A galvanometer shows the direction and strength of the current through it. A coil in a magnetic field experien

Question

NBAI=k\phi

(i) The current sensitivity of a galvanometer is given by: (A)

\dfrac{k}{NBA}

(B)

\dfrac{NBA}{k}

(C)

\dfrac{kBA}{N}

(D)

\dfrac{kNB}{A}

(ii) A galvanometer of resistance

6\ \Omega

shows full-scale deflection for a current of

0.2

A. The shunt to convert it into an ammeter of range

(0-5)

A is: (A)

0.25\ \Omega

(B)

0.30\ \Omega

(C)

0.50\ \Omega

(D)

6.0\ \Omega

(iii) The resistance of the ammeter in case

(ii) will be: (A)

0.20\ \Omega

(B)

0.24\ \Omega

(C)

6.0\ \Omega

(D)

6.25\ \Omega

(iv)(a) A galvanometer is converted into a voltmeter of range

(0-V)

by a series resistance

R_{1}

. If

R_{1}

is replaced by

R_{2}

, the range becomes

(0-2V)

. The galvanometer resistance is: (A)

(R_{2}-2R_{1})

(B)

(R_{2}-R_{1})

(C)

(R_{1}+R_{2})

(D)

(R_{1}-2R_{2})

(OR)

(iv)(b) A current of

5

mA flows through a galvanometer of 100 turns, each of area

18\ \mathrm{cm}^{2}

, in a magnetic field

0.20

T. The deflecting torque is: (A)

3.6\times10^{-3}

N m (B)

1.8\times10^{-4}

N m (C)

2.4\times10^{-3}

N m (D)

1.2\times10^{-4}

N m

CBSECBSE Class XII Board 2025Subjective· 4mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 — Torque on a Current Loop

Torque on a Current Loop

The Intuition First

Imagine a compass needle in the Earth's magnetic field. The needle always turns until it points north. Why? Because the needle itself is a tiny magnet, and the field exerts a twist — a torque — that tries to align it.

A current-carrying loop behaves exactly like that tiny magnet. It has a magnetic moment $\vec{m}$ , which is like its own internal compass arrow. When you place this loop in an external magnetic field $\vec{B}$ , the field pulls on one side of the loop and pushes on the other, creating a turning effect.

The loop doesn't feel a net force (if the field is uniform), but it does feel a torque. That torque always tries to rotate the loop so that its magnetic moment points along the field — just like a compass needle.

The Key Players

The magnetic moment of a planar current loop is:

\vec{m} = I \, A \, \hat{n}

where $I$ is the current, $A$ is the area of the loop, and $\hat{n}$ is a unit vector perpendicular to the plane of the loop (direction given by the right-hand rule: curl your fingers along the current, your thumb points along $\vec{m}$ ).

The external field $\vec{B}$ is uniform — same magnitude and direction everywhere in the region of the loop.

The Torque: Two Equivalent Forms

The torque on the loop is:

\tau = m B \sin\theta

where $\theta$ is the angle between $\vec{m}$ and $\vec{B}$ . The torque is maximum when $\vec{m}$ is perpendicular to $\vec{B}$ ( $\theta = 90^\circ$ ), and zero when they are parallel or antiparallel ( $\theta = 0^\circ$ or $180^\circ$ ).

The vector form captures both magnitude and direction:

\vec{\tau} = \vec{m} \times \vec{B}

\vec{\tau} = \vec{m} \times \vec{B}

The cross product tells you: the torque is perpendicular to both $\vec{m}$ and $\vec{B}$ , and its direction is given by the right-hand rule. This torque always rotates $\vec{m}$ toward $\vec{B}$ .

Why It Happens (The Physics)

Consider a rectangular loop of sides $a$ and $b$ , carrying current $I$ , placed in a uniform field $\vec{B}$ . Let the plane of the loop make an angle $\theta$ with the field.

The two sides of length $a$ are perpendicular to $\vec{B}$ . On each of these sides, the magnetic force is $F = I a B$ , but the forces on opposite sides are in opposite directions. These two forces form a couple — equal and opposite, not along the same line — which produces a torque.

The lever arm for each force is $(b/2) \sin\theta$ , so the net torque is:

\tau = 2 \times (I a B) \times \frac{b}{2} \sin\theta = I (ab) B \sin\theta = I A B \sin\theta

Since $m = I A$ , we get $\tau = m B \sin\theta$ .

Tip

For a rectangular loop, the torque comes only from the sides perpendicular to the field. The sides parallel to the field experience forces that are either zero or along the axis — they contribute nothing to the torque.

The Stable Equilibrium

When $\vec{m}$ is aligned with $\vec{B}$ ( $\theta = 0$ ), the torque is zero. This is a stable equilibrium — if you nudge the loop slightly, the torque brings it back. …

Start your 14-day free trial to unlock the full solution →

Using $NBAI=k\phi$ and the shunt/series-resistance rules: (i) B $\left(\tfrac{NBA}{k}\right)$ , (ii) A $(0.25\,\Omega)$ , (iii) B $(0.24\,\Omega)$ , (iv)(a) A $(R_2-2R_1)$ ; and (iv)(b) B $(1.8\times10^{-4}\ \text{N m})$ .

At equilibrium the deflecting torque balances the restoring torque: $NBAI=k\phi$ . Everything follows from this.

Part (a)

Current sensitivity is deflection per unit current: $\phi=\dfrac{NBA}{k}I\Rightarrow \dfrac{\phi}{I}=\dfrac{NBA}{k}$ . Larger $NBA$ (or smaller $k$ ) gives more deflection per unit current — option (B).
Shunt for ammeter conversion. To read up to $I=5$ A, a shunt $S$ in parallel with $G=6\,\Omega$ carries the excess current while the galvanometer still gets only $I_g=0.2$ A at full scale. Equal voltage across the parallel pair: $I_gG=(I-I_g)S\Rightarrow 0.2\times6=(5-0.2)S\Rightarrow S=\frac{1.2}{4.8}=0.25\ \Omega,$ option (A).
Ammeter resistance is $G$ and $S$ in parallel: $R_A=\frac{GS}{G+S}=\frac{6\times0.25}{6+0.25}=\frac{1.5}{6.25}=0.24\ \Omega,$ option (B) (an ammeter's resistance is very small, as it should be). (iv)(a) Galvanometer resistance from two voltmeter ranges. A voltmeter of range $V$ uses series resistance $R_1$ : $I_g=\dfrac{V}{G+R_1}$ . Range $2V$ uses $R_2$ : $I_g=\dfrac{2V}{G+R_2}$ . Equating $I_g$ : …

Start your 14-day free trial to unlock the full solution →

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 24 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: …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. If vector m be the magnetic moment of a magnetic dipole placed in a magnetic field of induction vector B, the torque experienced by the dipole will be$
(a) m . B
(b) |m| / |B|
(c) m x B
(d) |m| |B|
›Reveal solutionSolution

The torque on a magnetic dipole is $\vec{\tau} = \vec{m}\times\vec{B}$ , analogous to torque on an electric dipole $\vec{p}\times\vec{E}$ .

When a magnetic dipole of moment $\vec{m}$ is placed in a uniform magnetic field $\vec{B}$ , it experiences a torque that tends to align it with the field. This torque is given by the vector product:

$\vec{\tau} = \vec{m}\times\vec{B}$

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set SEM31 markMCQ
$Q. The ratio of the radii of two circular loops is 1 : 2. The ratio of their magnetic moments is 1 : 2. The ratio of currents flowing through them is$
(a) 1 : 1
(b) 2 : 1
(c) 4 : 1
(d) 1 : 4
›Reveal solutionSolution

Magnetic moment M = I·(πr²), so I = M/(πr²) ∝ M/r². Substituting the given ratios gives I₁ : I₂ = 2 : 1. Option (b).

Step 1 — magnetic moment of a current loop (NCERT/CBSE Class 12 Physics, Moving Charges and Magnetism): M = I·A = I·πr².

Step 2 — so current I = M/(πr²), i.e. I ∝ M/r².

…

Start your 14-day free trial to see the full solution →
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

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

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. Two circular loops having ratio of their radii 1 : 2 possess same magnetic moment. The ratio of their circulating currents will be$
(a) 4 : 1
(b) 1 : 4
(c) 2 : 1
(d) 1 : 2
›Reveal solutionSolution

Magnetic moment m = IA = Iπr²; equal m with r ratio 1:2 forces the current ratio to be 4:1 (inverse of the area ratio).

Magnetic moment of a current loop is $m = I\pi r^2$ . Let the radii be $r_1:r_2 = 1:2$ and the moments be equal, $m_1=m_2$ : …

Start your 14-day free trial to see the full solution →
CBSE 2024Set A11 mark
$Q.The torque on a rectangular current loop in a uniform magnetic field increases by ———————— the area of the loop. Fill in the blank choosing the appropriate answer from the bracket: (decreasing, interference, helium, greater, diffraction, increasing)$
›Reveal solutionSolution

increasing (the torque is directly proportional to the area of the loop). …

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
CBSE 2024Set A1 markMCQ
$Q. The value of torque (τ) experienced by current loop of magnetic moment (m) placed in magnetic field (B) is (A) τ = m \times B (B) τ = B \times m (C) τ = m/B (D) τ = B/m$
›Reveal solutionSolution

A current loop behaves like a magnetic dipole; the torque on it is the cross product of its magnetic moment and the field, τ = m × B.

A planar current loop carrying current I and enclosing area A has a magnetic (dipole) moment $\vec{m} = I\vec{A}$ , directed along the normal to the loop (right-hand rule).

When this dipole is placed in a uniform magnetic field $\vec{B}$ , the two sides of the loop carry equal and opposite forces that form a couple. The resulting torque is

$\vec{\tau} = \vec{m} \times \vec{B}, \qquad \tau = mB\sin\theta$

…

Start your 14-day free trial to see the full solution →
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:

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

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 mark
$Q. Write the formula for torque acting on rotating current carrying coil in terms of magnetic dipole moment, in vector form.$
›Reveal solutionSolution

Torque on a magnetic dipole (current loop) in a magnetic field.

A current-carrying coil of magnetic dipole moment $\vec{m}$ placed in a uniform magnetic field $\vec{B}$ experiences a torque that tends to align $\vec{m}$ with $\vec{B}$ :

$\vec{\tau} = \vec{m}\times\vec{B}$ …

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
CBSE 2023Set 55/4/11 markMCQ
$L$
›Reveal solutionSolution
For a fixed wire length $L$ , a circle encloses the maximum area, so the square loop has less area than the circular loop. Since torque $\tau = NIAB\sin\theta$ depends directly on area, the square loop experiences less torque — making Assertion false and Reason false as well. The correct choice is (D).

The Core Idea: Magnetic Torque and Area

When a current-carrying loop sits in a uniform magnetic field, the net force on it is zero, but the field exerts a torque that tries to rotate the loop. The magnitude of this torque is given by:
$\tau = N I A B \sin\theta$
Here $N$ is the number of turns (1 for a single loop), $I$ is the current, $A$ is the area enclosed by the loop, $B$ is the magnetic field strength, and $\theta$ is the angle between the loop’s normal and the field. For a fixed $I$ , $B$ , and $\theta$ , the torque is directly proportional to the area $A$ .

So the question boils down to: for a given wire length $L$ , which shape — square or circle — gives a larger area?

Step-by-Step Reasoning

The wire length is fixed. Both loops are made from the same wire of total length $L$ . This length becomes the perimeter of each loop. For the square, each side is $L/4$ , so its area is:

$A_{\text{square}} = \left(\frac{L}{4}\right)^2 = \frac{L^2}{16}$

For the circle, the circumference is $L = 2\pi r$ , so the radius is $r = L/(2\pi)$ . The area is:

$A_{\text{circle}} = \pi r^2 = \pi \left(\frac{L}{2\pi}\right)^2 = \frac{L^2}{4\pi}$

Compare the two areas. Which is larger? Compare $L^2/16$ and $L^2/(4\pi)$ . Since $L^2$ is positive, we compare the denominators: $16$ vs $4\pi \approx 12.57$ . A smaller denominator means a larger fraction, so:

$\frac{L^2}{4\pi} > \frac{L^2}{16}$
Therefore, $A_{\text{circle}} > A_{\text{square}}$ .

Tip

This is a classic isoperimetric result: for a given perimeter, the circle encloses the maximum possible area. Any other shape — square, rectangle, triangle — will have a smaller area. Memorising this saves you from recalculating every time. …
Start your 14-day free trial to see the full solution →

Part (a)

🎓Unlock everything free for 14 days

Unlock everything free for 14 days