(a) Write an expression of magnetic moment associated with a current (I) carrying circular coil of radius r ha

Question

I

(b) Consider the above mentioned coil placed in YZ plane with its centre at the origin. Derive expression for the value of magnetic field due to it at point

(x, 0, 0)

(OR)

(a) Define current sensitivity of a galvanometer. Write its expression.

(b) A galvanometer has resistance

G

and shows full scale deflection for current

I_g

(i) How can it be converted into an ammeter to measure current up to

I_0

(

I_0 > I_g

) ?

(ii) What is the effective resistance of this ammeter ?

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Part (a)Concept understanding — Magnetic Field on the Axis of a Loop

Magnetic Field on the Axis of a Current Loop

Imagine a circular wire carrying a steady current. You want to know the magnetic field not at the centre, but at some point along the line that passes through the centre and is perpendicular to the plane of the loop — that's the axis.

Why would the field be along the axis at all? Because of symmetry. For every tiny segment of the loop, there is an opposite segment on the other side. Their perpendicular components of the magnetic field cancel out, leaving only the component along the axis. So the net field points straight along the axis, either towards or away from the loop depending on the current direction.

The Intuition

At the centre of the loop ( $x = 0$ ), every segment is at the same distance $R$ from the centre, and the field is strongest. As you move away along the axis, two things happen: the distance from each current element to your observation point increases, and the angle at which the field points along the axis becomes less favourable. So the field drops off.

Far away from the loop, the loop looks like a tiny magnetic dipole — a small bar magnet. The field falls off as $1/x^3$ , exactly like a dipole field.

The Precise Statement

For a circular loop of radius $R$ , carrying a steady current $I$ , the magnitude of the magnetic field at a point on the axis at a distance $x$ from the centre is:

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

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

The direction of $\mathbf{B}$ is along the axis, given by the right-hand rule: curl the fingers of your right hand in the direction of the current, and your thumb points in the direction of the magnetic field on the axis.

Special Cases

At the centre ( $x = 0$ ):

B_{\text{centre}} = \frac{\mu_0 I}{2R}

Far away ( $x \gg R$ ):

The denominator $(R^2 + x^2)^{3/2} \approx x^3$ , so

B \approx \frac{\mu_0 I R^2}{2 x^3}

This is exactly the field of a magnetic dipole of moment $m = I \cdot (\pi R^2) = I A$ , where $A$ is the area of the loop. So a current loop behaves like a magnetic dipole at large distances.

Watch out

The formula $B = \mu_0 I R^2 / 2(R^2 + x^2)^{3/2}$ is valid only on the axis. Off-axis, the field is much more complicated and cannot be written in such a simple closed form.

Why the $3/2$ Power? …

Part (b)Concept understanding — Galvanometer to Ammeter Conversion (Shunt)

Galvanometer to Ammeter Conversion (Shunt)

A moving-coil galvanometer carries only a tiny full-scale current $I_g$ (and has resistance $G$ ), so on its own it can measure at most $I_g$ . To read a much larger current $I$ , we connect a small resistance $S$ (the shunt) in parallel with the galvanometer. The shunt diverts most of the current, letting only $I_g$ pass through the coil.

Key idea: the galvanometer and shunt are in parallel, so they share the same potential difference:

I_g\,G = (I - I_g)\,S \quad\Longrightarrow\quad S = \frac{I_g}{I - I_g}\,G

Because $I \gg I_g$ , the shunt $S$ is very small, and the combined resistance of the ammeter is even smaller than $S$ — ideal, since an ammeter must not disturb the circuit it measures. …

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Part (a): $\vec{m}=NI\pi r^2\,\hat{n}$ and the axial field of the coil is $B=\dfrac{\mu_0 NI r^2}{2(r^2+x^2)^{3/2}}$ .

Part (b): current sensitivity $S_I=\dfrac{NBA}{k}$ ; a shunt $S=\dfrac{I_g G}{I_0-I_g}$ converts the galvanometer to an ammeter of effective resistance $R_{\text{eff}}=\dfrac{I_g G}{I_0}$ .

Part (a): Magnetic Moment and Axial Field of a Circular Coil

(i) Magnetic moment. The magnetic moment measures the strength and orientation of a current loop. For one turn it is $I\times(\text{area})$ ; with $N$ turns:

\vec{m} = N I A\,\hat{n} = N I (\pi r^2)\,\hat{n}

where $\hat{n}$ is the unit normal to the coil (curl fingers along the current, thumb gives $\vec{m}$ ).

(ii) Field at $(x,0,0)$ . The coil is in the YZ-plane, centre at the origin, so its axis is the x-axis. Take a current element $d\vec{l}$ on the loop; the vector from it to the axial point has constant magnitude $R=\sqrt{r^2+x^2}$ .

Biot–Savart law: $d\vec{B}=\dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec{l}\times\vec{R}}{R^3}$ .
By symmetry the components perpendicular to the axis cancel when integrated around the loop; only $dB_x$ survives, and $(d\vec{l}\times\vec{R})_x = r\,dl$ (the radial component of $\vec R$ is $r$ ).
Integrating $\oint dl = 2\pi r$ and including $N$ turns: $B_x = \frac{\mu_0 I}{4\pi}\,\frac{r}{R^3}\,(2\pi r)\,N = \frac{\mu_0 N I r^2}{2\,(r^2+x^2)^{3/2}}$ …

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

CBSE 2026Set 55/2/11 markMCQ
$27\ \Omega$
›Reveal solutionSolution
A galvanometer is converted to an ammeter by connecting a small shunt resistor in parallel. Using the current division rule, the full-scale deflection current of the galvanometer is found to be $1\ \text{mA}$ , which corresponds to option (C).

When a galvanometer is converted into an ammeter, a small resistance (shunt) is connected in parallel with it. The purpose is to allow most of the current to bypass the delicate galvanometer coil, so only a small fraction passes through the meter itself. The galvanometer shows full-scale deflection when the current through its coil reaches its maximum rated value, say $I_g$ . The shunt carries the remaining current.

Here, the galvanometer resistance is $G = 27\ \Omega$ , the shunt resistance is $S = 3\ \Omega$ , and the ammeter range is $0$ to $10\ \text{mA}$ — meaning the total current that produces full-scale deflection in the ammeter is $I = 10\ \text{mA}$ .

Let’s work through the reasoning step by step.

Understand the parallel connection In an ammeter, the galvanometer and shunt are in parallel. So the voltage across both is the same. If $I_g$ is the current through the galvanometer at full deflection, and $I_s$ is the current through the shunt, then:

$I_g \cdot G = I_s \cdot S$
Also, the total current entering the ammeter is:
$I = I_g + I_s$

Express $I_s$ in terms of $I_g$ From the voltage equality:

$I_s = I_g \cdot \frac{G}{S}$
Substitute into the total current equation:
$I = I_g + I_g \cdot \frac{G}{S} = I_g \left(1 + \frac{G}{S}\right)$

Plug in the given values $G = 27\ \Omega$ , $S = 3\ \Omega$ , $I = 10\ \text{mA}$ :

$10 = I_g \left(1 + \frac{27}{3}\right) = I_g \left(1 + 9\right) = I_g \cdot 10$
Therefore:

$I_g = \frac{10}{10} = 1\ \text{mA}$ …
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CBSE 2025Set 55/6/11 markMCQ
$Q. A galvanometer can be converted into an ammeter of desired range by connecting a: (A) small resistance in series (B) large resistance in series (C) small resistance in parallel (D) large resistance in parallel$
›Reveal solutionSolution
To convert a galvanometer into an voltmeter, a large resistance is connected in series with it. For an ammeter, a small resistance is connected in parallel. The question asks for ammeter conversion, so the correct choice is (C) small resistance in parallel.

The key idea is that a galvanometer is a sensitive current-measuring device that deflects fully for a small current (its full-scale deflection current, $I_g$ ). To measure larger currents (as an ammeter does), we need to bypass most of the current around the galvanometer coil, protecting it from burning out. This is done by connecting a shunt — a small resistance — in parallel.

Why parallel? Because a parallel path divides the current. The galvanometer still sees only $I_g$ at full deflection, while the shunt carries the excess current $(I - I_g)$ . The shunt resistance $S$ is chosen so that at the desired maximum current $I$ , exactly $I_g$ flows through the galvanometer. Since the voltage across parallel branches is equal:
$I_g \cdot G = (I - I_g) \cdot S$
where $G$ is the galvanometer resistance. Solving:
$S = \frac{I_g G}{I - I_g}$
For a large range ( $I \gg I_g$ ), $S$ becomes very small — hence a small resistance in parallel.

Watch out

A common mistake is confusing ammeter and voltmeter conversion. For a voltmeter, you add a large series resistance to limit voltage. For an ammeter, you add a small parallel resistance to shunt current. Mixing them up leads to wrong answers.

Now, let's work through the reasoning step by step:

Understand the galvanometer's limitation: A galvanometer is essentially a sensitive moving-coil meter with resistance $G$ (typically 10–100 $\Omega$ ) and full-scale deflection current $I_g$ (often a few mA). It cannot handle large currents directly — passing a large current through it would permanently damage the coil.

Goal of an ammeter: An ammeter must measure a wide range of currents (say 0–1 A or more) while offering very low resistance to the circuit, so it doesn't disturb the current being measured. The galvanometer alone has too high a resistance and too low a current capacity.

Why parallel (shunt) works: Connecting a small resistance $S$ in parallel creates a current divider. At full-scale deflection, the total current $I$ entering the ammeter splits: $I_g$ through the galvanometer and $(I - I_g)$ through the shunt. The shunt "steals" the excess current. The parallel combination also reduces the overall ammeter resistance to $\frac{G S}{G+S}$ , which is very small — ideal for an ammeter.

Derive the shunt value: Using the voltage equality across parallel branches: …
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CBSE 2025Set D1 markMCQ
$Q. When an ammeter is shunted then its measurement limit (A) increases (B) decreases (C) remains unchanged (D) none of these$
›Reveal solutionSolution

A shunt is a small resistance placed in parallel with the ammeter; it bypasses most of the current so the meter reads only a fraction, extending (increasing) its range.

An ammeter (galvanometer) can carry only a small full-scale current $I_g$ . To measure a larger current I, a low resistance shunt S is connected in parallel:

$S = \frac{I_g}{I - I_g}\,G$

…

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CBSE 2025Set ANNUAL1 mark
$Q. To convert a galvanometer into an ammeter ____________ is connected in parallel to it.$
›Reveal solutionSolution

A galvanometer becomes an ammeter by adding a low-resistance shunt in parallel, so most of the current bypasses the sensitive galvanometer coil.

A galvanometer is a sensitive device with a resistance $G$ that can only safely carry a small current $I_g$ for full-scale deflection. To measure large currents, a small resistance called a shunt ( $S$ ) is connected in parallel with the galvanometer. Most of the current flows through the low-resistance shunt, and only the small fraction $I_g$ flows through the galvanom …

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CBSE 2024Set 55/1/11 markMCQ
$G$
›Reveal solutionSolution

When a galvanometer is converted to an ammeter using a shunt, only 0.1% of the total current flows through the galvanometer coil. Using the parallel-resistance formula and the current-division condition, the net resistance of the ammeter is $\boxed{\frac{G}{1000}}$ .

Why a shunt converts a galvanometer into an ammeter

A galvanometer is a sensitive current-measuring device with high resistance $G$ that can only handle a small current $I_g$ before its coil deflects fully. To measure larger currents, we place a low-resistance shunt $S$ in parallel with the galvanometer. Most of the current bypasses the galvanometer through this shunt, while a small fraction flows through the coil to produce the deflection.

The ammeter's effective resistance is the parallel combination of $G$ and $S$ , which must be very small so that inserting the ammeter into a circuit doesn't significantly alter the current being measured.

Step-by-step solution

1. Identify what flows through the galvanometer

The problem states that when the ammeter reads its full-scale value $I$ , the current through the galvanometer is $0.1\%$ of $I$ :
$I_g = 0.001 \, I = \frac{I}{1000}$
2. Find the current through the shunt

Since the galvanometer and shunt are in parallel, the total current splits between them:
$I_s = I - I_g = I - \frac{I}{1000} = \frac{999I}{1000}$
3. Apply the voltage-equality condition

Both the galvanometer and shunt have the same potential difference across them (parallel connection). Using Ohm's law:
$V_g = V_s$ $I_g \cdot G = I_s \cdot S$
Substituting the currents:
$\frac{I}{1000} \cdot G = \frac{999I}{1000} \cdot S$
Simplifying:
$G = 999 \, S$ $S = \frac{G}{999}$
4. Calculate the ammeter's net resistance …
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CBSE 2024Set A1 markMCQ
$Q. A galvanometer is converted into ammeter by adding (A) low resistance in parallel (B) high resistance in series (C) low resistance in series (D) high resistance in parallel$
›Reveal solutionSolution

An ammeter = galvanometer + a small shunt resistance in parallel.

An ammeter must (i) read large currents and (ii) have very low resistance so it does not disturb the circuit. A galvanometer is a sensitive, high-resistance device that can carry only a tiny current.

To convert it, a low resistance (shunt) S is connected in parallel with the galvanometer. Most of the current then passes through the shunt and only a small fixed fraction through the coil:

$S = \frac{I_g}{I - I_g}\,G.$

…

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CBSE 2023Set F1 markMCQ
$Q. If any ammeter is shunted, then the total resistance of the circuit (A) increases (B) decreases (C) remains same (D) none of these$
›Reveal solutionSolution

A shunt is a small resistance in parallel with the ammeter, so the net resistance decreases.

An ammeter is shunted by connecting a low-value resistance (the shunt) in parallel with it, so that most of the current bypasses the meter coil. Two resistances in parallel give an equivalent resistance smaller than either one:

…

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CBSE 2023Set B1 mark
$Q. Fill in the blank: The ______ of the galvanometer is reduced by the use of shunt.$
›Reveal solutionSolution

A shunt is a small resistance connected in parallel with the galvanometer coil to reduce its effective resistance (and to bypass most of the current), converting it into an ammeter.

A galvanometer coil itself has a certain internal resistance $G$ . When a low-value resistance $S$ (the shunt) is connected in parallel with it, the combination's effective (equivalent) resistance $R_{eq} = \dfrac{GS}{G+S}$ is always smaller than either $G$ or $S$ alone. This lets most of the current bypass the sensitive coil through the shunt (protecting the galvanometer and allowing it to measure la …

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CBSE 2020Set OC1 markMCQ
$Q. To convert a given galvanometer into an ammeter of desired range$
(a) low resistance is connected in series
(b) low resistance is connected in parallel
(c) high resistance is connected in series
(d) high resistance is connected in parallel
›Reveal solutionSolution

An ammeter needs near-zero resistance so it barely disturbs the circuit; shunting the galvanometer with a low parallel resistance diverts most of the current around the (relatively high-resistance) coil.

Reasoning

To convert a galvanometer (which has a small full-scale current capacity and non-negligible resistance) into an ammeter of a larger desired range, a low resistance (called a shunt) is connected in parallel with the galvanometer. This shunt diverts most of the current around the galvanometer coil, letting only a small, fixed fraction pass through the coil itself, so the combination can measure mu …

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CBSE 2018Set ANNUAL1 markMCQ
$Q. A circular coil of radius R carries a current I. The magnetic field at its center is B. At what distance from the center, on the axis of the coil, the magnetic field will be B/8 :-$
(a) √2R
(b) 2R
(c) √3R
(d) 3R
›Reveal solutionSolution

$\dfrac{B_{axis}}{B_{centre}} = \dfrac{R^3}{(R^2+x^2)^{3/2}} = \dfrac18 \Rightarrow x = \sqrt3\,R$ .

Field at centre: $B = \dfrac{\mu_0 I}{2R}$ .

Field on axis at distance x: $B_x = \dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$ .

Take the ratio:

$\dfrac{B_x}{B} = \dfrac{R^3}{(R^2+x^2)^{3/2}} = \dfrac{1}{8}.$

…

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Q.(a) Write an expression of magnetic moment associated with a current (III) carrying circular coil of radius rrr having NNN turns.

Part (a): Magnetic Moment and Axial Field of a Circular Coil

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Q.(a) Write an expression of magnetic moment associated with a current ( $I$ ) carrying circular coil of radius $r$ having $N$ turns.

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