(a) (i) Write the principle and explain the working of a moving coil galvanometer. A galvanometer as such cann

Question

Q. (a)

(i) Write the principle and explain the working of a moving coil galvanometer. A galvanometer as such cannot be used to measure the current in a circuit. Why ?

(ii) Why is the magnetic field made radial in a moving coil galvanometer ? How is it achieved ?

(OR)

(b)

(i) Derive an expression for the magnetic field on the axis of a current-carrying circular loop.

(ii) Write any two points of difference between a diamagnetic and a paramagnetic substance.

CBSECBSE Class XII Board 2023Subjective· 5mImportance★★★★★

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Part (a)Concept understanding — Moving Coil Galvanometer

Moving Coil Galvanometer: From Intuition to Formula

Imagine you have a tiny, lightweight coil of wire, suspended so it can rotate freely. If you pass a current through it, that coil becomes an electromagnet. Now place it between the poles of a strong permanent magnet. The coil will try to twist — it experiences a torque. The bigger the current, the harder it twists. That is the entire physical idea behind a moving coil galvanometer: use a current to produce a rotation, and measure the rotation to know the current.

But a freely spinning coil would just keep turning. To get a useful measurement, you need something that opposes that rotation — a restoring force that grows as the coil turns further. That is the job of a spring (usually a fine phosphor-bronze strip called a torsion fibre). The spring twists as the coil rotates, producing a restoring torque that exactly balances the magnetic torque at some angle. That equilibrium angle is your reading.

The Radial Magnetic Field — The Key Trick

Here is the clever part. If the magnetic field were uniform and the coil rotated out of alignment, the torque would change with angle — making the scale non-linear. To avoid that, the poles of the magnet are shaped into concave cylindrical surfaces, and a soft iron cylinder is placed inside the coil. This creates a radial magnetic field: the field lines always point radially outward (or inward), so the plane of the coil is always parallel to the field as it rotates.

Important

In a radial field, the magnetic torque on the coil is independent of the coil's angular position. The torque depends only on the current.

That is what makes the deflection directly proportional to current — a linear scale.

The Physics in Equations

Let the coil have $N$ turns, each of area $A$ . A current $I$ flows through it. The magnetic field strength is $B$ (radial). The torque due to the magnetic field on a single turn is:

\tau_m = N I A B

This is because the force on each vertical side of the coil is $I L B$ (where $L$ is the length of the side), and the lever arm is the width of the coil, so the product gives $I \times (\text{area}) \times B$ per turn.

The spring provides a restoring torque proportional to the twist angle $\theta$ :

\tau_s = k \theta

where $k$ is the torsion constant of the spring (unit: N·m/rad).

At equilibrium, the two torques balance:

N I A B = k \theta

So the deflection is:

\theta = \frac{N A B}{k} \, I

The quantity $\frac{N A B}{k}$ is called the current sensitivity of the galvanometer. It tells you how many radians of deflection you get per ampere of current.

\theta = \left( \frac{N A B}{k} \right) I

What This Means for a Student

Larger $N$ , $A$ , or $B$ makes the galvanometer more sensitive — more deflection for the same current. …

Part (b)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? …

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Part (a): a moving-coil galvanometer works because the torque on the coil is proportional to current ( $I=\tfrac{k}{NAB}\theta$ ); it cannot read large currents directly (high resistance, small rating); a radial field (soft-iron core + cylindrical poles) keeps the torque constant for a linear scale. Part (b): the axial field of a circular loop is $B=\dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$ , and diamagnetic/paramagnetic substances differ in the sign of $\chi$ and their response to a field.

Part (a)

(i) Principle and working

Principle: a current-carrying coil placed in a magnetic field experiences a torque; the deflection is proportional to the current.

A rectangular coil of $N$ turns, each of area $A$ , is suspended in a radial magnetic field $B$ . When current $I$ flows, each side of length $l$ feels a force $F=BIl$ ; the two forces on opposite sides (separated by width $b$ ) form a couple of torque

\tau=NBIl\,b=NBIA.

The coil turns until this is balanced by the restoring torque of the phosphor-bronze suspension, $\tau_r=k\theta$ :

NBIA=k\theta\;\Rightarrow\;\theta=\frac{NBA}{k}I,\qquad I=\frac{k}{NAB}\theta.

So $\theta\propto I$ and a pointer reads the current on a scale.

Why not directly? The galvanometer is very sensitive — it has a comparatively high resistance and gives full-scale deflection at only microamperes/milliamperes. Placed in series in an ordinary circuit it would (i) alter the current substantially and (ii) risk burning out the delicate coil. It is converted to an ammeter by a low-resistance shunt in parallel.

(ii) Why the field is made radial, and how

In a radial field the field lines are always in the plane of the coil, so the angle between $\vec B$ and the coil plane is fixed and

\tau=NBIA\sin90^\circ=NBIA

is independent of the deflection $\theta$ . This makes the deflection strictly proportional to the current — a linear, evenly divided scale.

It is produced by:

Concave cylindrical pole pieces of the permanent magnet, and
a cylindrical soft-iron core placed inside the coil. Together they concentrate the flux and make it radial in the narrow annular gap in which the coil moves. …

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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 55/3/11 markMCQ
$Q. Assertion (A) : The cylindrical soft iron core in a moving coil galvanometer only makes the magnetic field radial and does not affect the strength of the magnetic field. Reason (R) : In a moving coil galvanometer, the plane of the coil is always perpendicular to the magnetic field. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Both Assertion (A) and Reason (R) are false.$
›Reveal solutionSolution

The soft iron core makes the field radial AND, being ferromagnetic, concentrates the magnetic flux — it increases the field strength, so the Assertion is false. In a radial field the plane of the coil is always parallel to the field lines (the coil's normal is perpendicular to $\vec B$ ), so the Reason is also false. The correct option is (D).

The question tests two separate facts about the moving coil galvanometer: what the cylindrical soft iron core actually does, and how the coil sits relative to the magnetic field.

1. Role of the soft iron core — is the Assertion true?

The core, together with the concave pole pieces, shapes the field in the air gap so that it is radial: at every angular position of the coil, the field lines point along the radius. This makes the deflecting torque independent of the coil's position, which is what gives the galvanometer its linear scale $\theta \propto I$ .

But that is not all the core does. Soft iron is ferromagnetic, with a very high relative permeability ( $\mu_r \gg 1$ ). It provides a low-reluctance path for the magnetic flux, so the flux crowds through the core and the field strength $B$ in the narrow air gap becomes much larger than it would be without the core. The claim that the core "only makes the field radial and does not affect the strength" is therefore false — the Assertion is false.

2. Orientation of the coil — is the Reason true?

In the radial field the field lines run along the radius, and the plane of the rectangular coil (tangential to the cylindrical core) always contains those field lines. So the plane of the coil is always parallel to the magnetic field — equivalently, the coil's normal is perpendicular to $\vec B$ in every position. That is exactly what keeps the torque at its maximum value throughout the rotation: …
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CBSE 2025Set D1 markMCQ
$Q. If the number of turns is increased in any moving coil galvanometer, then its sensitivity (A) increases (B) decreases (C) remains unchanged (D) may increase or may decrease$
›Reveal solutionSolution

A galvanometer's current sensitivity is (NAB/k), directly proportional to the number of turns N, so more turns → higher sensitivity.

The deflection of a moving-coil galvanometer is

$\theta = \frac{NAB}{k}\,I$

so its current sensitivity is

$\frac{\theta}{I} = \frac{NAB}{k}$

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. In any electric circuit, galvanometer in its original form is used to -$
(a) detect the current
(b) measure the current
(c) measure the voltage
(d) measure the resistance
›Reveal solutionSolution

A galvanometer in its basic form is a sensitive current-detecting device, not a calibrated measuring instrument.

A galvanometer is a sensitive instrument used to detect the presence (and direction) of a small current in a circuit through the deflection of a coil/needle. In its original form it is not calibrated to read numerical values of current, voltage o …

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CBSE 2024Set A1 markMCQ
$Q. The value of current obtained in a moving coil galvanometer is proportional to (A) deflection (θ) (B) resistance (R) (C) magnetic field (B) (D) none of these$
›Reveal solutionSolution

A moving-coil galvanometer is linear: I ∝ θ (the deflection).

In a moving-coil galvanometer, the current-carrying coil in the radial magnetic field experiences a deflecting torque $NBIA$ , balanced by the restoring torque $k\theta$ of the suspension:

$NBIA = k\theta \;\Rightarrow\; I = \frac{k}{NBA}\,\theta.$

…

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CBSE 2024Set ANNUAL1 mark
$Q. Why is it necessary to introduce a cylindrical soft iron core inside the coil of a galvanometer ?$
›Reveal solutionSolution

The soft-iron core makes the field radial, ensuring the deflecting torque (and hence the scale) is uniform.

In a moving-coil galvanometer, concave pole pieces together with a cylindrical soft-iron core placed inside the coil make the magnetic field radial — i.e. $\vec B$ is always along the plane of the coil, no matter what angle the coil has turned through. Because of this, the angle between the field and the normal to the coil stays 90° throughout the motion, so the deflecting torque $\tau = NBIA$ has no $\sin\theta$ dependence and stays proportional to the current $I$ alone. This gives a uniform torque for a given current at every deflection, so the pointer's deflection is directly proportional to the cu …

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CBSE 2023Set ANNUAL1 mark
$Q. What happens to the voltage sensitivity of the galvanometer when the current sensitivity of a moving coil galvanometer is doubled by doubling the number of turns of the coil?$
›Reveal solutionSolution

Doubling the turns doubles both the current sensitivity and the coil's resistance, leaving voltage sensitivity the same.

Current sensitivity of a galvanometer is $I_s = \dfrac{NBA}{k}$ , which is directly proportional to the number of turns $N$ ; doubling $N$ doubles $I_s$ , as stated. Voltage sensitivity is defined as $V_s = \dfrac{I_s}{R} = \dfrac{NBA}{kR}$ , where $R$ is the resistance of the galvanometer coil. Since doubling the number of turns (with the same wire, same coil geometry) also roughly doubles the length of wire used and hence doubles the coil's resistance $R$ , the factor of 2 in the numerator ( $N$ ) is cancelled by the factor of 2 in the den …

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CBSE 2022Set I1 markMCQ
$Q. Which of the following is correct for galvanometer constant? (A) C/(NAB) (B) C(B)/(NA) (C) C(A/(NB)) (D) C(N/(AB))$
›Reveal solutionSolution

Galvanometer constant $k = \dfrac{C}{NAB}$ .

In a moving-coil galvanometer, the current $I$ produces a torque $NIAB$ that is balanced at equilibrium by the restoring torque $C\theta$ of the suspension (torsional constant $C$ ):

$NIAB = C\theta \;\Rightarrow\; I = \left(\frac{C}{NAB}\right)\theta.$

The proportionality factor relating current to deflection is the galvanometer constant …

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CBSE 2022Set ANNUAL1 mark
$Q. Why does a moving coil galvanometer use concave magnetic poles?$
›Reveal solutionSolution

Concave poles give a radial field -> uniform torque per unit current -> linear scale.

In a moving-coil galvanometer, the deflecting torque on the coil is $\tau = NIAB\sin\phi$ , where $\phi$ is the angle between the plane of the coil and the field $B$ . If $B$ were uniform (as between flat pole pieces), $\phi$ would change as the coil rotates, making the torque, and hence the deflection, a non-linear function of the rotation angle — the scale would be cramped/uneven. By using concave (curved) pole pieces together with a soft-iron cylindrical core, the magnetic field lines are made radial, i.e. always directed along the plane of the coil no matter what angle the coil has turned to. This keeps $\phi = 90°$ at all deflections, so $\sin\phi = 1$ always …

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CBSE 2021Set NC1 mark
$Q. Why are pole pieces of a magnet within a galvanometer made concave?$
›Reveal solutionSolution

Concave pole pieces bend the field lines so they always point along the coil's plane no matter how far it has rotated, keeping the torque formula free of any angle-dependence.

Reasoning

If the field between the poles were uniform, the deflecting torque on the coil would be $\tau=nBIA\sin\theta$ , which depends on the coil's instantaneous angle $\theta$ — making the current-vs-deflection relationship non-linear (a non-uniform scale).

Concave pole pieces (curved to match the coil's cylindrical geometry, often combined with a soft-iron cylindrical core) shape the field so that it is always radial — i.e. the field lines are always directed along the plane of the coil, regardless of its angular position. This eliminates the $\sin\theta$ dependence entirely:

$\tau = nBIA \quad (\text{no}\ \sin\theta \text{ factor})$

…

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CBSE 2020Set XS1 mark
$Q. Define current sensitivity of a moving coil galvanometer.$
›Reveal solutionSolution

Deflection per unit current, $S_I=\theta/I=NBA/k$ .

Definition. The current sensitivity of a moving-coil galvanometer is the deflection $\theta$ produced per unit current $I$ passing through it:

$S_I=\frac{\theta}{I}.$

At equilibrium the deflecting torque $NBIA$ balances the restoring torque $k\theta$ , so $NBIA=k\theta$ , giving

$S_I=\frac{\theta}{I}=\frac{NBA}{k},$ …

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CBSE 2019Set HE2341 mark
$Q. Match the following. Column A item: Moving coil Galvanometer. Select the matching option from Column B:$
(i) Maximum deviation by prism
(ii) Work function
(iii) Law of conservation of energy
(iv) Magnetic effect of current
(v) Maximum speed in glass.
›Reveal solutionSolution

A moving-coil galvanometer works because a current-carrying coil placed in a magnetic field experiences a torque — the magnetic effect of current.

A moving coil galvanometer consists of a rectangular coil suspended in a strong, radial magnetic field of a permanent magnet. When a current is passed through the coil, each current-carrying side of the coil experiences a force $F = BIl$ in the magnetic field, and the pair of oppositely-directed forces on the two sides produces a torque that deflects the coil. This is a direct application of the magnetic effect of …

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CBSE 2019Set ANNUAL1 markMCQ
$Q. Phosphor-bronze wire is used for suspension in a moving coil galvanometer, because it has :$
(a) large couple per unit twist
(b) small couple per unit twist
(c) high conductivity
(d) high resistivity
›Reveal solutionSolution

Phosphor-bronze is chosen for the suspension because its low torsional (restoring) constant gives the galvanometer high current sensitivity.

In a moving coil galvanometer, the coil is suspended by a thin wire (or fibre) that also acts as one of the current leads and provides the restoring torque against the deflecting torque produced by the magnetic field acting on the current-carrying coil.

At equilibrium, the deflecting torque equals the restoring torque: $NIAB = k\theta$ , where $k$ is the torsional constant (restoring couple per unit twist/radian) of the suspension.

The current sensitivity of the galvanometer is $\dfrac{\theta}{I} = \dfrac{NAB}{k}$

This shows sensitivity is inversely proportional to $k$ : the smaller the torsional constant of the suspension, the larger the deflection $\theta$ produced by a given small current $I$ , i.e. the more sensitive the instrument.

…

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Q.(a)

Part (a)

(i) Principle and working

(ii) Why the field is made radial, and how

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