Consider a circular current-carrying loop of radius R in the x-y plane with centre at origin. Consider the lin

NCERT Exemplar · Q25

R

(a) Show that

\Im(L)

monotonically increases with

L

(b) Use an appropriate Amperian loop to show that

\Im(\infty) = \mu_0 I

, where

I

is the current in the wire.

(d) Suppose we replace the circular coil by a square coil of sides

R

carrying the same current

I

. What can you say about

\Im(L)

and

\Im(\infty)

Meghalaya MboseSubjective· 5mImportance★★★★★est

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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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The line integral $\Im(L)$ of the magnetic field along the $z$ -axis for a circular loop increases monotonically with $L$ because the field is always positive and decays. Using a large semicircular Amperian loop, $\Im(\infty) = \mu_0 I$ follows from Ampère's law. For a square coil, $\Im(\infty)$ remains $\mu_0 I$ , but $\Im(L)$ differs in shape.

Why Ampère's Circuital Law?

The problem asks about the line integral of $\vec{B}$ along the $z$ -axis, not the field itself. The key insight is that $\vec{B}$ on the axis of a circular loop points purely along $z$ (by symmetry), so $\vec{B}\cdot d\vec{l} = B_z \, dz$ . The integral $\Im(L)$ is just the accumulated area under $B_z(z)$ from $-L$ to $L$ . Since $B_z(z)$ is an even, positive function that decays to zero at infinity, the integral grows as $L$ increases -- that's the monotonicity.

For part (b), the trick is to connect this line integral to Ampère's law. The $z$ -axis is a straight line, but Ampère's law applies to closed loops. So we close the path with a large semicircle at infinity, where $\vec{B}$ vanishes, making the closed loop integral equal to $\Im(\infty)$ . Then Ampère's law gives $\mu_0 I$ times the number of times the loop encloses the current.

Step-by-step solution

1. Magnetic field on the axis of a circular loop

For a circular loop of radius $R$ carrying current $I$ , the field at a point $(0,0,z)$ on the axis is:

\vec{B}(z) = \frac{\mu_0 I R^2}{2(R^2 + z^2)^{3/2}} \, \hat{z}

This is a standard result derived from the Biot–Savart law. The field is purely along $z$ and symmetric about $z=0$ .

Note

The field is maximum at $z=0$ ( $B_0 = \mu_0 I / (2R)$ ) and falls off as $1/|z|^3$ for large $z$ .

2. Express $\Im(L)$ explicitly

Since $d\vec{l} = dz \, \hat{z}$ along the $z$ -axis:

\Im(L) = \left| \int_{-L}^{L} B_z(z) \, dz \right| = \int_{-L}^{L} \frac{\mu_0 I R^2}{2(R^2 + z^2)^{3/2}} \, dz

The absolute value is unnecessary because $B_z > 0$ everywhere. The integrand is even, so:

\Im(L) = \mu_0 I R^2 \int_{0}^{L} \frac{dz}{(R^2 + z^2)^{3/2}}

3. Evaluate the integral

Let $z = R \tan\theta$ , so $dz = R \sec^2\theta \, d\theta$ and $R^2 + z^2 = R^2 \sec^2\theta$ . Then:

\int \frac{dz}{(R^2 + z^2)^{3/2}} = \int \frac{R \sec^2\theta \, d\theta}{R^3 \sec^3\theta} = \frac{1}{R^2} \int \cos\theta \, d\theta = \frac{\sin\theta}{R^2}

Since $\sin\theta = z / \sqrt{R^2 + z^2}$ , we get:

\Im(L) = \mu_0 I R^2 \left[ \frac{z}{R^2 \sqrt{R^2 + z^2}} \right]_{0}^{L} = \mu_0 I \, \frac{L}{\sqrt{R^2 + L^2}}

\Im(L) = \mu_0 I \, \frac{L}{\sqrt{R^2 + L^2}}

4. Show monotonic increase (part a)

The function $f(L) = L / \sqrt{R^2 + L^2}$ has derivative:

f'(L) = \frac{\sqrt{R^2 + L^2} - L \cdot \frac{L}{\sqrt{R^2 + L^2}}}{R^2 + L^2} = \frac{R^2}{(R^2 + L^2)^{3/2}} > 0

So $f(L)$ is strictly increasing for all $L > 0$ . Since $\Im(L) = \mu_0 I \, f(L)$ , it too increases monotonically with $L$ .

Watch out

A common mistake is to think $\Im(L)$ increases because $B_z$ is positive. That's necessary but not sufficient -- you need the integral to grow, which requires the integrand to not decay too fast. Here it does grow because $B_z \sim 1/z^3$ integrates to a finite limit.

5. Use an Amperian loop to find $\Im(\infty)$ (part b)

Consider a closed Amperian loop consisting of:

The segment along the $z$ -axis from $-L$ to $L$
A large semicircle of radius $L$ in the $x$ - $z$ plane (or any plane containing the $z$ -axis) that closes the path

As $L \to \infty$ , the semicircle goes to infinity where $B \to 0$ , so its contribution to the line integral vanishes. The closed loop integral then equals $\Im(\infty)$ . …

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Method: Evaluating an Open Line Integral of $\vec B$ by Closing the Amperian Loop

This method applies whenever you're asked for the line integral of $\vec B$ along an OPEN path (like a single axis, not a full loop), and you need to connect it to the enclosed current via Ampère's Circuital Law, which strictly applies only to CLOSED loops.

Steps

Step 1: Note what Ampère's law actually requires

\oint \vec B\cdot d\vec l=\mu_0 I_{\text{enc}}

holds only for a closed loop. An open path's line integral cannot be found from Ampère's law directly — it must first be turned into (or related to) a closed loop.

Step 2: Close the loop with a return path where $\vec B$ is known or vanishes

Extend the open path with an additional segment that brings it back to its start — chosen so its contribution to the line integral is either zero (e.g. an arc taken out to infinity, where $B\to0$ for a localized current distribution) or otherwise computable. This turns the open-path integral into (part of) a genuine closed-loop integral.

Step 3: Apply Ampère's law to the now-closed loop

\oint \vec B\cdot d\vec l=\mu_0 I_{\text{enc}}

where $I_{\text{enc}}$ is the current threading the loop, counted with sign, according to how many times (and in which sense) the loop links the current-carrying circuit.

Step 4: Subtract off the known return-path contribution

Since the return path's contribution was zero (or known) by construction in Step 2, the original open-path integral equals the full closed-loop result from Step 3 (minus that known piece).

Step 5: Cross-check with a direct integration when possible

If the explicit field expression along the path is known (e.g. from the Biot–Savart law), integrate it directly and confirm it agrees with the Ampère's-law shortcut — this is both a check and, for questions that ask for the integral at a FINITE (not infinite) distance, often the only way to get that finite- $L$ value, since Ampère's law by itself only fixes the infinite-limit result. …

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Q.Consider a circular current-carrying loop of radius RRR in the xxx-yyy plane with centre at origin. Consider the line integral ℑ(L)=∣∫−LLB⃗⋅dl⃗∣\Im(L) = \left|\int_{-L}^{L} \vec{B}\cdot d\vec{l}\right|ℑ(L)=​∫−LL​B⋅dl​ taken along the zzz-axis.

Why Ampère's Circuital Law?

Step-by-step solution

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Q.Consider a circular current-carrying loop of radius $R$ in the $x$ - $y$ plane with centre at origin. Consider the line integral $\Im(L) = \left|\int_{-L}^{L} \vec{B}\cdot d\vec{l}\right|$ taken along the $z$ -axis.

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