(a) A circular coil of $30$ turns and radius $8.0$ cm carrying a current of $6$ A is suspended vertically in a uniform horizontal magnetic field of $1.0$ T. The field lines make an angle of $30^\circ$ with the plane of the coil. Calculate the magnitude of the external torque that must be applied to prevent the coil from turning. What would happen if the circular coil is replaced by a planar coil of irregular shape that encloses the same area, keeping other parameters unchanged ? OR (b) An alpha particle (mass $6.4\times10^{-27}$ kg and charge $3.2\times10^{-19}$ C) having $8.0$ MeV energy enters a region of a uniform magnetic field of $0.5$ T. If the field is directed perpendicular to the velocity of the particle, find the radius of the circular path described by the particle. Mention the condition under which the particle in this region (i) describes a helical path, and (ii) goes straight undeviated.

(a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6 A is suspended vertically in a unifo

Question

30

(OR)

(b) An alpha particle (mass

6.4\times10^{-27}

kg and charge

3.2\times10^{-19}

C) having

8.0

MeV energy enters a region of a uniform magnetic field of

0.5

T. If the field is directed perpendicular to the velocity of the particle, find the radius of the circular path described by the particle. Mention the condition under which the particle in this region

(i) describes a helical path, and

(ii) goes straight undeviated.

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Part (a)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. …

Part (b)Concept understanding — Cyclotron Motion Radius

Cyclotron Motion Radius – From Intuition to Formula

Imagine you're pushing a charged ball on a frictionless table, and there's a giant magnet underneath. The moment that ball starts moving, the magnet doesn't pull it or push it forward — it turns it. The force from the magnet always acts sideways, perpendicular to the ball's velocity. So the ball never speeds up or slows down; it just keeps changing direction. If the magnetic field is uniform and the ball keeps moving, it will trace out a perfect circle.

That circle is called cyclotron motion, and the radius of that circle is what we're after.

Why does it curve at all?

The magnetic force on a moving charge is given by:

\vec{F} = q (\vec{v} \times \vec{B})

The cross product means the force is always perpendicular to both the velocity $\vec{v}$ and the magnetic field $\vec{B}$ . For a charge moving perpendicular to a uniform field, this force acts as a centripetal force — it constantly pulls the charge toward the centre of a circle, without doing any work (since force is perpendicular to displacement).

So the charge moves in uniform circular motion. The magnetic force provides the necessary centripetal acceleration.

Deriving the radius

For circular motion, the centripetal force required is:

F_{\text{centripetal}} = \frac{m v^2}{r}

where $m$ is the mass of the particle, $v$ is its speed, and $r$ is the radius of the circle.

The magnetic force (for $\vec{v} \perp \vec{B}$ ) has magnitude:

F_B = |q| v B

Set them equal:

|q| v B = \frac{m v^2}{r}

Cancel one factor of $v$ (assuming $v \neq 0$ ):

|q| B = \frac{m v}{r}

Solve for $r$ :

r = \frac{m v}{|q| B}

That's the cyclotron motion radius (also called the Larmor radius or gyroradius).

What the formula tells you

Faster particle → larger radius (it's harder to turn something moving fast).
Heavier particle → larger radius (more inertia resists the turn).
Stronger magnetic field → smaller radius (the turning force is stronger).
Larger charge → smaller radius (more force for the same field).

Tip

If the particle's velocity has a component parallel to $\vec{B}$ , it doesn't feel any magnetic force in that direction. So the particle moves in a helix — circular motion in the plane perpendicular to $\vec{B}$ , plus constant speed along $\vec{B}$ . The radius formula above still applies using only the perpendicular component of velocity, $v_\perp$ .

A quick example

A proton ( $m = 1.67 \times 10^{-27}$ kg, $q = 1.6 \times 10^{-19}$ C) moves at $2.0 \times 10^6$ m/s perpendicular to a 0.50 T magnetic field. …

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Part (a): balancing torque $\tau=NIAB\sin\alpha\approx 3.13\text{ N·m}$ , and it is the same for any planar coil of equal area. Part (b): $r=\dfrac{mv}{qB}=0.8\text{ m}$ ; helical path when $\vec v$ has a component along $\vec B$ , undeviated when $\vec v\parallel\vec B$ .

Part (a)

A current loop of moment $\vec m=NI A\hat n$ in a field $\vec B$ feels a torque $\vec\tau=\vec m\times\vec B$ of magnitude $\tau=NIAB\sin\alpha$ , where $\alpha$ is the angle between the coil normal and $\vec B$ . To hold the coil stationary, the applied external torque must equal this.

The field makes $30^\circ$ with the plane of the coil, so with the normal it makes $\alpha=90^\circ-30^\circ=60^\circ$ .

A=\pi r^2=\pi(0.08)^2=0.0201\text{ m}^2,

$\tau=NIAB\sin\alpha=30\times6\times0.0201\times1.0\times\sin60^\circ=3.619\times0.866\approx3.13\text{ N·m}.$ …

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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 22 on this concept.

CBSE 2026Set DS1 mark
$10^{-5}$
›Reveal solutionSolution

Using $r = \dfrac{\sqrt{2mE}}{eB}$ , the radius comes out to about $1.07$ m.

Concept. A charged particle moving perpendicular to a magnetic field goes in a circle whose radius is set by balancing the magnetic force against the centripetal requirement: $qvB = \dfrac{mv^2}{r}$ , giving $r = \dfrac{mv}{qB} = \dfrac{p}{qB}$ . The momentum is found from the kinetic energy, $p = \sqrt{2mE}$ .

Calculation. Kinetic energy $E = 10\text{ eV} = 10\times1.6\times10^{-19} = 1.6\times10^{-18}$ J. …

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CBSE 2026Set A1 markMCQ
$Q. If a charged particle of mass m and charge q enters a uniform magnetic field B at an angle θ in the direction of field with velocity v, then the path of the particle is helical. The radius of circular path of the helix will be (A) mv/qB (B) mv cosθ/qB (C) mv sinθ/qB (D) 2πmv/qB$
›Reveal solutionSolution
The perpendicular component v sinθ gives circular motion; r = m v sinθ / qB.

When a charge enters a magnetic field at angle θ to B, resolve its velocity:

Component along B: $v\cos\theta$ — unaffected by the field, gives uniform motion along the axis (the pitch of the helix).

Component perpendicular to B: $v\sin\theta$ — experiences the magnetic force $qvB\sin\theta$ , producing circular motion. …
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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:

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

…

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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.

…

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CBSE 2025Set 55/4/11 markMCQ
$+q$
›Reveal solutionSolution

The charge enters with velocity along $+Y$ and the field is into the page ( $-Z$ ), so the magnetic force keeps it in the $XY$ plane — the path is a semicircle in the $XY$ plane. Option (B).

Reading the figure.

Figure — 55/4/1 Q3

The axes are $X$ (right), $Y$ (up) and $Z$ (out of the page, toward the viewer). The $\times$ grid marks a uniform field into the page, i.e. $\vec{B} = -B\,\hat{k}$ , filling the upper region. The charge $+q$ sits on the $+X$ axis and enters moving along $+Y$ , so $\vec{v} = v\,\hat{j}$ .

Force direction.
$\vec{F} = q\,\vec{v}\times\vec{B} = q\,(v\,\hat{j})\times(-B\,\hat{k}) = -qvB\,(\hat{j}\times\hat{k}) = -qvB\,\hat{i}$
The force is along $-X$ , i.e. it lies in the $XY$ plane, perpendicular to $\vec{v}$ . …
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CBSE 2025Set 55/5/11 markMCQ
$10^{6}\ \mathrm{ms^{-1}}$
›Reveal solutionSolution

The radius of cyclotron motion is $r = \frac{mv}{qB}$ . Using the kinetic energy gained from the potential difference, we find the charge-to-mass ratio, then substitute into the radius formula to get 5 cm.

Why this works — the physics of circular motion in a magnetic field

When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force acts as a centripetal force, bending the path into a circle. The key relation comes from equating:
$q v B = \frac{m v^2}{r}$
which simplifies to the cyclotron radius:
$r = \frac{mv}{qB}$
The problem gives us $v = 10^6\ \mathrm{m/s}$ and $B = 0.4\ \mathrm{T}$ , but we don't directly know $m/q$ — the mass-to-charge ratio of the particle. However, we can find it from the acceleration step: the particle was accelerated from rest through a potential difference of $10\ \mathrm{kV} = 10^4\ \mathrm{V}$ .

Step-by-step solution

1. Find the kinetic energy gained

When a charge $q$ is accelerated through a potential difference $V$ , it gains kinetic energy equal to the work done by the electric field:
$\frac{1}{2} m v^2 = qV$
We know $v = 10^6\ \mathrm{m/s}$ and $V = 10^4\ \mathrm{V}$ . This gives us a direct relation between $m$ and $q$ .

2. Extract the charge-to-mass ratio

From $\frac{1}{2} m v^2 = qV$ , rearrange:
$\frac{q}{m} = \frac{v^2}{2V}$
Plug in the numbers:
$\frac{q}{m} = \frac{(10^6)^2}{2 \times 10^4} = \frac{10^{12}}{2 \times 10^4} = \frac{10^8}{2} = 5 \times 10^7\ \mathrm{C/kg}$
Tip

You don't need to identify the particle — the ratio $q/m$ is all that matters for the radius. This is a common exam trick: they give you $v$ and $V$ so you can find $q/m$ without needing the particle's identity.

3. Write the radius formula in terms of known quantities

From $r = \frac{mv}{qB}$ , we can write:

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CBSE 2025Set 55/6/11 markMCQ
$\alpha$
›Reveal solutionSolution
When charged particles enter perpendicular to a magnetic field, the radius depends on momentum and charge: $r = \frac{mv}{qB}$ . Since the proton and α-particle have the same velocity but different mass-to-charge ratios, we find $r_p : r_\alpha = 1 : 2$ .

Why the radius depends on mass and charge

When a charged particle moves perpendicular to a magnetic field, the Lorentz force acts as a centripetal force, bending the particle into a circular path. The magnetic force $qvB$ must equal the centripetal force $\frac{mv^2}{r}$ , which immediately tells us that heavier particles or those with less charge will trace larger circles.

The key insight: the radius is proportional to the particle's momentum-to-charge ratio. Two particles with the same velocity will have radii in the ratio of their $\frac{m}{q}$ values.

Step-by-step calculation

Write the force balance equation The magnetic force provides the centripetal acceleration:

$qvB = \frac{mv^2}{r}$

Solve for the radius Canceling one factor of $v$ from both sides:

$r = \frac{mv}{qB}$ $r = \frac{mv}{qB}$

Identify the particle properties

Proton: mass $m_p$ , charge $q_p = e$

α-particle (helium nucleus): mass $m_\alpha = 4m_p$ , charge $q_\alpha = 2e$

Write the radius for each particle

For the proton:

$r_p = \frac{m_p v}{eB}$
For the α-particle: …
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CBSE 2025Set IMPROVEMENT1 markMCQ
$Q. If the velocity of a charged particle moving perpendicular to the direction of a uniform magnetic field is doubled and the value of the magnetic field is halved, then the radius of the path of the charged particle will become:$
(a) 8 times
(b) Double
(c) 4 times
(d) 3 times
›Reveal solutionSolution

The radius of a charged particle's circular path in a magnetic field is $r = \dfrac{mv}{qB}$ ; doubling $v$ and halving $B$ makes $r$ four times larger.

For a charged particle moving perpendicular to a uniform magnetic field, the magnetic force provides the centripetal force: $qvB = \dfrac{mv^2}{r} \implies r = \dfrac{mv}{qB}$ .

…

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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, $m_1=m_2$ : …

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CBSE 2024Set 55/5/11 markMCQ
$m$
›Reveal solutionSolution
The radius of cyclotron motion depends on the ratio $m/q$ . When mass doubles and charge halves, the ratio quadruples, so the new radius is $4R$ .

The key idea here is that a charged particle moving perpendicular to a uniform magnetic field experiences a centripetal force provided by the magnetic Lorentz force. The radius of the circular path is not an independent quantity — it emerges from balancing these two forces.

For any such problem, always start from the force balance equation. The magnetic force is $qvB$ , and the centripetal force required for circular motion is $mv^2/R$ . Setting them equal gives the radius directly.

Write the force balance The magnetic force provides the centripetal force:

$qvB = \frac{mv^2}{R}$
Cancel one factor of $v$ (assuming $v \neq 0$ ):
$qB = \frac{mv}{R}$

Solve for the radius Rearranging:

$R = \frac{mv}{qB}$
This is the standard formula for the cyclotron radius (also called the gyroradius or Larmor radius). Notice that $R$ depends on the ratio $m/q$ , not on $m$ or $q$ individually.
$R = \frac{mv}{qB}$

Identify what changes

The problem states:

New mass: $m' = 2m$

New charge: $q' = q/2$ The magnetic field $B$ and the speed $v$ are not mentioned as changing, so we assume they remain the same. (This is a standard assumption in such problems unless stated otherwise.)

Find the new radius

Substitute the new values into the formula: …
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CBSE 2024Set 55/2/11 markMCQ
$\vec{p}$
›Reveal solutionSolution
The radius of circular motion in a perpendicular magnetic field depends on momentum, not mass or velocity separately. Since both particles have the same momentum, they trace the same radius — the assertion is true, and the reason correctly explains it.

Why this works — the core idea

When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force $\vec{F} = q(\vec{v} \times \vec{B})$ acts as a centripetal force. The particle is forced into a circular path whose radius depends on how much "oomph" (momentum) it has versus how strongly the field tries to bend it. The key insight: the radius formula $r = \frac{mv}{qB}$ is really $r = \frac{p}{qB}$ — it's momentum that matters, not mass or speed alone.

Start with the force balance. For a particle of charge $q$ , mass $m$ , and speed $v$ moving perpendicular to $\vec{B}$ , the magnetic force provides the centripetal force:

$qvB = \frac{mv^2}{r}$
Cancel one $v$ (valid since $v \neq 0$ ):
$qB = \frac{mv}{r} \quad \Rightarrow \quad r = \frac{mv}{qB}$

Rewrite in terms of momentum. Linear momentum $p = mv$ , so:

$r = \frac{p}{qB}$
This is the cleaner, more revealing form. The radius depends only on the magnitude of momentum, the charge magnitude, and the field strength — not on mass or velocity individually.

Apply to the given situation. Both the electron and the proton have the same momentum $\vec{p}$ (same magnitude and direction), and both have the same magnitude of charge $|q| = e$ (ignoring sign, which only affects direction of rotation, not radius). They enter the same magnetic field $\vec{B}$ with $\vec{p} \perp \vec{B}$ . Therefore:

$r_{\text{electron}} = \frac{p}{eB} = r_{\text{proton}}$
Both paths have identical radii.

Check the reason statement. …
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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). …

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Part (a)

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