(a) A circular loop of radius R carries a current I. Obtain an expression for the magnetic field at a point on

Question

x

(b) A conducting rod of length 2 m is placed on a horizontal table in north-south direction. It carries a current of 5 A from south to north. Find the direction and magnitude of the magnetic force acting on the rod. Given that the Earth's magnetic field at the place is

0\cdot6 \times 10^{-4}

T and angle of dip is

\dfrac{\pi}{6}

(OR)

(a) Obtain the expression for the deflecting torque acting on the current carrying rectangular coil of a galvanometer in a uniform magnetic field. Why is a radial magnetic field employed in the moving coil galvanometer ?

(b) Particles of mass

1\cdot6 \times 10^{-27}

kg and charge

1\cdot6 \times 10^{-19}

C are accelerated in a cyclotron of dee radius 40 cm. It employs a magnetic field

0\cdot4

T. Find the kinetic energy (in MeV) of the particle beam imparted by the accelerator.

CBSECBSE Class XII Board 2020Subjective· 5mImportance★★★★★

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Part (a)Concept understanding — Magnetic Force on a Current-Carrying Conductor

Since an electric current is physically many moving charges (conduction electrons) travelling together, a current-carrying wire in a magnetic field experiences a force that is the sum of the tiny Lorentz forces on every individual moving charge. For a straight wire of length L carrying current I in a field B applied perpendicular to it, this sum works out (after the drift speed conveniently cancels out of the calculation) to $F=BIL$ ; more generally, with a length vector $\vec{L}$ pointing along the current, $\vec{F}=I\vec{L}\times\vec{B}$ , valid at any angle between the wire and the field. …

Part (b)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. …

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Part (a): $B=\dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$ on the axis; the N–S rod feels $F=3\times10^{-4}\text{ N}$ (west). Part (b): galvanometer torque $\tau=NBIA$ (radial field gives a linear scale); cyclotron beam KE $\approx 1.28\text{ MeV}$ .

Part (a)

Magnetic field on the axis of a circular loop. Take a loop of radius $R$ carrying current $I$ , and a point P on the axis a distance $x$ from the centre. A current element $I\,d\vec l$ is at distance $r=\sqrt{R^2+x^2}$ from P and is perpendicular to $\vec r$ , so by the Biot–Savart law $dB=\frac{\mu_0}{4\pi}\frac{I\,dl}{R^2+x^2}.$ Resolving $d\vec B$ into an axial part ( $dB\sin\theta$ , with $\sin\theta=R/\sqrt{R^2+x^2}$ ) and a radial part: by symmetry the radial parts of diametrically opposite elements cancel, leaving only the axial parts. Integrating over the loop ( $\oint dl=2\pi R$ ): $B=\frac{\mu_0}{4\pi}\frac{I(2\pi R)}{R^2+x^2}\cdot\frac{R}{\sqrt{R^2+x^2}}=\frac{\mu_0 I R^2}{2\,(R^2+x^2)^{3/2}},$ directed along the axis (by the right-hand rule).
Force on the conducting rod. $L=2\text{ m}$ , $I=5\text{ A}$ south-to-north, Earth's field $B_E=0.6\times10^{-4}\text{ T}$ , dip $\delta=\pi/6=30^\circ$ . Split $\vec B_E$ into:

horizontal $B_H=B_E\cos\delta$ — this lies along the N–S line, parallel to the current, so it exerts no force;
vertical $B_V=B_E\sin\delta$ — perpendicular to the rod, so it produces the force.

$F=ILB_V=ILB_E\sin\delta=5\times2\times(0.6\times10^{-4})\times\sin30^\circ=5\times2\times0.3\times10^{-4}=3\times10^{-4}\text{ N}.$ …

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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 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: …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. In a uniform magnetic field B, a conductor of length l is placed parallel to the magnetic field. When a current I is passed through the conductor, the force on the conductor will be$
(a) IlB
(b) IB/l
(c) Il/B
(d) zero
›Reveal solutionSolution

The magnetic force on a current-carrying wire depends on sin(theta) between the current direction and B; when the wire is parallel to B, that force is zero.

The force on a straight current-carrying conductor of length l in a uniform magnetic field B is given by

F = BIl*sin(theta)

…

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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 2025Set ANNUAL1 markMCQ
$Q. A straight current carrying wire kept in a uniform magnetic field will experience a maximum force when it is :$
(a) perpendicular to the magnetic field
(b) parallel to the magnetic field
(c) at an angle of 45° to the magnetic field
(d) at an angle of 60° to the magnetic field
›Reveal solutionSolution
The magnetic force on a current-carrying wire is $F = BIL\sin\theta$ , which is maximum when $\sin\theta = 1$ , i.e. when the wire is perpendicular to $B$ .

The force on a straight wire of length $L$ carrying current $I$ in a uniform field $B$ is

$F = BIL\sin\theta$

where $\theta$ is the angle between the current direction and $B$ .

If the wire is parallel to $B$ ( $\theta = 0^\circ$ ), $\sin\theta = 0$ , so $F = 0$ . …
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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
$Q. Which one of the following is not a unit of magnetic field? (A) tesla (B) weber/metre^2 (C) newton/ampere-metre (D) newton/ampere^2$
›Reveal solutionSolution
B has units tesla = Wb/m² = N/(A·m); newton/ampere² is NOT a unit of B.

Magnetic field B can be expressed as:

tesla (T),

weber/metre² (Wb/m²), since 1 T = 1 Wb/m²,

newton/(ampere·metre), from F = BIL ⇒ B = F/(IL) = N/(A·m). …
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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 markMCQ
$Q. A current carrying long erect wire is kept at an angle θ with an external uniform magnetic field. The wire experiences highest force if$
(a) θ = 0°
(b) θ = 30°
(c) θ = 60°
(d) θ = 90°.
›Reveal solutionSolution

The force on a current-carrying wire in a magnetic field depends on $\sin\theta$ , which is maximum (=1) at θ = 90°.

A straight current-carrying conductor of length $L$ carrying current $I$ , placed at angle $\theta$ to a uniform magnetic field $B$ , experiences a force

$F = BIL\sin\theta$

…

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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 markMCQ
$Q. The magnetic force F (vector) on a current carrying conductor of length l (vector) in an external magnetic field B (vector) is given by$
(1) (I x B) / l [I=current scalar; l and B vectors]
(2) (l x B) / I
(3) I(l x B)
(4) I^2 (l x B)
›Reveal solutionSolution

Summing the Lorentz force qv x B over all the moving charges in a straight conductor of length l carrying current I gives F = I l x B.

…

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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 2023Set ANNUAL1 mark
$Q. What is the value of force on a closed circuit in a magnetic field?$
›Reveal solutionSolution

The net force on any closed current loop in a uniform field is always zero.

For a closed circuit of current $I$ in a uniform magnetic field $\vec{B}$ , the total force is

$\vec{F} = I \oint d\vec{l} \times \vec{B} = I\left(\oint d\vec{l}\right)\times \vec{B}$ …

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Q.(a) A circular loop of radius R carries a current I. Obtain an expression for the magnetic field at a point on its axis at a distance xxx from its centre.

Part (a)

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Q.(a) A circular loop of radius R carries a current I. Obtain an expression for the magnetic field at a point on its axis at a distance $x$ from its centre.

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