $2\times10^{-7}\,\text{N}$ per metre of length.

Case 3: balancing on a beam

A current-carrying coil arm hanging from one pan of a beam balance feels an extra force $F=NBIl$ when only that arm sits in an external field. Re-balancing the beam means adding a mass $m$ so that $mg = NBIl$ .

Watch out

Always check which length enters the formula - for a coil of $N$ turns the force multiplies by $N$ ; for a single suspended straight wire it doesn't. …

Why this formula?

Magnetic Force Balance: Why the Key Formulas Hold

The Magnetic Force Balance describes when the magnetic force on a charged particle or current-carrying conductor is exactly balanced by another force (gravity, electric force, or tension). Let's build the reasoning step-by-step.

1. The Core Idea: What Does "Balance" Mean?

A force balance means the net force on an object is zero:

\vec{F}_{\text{net}} = 0

For magnetic forces we use the Lorentz force law:

On a moving charge: $\vec{F}_m = q (\vec{v} \times \vec{B})$
On a current-carrying wire: $\vec{F}_m = I (\vec{L} \times \vec{B})$

When this is balanced by another force (say gravity $\vec{F}_g = m\vec{g}$ ):

\vec{F}_m + \vec{F}_{\text{other}} = 0

2. Case 1: Charged Particle in Crossed Fields (Velocity Selector)

A charged particle moves perpendicular to both electric field $\vec{E}$ and magnetic field $\vec{B}$ .

Electric force: $\vec{F}_e = q\vec{E}$ (along $\vec{E}$ )
Magnetic force: $\vec{F}_m = q(\vec{v} \times \vec{B})$ (perpendicular to both $\vec{v}$ and $\vec{B}$ )

For straight-line motion (no deflection), the two forces must cancel:

qE = qvB \quad \Rightarrow \quad \boxed{v = \frac{E}{B}}

Key insight: Only particles with this exact speed pass undeflected — this is how velocity selectors work in mass spectrometers.

3. Case 2: Current-Carrying Wire Balanced by Gravity

A horizontal wire carrying current $I$ sits in a perpendicular magnetic field $\vec{B}$ , suspended by strings.

The magnetic force on a straight wire is $F_m = I L B \sin\theta$ ; for a wire perpendicular to the field ( $\theta = 90^\circ$ ), $F_m = ILB$ . Setting this equal to the weight $F_g = mg$ for equilibrium:

\boxed{I L B = mg}

Key insight: This balance lets you measure $B$ if $I$ , $L$ , and $m$ are known — the principle behind a current balance experiment.

4. Case 3: Circular Motion of a Charged Particle …

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Constant velocity needs zero net Lorentz force, $q(\vec{E}+\vec{v}\times\vec{B})=0$ . That is possible for (a) $E=0,B\neq0$ with $\vec{v}\parallel\vec{B}$ ; for (b) $E\neq0,B\neq0$ with $\vec{E}=-\vec{v}\times\vec{B}$ ; and for (d) $E=0,B=0$ trivially. It is impossible for (c) $E\neq0,B=0$ .

Condition for constant velocity

Constant velocity means zero acceleration, hence zero net force. The only force acting is the Lorentz force, so

q(\vec{E}+\vec{v}\times\vec{B})=0\quad\Longrightarrow\quad \vec{E}+\vec{v}\times\vec{B}=0.

(a) $E=0,\ B\neq0$ . The condition reduces to $\vec{v}\times\vec{B}=0$ , satisfied when $\vec{v}$ is parallel (or anti-parallel) to $\vec{B}$ . The magnetic force is then zero and the velocity stays unchanged. Possible.

(b) $E\neq0,\ B\neq0$ . Possible when the electric and magnetic forces exactly cancel: $\vec{E}=-\vec{v}\times\vec{B}$ (with $\vec{E}\perp\vec{B}$ and speed $v=E/B$ ). This is precisely the working principle of a velocity selector. Possible. …

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

AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
$20\pi$
›Reveal solutionSolution
This tests superposition of the magnetic field of a circular coil (at its centre) and of a long straight wire, set to cancel — solve for $N$ by equating magnitudes.

Concept and Intuition

A circular coil of $N$ turns carrying current $I$ produces a field at its own centre of $B=\dfrac{\mu_0 N I}{2r}$ , directed along the coil's axis (direction fixed by the right-hand rule for the shown current sense). A long straight wire carrying current $I$ produces, at perpendicular distance $d$ , a field $B=\dfrac{\mu_0 I}{2\pi d}$ , circling the wire (again right-hand rule). The problem is engineered so that, given the current directions in the figure, these two fields point in opposite directions at the coil's centre. "Resultant field is zero" therefore just means the two magnitudes are equal — the geometry/direction part is already built into the problem statement, so we only need magnitude balance.

Step-by-Step Solution

Field due to the coil at its centre: $B_{coil}=\dfrac{\mu_0 N I_{coil}}{2r}=\dfrac{\mu_0 N (2)}{2(0.1\,\text{m})}=10\mu_0 N$ . …
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AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$0.12\ \mathrm{kg\,m^{-1}}$
›Reveal solutionSolution
A current-carrying wire floats in a horizontal magnetic field when the magnetic force exactly cancels gravity; solving gives $I = 2.4\ \text{A}$ .

Concept and Intuition

A straight wire carrying current $I$ in a magnetic field $B$ (perpendicular to the wire) experiences a force per unit length $F/L = BI$ . For the wire to be suspended in mid-air (in equilibrium), this magnetic force must balance the weight per unit length of the wire, which is $\lambda g$ where $\lambda$ is the linear mass density.

Step-by-Step Solution

Force balance per unit length: $BI = \lambda g$ .

Solve for current: $I = \dfrac{\lambda g}{B}$ .

Substitute values: $\lambda = 0.12\ \text{kg m}^{-1}$ , $g = 10\ \text{m s}^{-2}$ , $B = 0.5\ \text{T}$ . …
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AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
$\vec{B}$
›Reveal solutionSolution
Two parallel wires symmetric about the x-axis add their fields at the origin constructively along one direction; the resultant is $\sqrt2$ times the field of either wire alone.

Concept and Intuition

An infinite straight wire carrying current $I$ produces a field of magnitude $\mu_0 I/(2\pi d)$ at perpendicular distance $d$ , circling the wire (direction given by $\hat z\times \hat r$ , where $\hat r$ points from the wire towards the field point, for current along $+\hat z$ ). With two wires we must add the two field vectors, not just their magnitudes.

Step-by-Step Solution

Wire 1 is at $(1,1)$ cm, wire 2 at $(1,-1)$ cm; both distances from the origin are $d=\sqrt{1^2+1^2}=\sqrt2$ cm, so each alone gives a field of magnitude $B_0=\dfrac{\mu_0 I}{2\pi\sqrt2}$ .

Vector from wire 1 to origin: $\vec r_1=(-1,-1)$ . Field direction $\propto \hat z\times \vec r_1 = (1,-1,0)$ (up to normalization).

Vector from wire 2 to origin: $\vec r_2=(-1,1)$ . Field direction $\propto \hat z\times \vec r_2 = (-1,-1,0)$ . …
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AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
$Q. Two long parallel straight metal wires A and B carrying currents 12 A and 36 A respectively, in the same direction are separated by 50 cm. The point relative to A, where the resultant magnetic induction between the two wires due to the currents is zero, will be (A) 90 cm (B) 7.5 cm (C) 28 cm (D) 12.5 cm$
›Reveal solutionSolution
With both currents in the same direction, the magnetic fields cancel only in the region between the two wires. Equating $B_A=B_B$ and solving gives the null point at 12.5 cm from wire A (closer to the weaker current, as expected).

Concept and Intuition

Each long straight wire produces a field $B=\dfrac{\mu_0 I}{2\pi d}$ circling around it. Between two wires carrying current in the same direction, the two fields point in opposite directions (one wire's field goes into the page there, the other's comes out), so they can cancel at some point between them. The cancellation point sits closer to the wire with the smaller current (since a weaker source needs to be closer to match the stronger one's field at the same magnitude).

Step-by-Step Solution

Let the null point be at distance $x$ from wire A, so it is at $(50-x)$ cm from wire B.

Equate the magnitudes: $\dfrac{\mu_0 (12)}{2\pi x} = \dfrac{\mu_0(36)}{2\pi(50-x)}$ .

Cancel common factors: $\dfrac{12}{x} = \dfrac{36}{50-x}$ . …
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AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
$cm$
›Reveal solutionSolution
At a 45° deflection the tangent law gives $\tan45°=1$ , so the coil's magnetic field exactly equals Earth's horizontal field, letting us solve directly for the current: $I=\dfrac{3}{5\pi}\,A$ .

Concept and Intuition

A tangent galvanometer balances the magnetic field of its coil against Earth's horizontal field $H$ ; the needle's deflection $\theta$ obeys $\tan\theta=\dfrac{B_{coil}}{H}$ , where $B_{coil}=\dfrac{\mu_0 nI}{2r}$ .

Step-by-Step Solution

Tangent law: $B_{coil}=H\tan\theta$ . At $\theta=45°$ , $\tan45°=1$ , so $B_{coil}=H$ .

$B_{coil}=\dfrac{\mu_0 nI}{2r}$ , so $\dfrac{\mu_0 nI}{2r}=H\Rightarrow I=\dfrac{2rH}{\mu_0 n}$ .

Substitute $r=0.2\,m$ , $H=3\times10^{-5}\,T$ , $n=50$ , $\mu_0=4\pi\times10^{-7}$ : …
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Q.A charged particle would continue to move with a constant velocity in a region wherein,

Concept understanding — Magnetic Force Balance

Magnetic Force Balance

The balance condition

Case 1: a wire suspended against gravity

Case 2: two wires balancing each other

Case 3: balancing on a beam

Why this formula?

Magnetic Force Balance: Why the Key Formulas Hold

1. The Core Idea: What Does "Balance" Mean?

2. Case 1: Charged Particle in Crossed Fields (Velocity Selector)

3. Case 2: Current-Carrying Wire Balanced by Gravity

4. Case 3: Circular Motion of a Charged Particle …

Condition for constant velocity

Method: Testing Whether a Charged Particle Can Move With Constant Velocity

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

Concept and Intuition

Step-by-Step Solution

Q.A charged particle would continue to move with a constant velocity in a region wherein,

Condition for constant velocity

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