Moving Charges and Magnetism — Class 12 Physics practice questions

Question 1

A school science club is designing a simple current detector for their project. They use a circular coil of 50 turns, each of radius 0.07 m, wound on a cylindrical soft-iron core. The coil carries a current of 2 A and is placed in a radial magnetic field of strength 0.5 T, so that its plane is always parallel to the field.

(i) What is the area of one turn of the coil?
(ii) Calculate the magnetic moment of the coil.
(iii) Find the torque acting on the coil in this position.
(iv) The students observe that doubling the current doubles the deflection. Name the law/principle that explains this proportionality between current and deflection in a moving coil galvanometer.

Accepted Answer

(i) Area of one turn of the coil:

Formula: A = π r²

A = π × (0.07)²
→ A = 3.14159 × 0.0049

∴ A = 1.54 × 10⁻² m²

(1 mark)

——————————————————

(ii) Magnetic moment of the coil:

The magnetic moment of a current-carrying coil is given by:

Formula: m = N I A

where N = number of turns, I = current, A = area of one turn.

Substituting:
m = 50 × 2 × 1.54 × 10⁻²

∴ m = 1.54 A m²

(1 mark)

——————————————————

(iii) Torque acting on the coil:

The torque on a magnetic dipole in a uniform magnetic field is given by:

Formula: τ = m B sin θ

In a moving coil galvanometer with a radial field and soft-iron core, the plane of the coil is always parallel to B⃗, so the angle between m⃗ and B⃗ is θ = 90°, giving sin 90° = 1.

τ = m × B × sin 90°
→ τ = 1.54 × 0.5 × 1

∴ τ = 0.77 N m

(1 mark)

——————————————————

(iv) The proportionality between current and deflection:

In a moving coil galvanometer at equilibrium, the deflecting torque τ = NIBA equals the restoring torque τ = kφ (where k is the torsion constant of the suspension).

This gives: φ = (NBA/k) × I, i.e., φ ∝ I.

This linear relationship — that deflection is directly proportional to the current — is the working principle of the moving coil galvanometer, which follows from the law that torque on a current loop in a magnetic field is τ = NIBA (since the field is always radial, sin θ = 1 always).

∴ The principle is: In a moving coil galvanometer, the deflection is directly proportional to the current flowing through it (φ ∝ I), made possible by the radial magnetic field maintained by the cylindrical soft-iron core.

(1 mark)

Question 2

A proton moving with velocity v enters a uniform magnetic field B⃗ directed perpendicular to its velocity. Which of the following correctly describes the resulting motion of the proton?

Accepted Answer

Option (B) is correct.

Explanation: When a charged particle moves perpendicular to a uniform magnetic field, the Lorentz force F⃗ = q(v⃗ × B⃗) acts always perpendicular to the velocity. Since the force does no work, the speed remains constant. This perpendicular force provides the centripetal acceleration, causing the proton to move in a circular path.

By Newton's second law: qvB = mv²/r  →  r = mv/qB = mv/eB.

∴ The proton moves in a circular path of radius r = mv/eB.

Question 3

A straight long current-carrying conductor carries a current I. At a perpendicular distance r from the conductor, the magnitude of the magnetic field B is given by:

Accepted Answer

Option (b) is correct.
Explanation: By Ampere's Circuital Law, ∮B⃗·dL⃗ = μ₀I_enc. For a circular Amperian loop of radius r coaxial with the wire, B(2πr) = μ₀I, giving B = μ₀I / 2πr. The field varies as 1/r (not 1/r²), so (b) is the correct expression.

Question 4

A galvanometer of resistance G = 30 Ω gives full-scale deflection for a current Ig = 3 mA. What is the value of the shunt resistance required to convert it into an ammeter of range 0 – 3 A?

Accepted Answer

Option (A) is correct.

Explanation: For conversion to an ammeter, shunt S = IgG / (I − Ig).

Here Ig = 3 mA = 3×10⁻³ A, G = 30 Ω, I = 3 A.

∴ S = (3×10⁻³ × 30) / (3 − 3×10⁻³) = 0.09 / 2.997 ≈ 0.03 Ω.

Question 5

A galvanometer of resistance G = 40 Ω gives full-scale deflection for a current of 2 mA. What is the value of the shunt resistance required to convert it into an ammeter of range 0 – 2 A?

Accepted Answer

Option (A) is correct.

Explanation: For converting a galvanometer into an ammeter, the shunt resistance S is given by:

S = IgG / (I − Ig)

Substituting Ig = 2 mA = 2×10−3 A, G = 40 Ω, I = 2 A:

S = (2×10−3 × 40) / (2 − 2×10−3)

S = 0.08 / 1.998

∴ S ≈ 0.04 Ω

Question 6

Assertion (A): A proton moving with velocity v⃗ along the positive x-axis enters a uniform magnetic field B⃗ directed along the positive z-axis. The proton moves in a circular path in the x-y plane.

Reason (R): The magnetic force on a moving charged particle is always perpendicular to its velocity.

Reason (R): The magnetic force on a moving charged particle is always perpendicular to its velocity.

Accepted Answer

Option (a) is correct.

Explanation: The magnetic force on a charged particle is given by F⃗ = q(v⃗ × B⃗). Since F⃗ ⟂ v⃗ always (Reason R is TRUE), the speed — and hence kinetic energy — of the particle remains constant. With v⃗ along x̂ and B⃗ along ẑ, the force F⃗ = q(v⃗ × B⃗) points along ŷ, continuously deflecting the proton within the x-y plane, producing a circular path (Assertion A is TRUE). Because the perpendicularity of F⃗ to v⃗ is precisely the reason no component of velocity develops along B⃗, and the particle is confined to a circular — not helical — orbit in the x-y plane, Reason (R) is the correct explanation of Assertion (A).

Question 7

A circular loop of radius 5 cm carries a current of 2 A. What is the magnitude of the magnetic moment of the loop?

Accepted Answer

Option (A) is correct.

Explanation: The magnetic moment of a current loop is given by m = IA, where A = πr² is the area of the loop.

Here I = 2 A, r = 5 cm = 5 × 10⁻² m.

∴ m = I × πr² = 2 × π × (5 × 10⁻²)² = 2 × π × 25 × 10⁻⁴ = π × 10⁻² A m².

Question 8

A straight wire of length L carrying current I is bent into a semicircular loop of radius R. Write the expression for the magnetic field at the centre of the semicircular loop. How does the field change if the same wire is bent into a quarter-circle?

Accepted Answer

By the Biot-Savart law, the magnetic field at the centre of a circular arc of radius R subtending angle θ (in radians) at the centre, carrying current I, is:

B = (μ₀ I θ) / (4π R)

For a semicircular loop, θ = π, and R = L/π (since the arc length L = Rθ = Rπ):

B_semi = (μ₀ I · π) / (4π R) = μ₀ I / (4R)

∴ B_semi = μ₀ I / 4R

For a quarter-circle, the same wire of length L is bent so that θ = π/2, giving R = L/(π/2) = 2L/π.

Substituting in B = (μ₀ I θ) / (4π R):

B_quarter = (μ₀ I · (π/2)) / (4π · (2L/π)) = (μ₀ I π) / (16L)

Since R_quarter = 2R_semi (the radius doubles for the same wire length), and θ halves:

∴ B_quarter = μ₀ I / (8R_quarter) = half the value of B_semi

The magnetic field at the centre is halved when the wire is bent into a quarter-circle instead of a semicircle.

Question 9

A straight conductor of length L carrying current I is placed in a uniform magnetic field B⃗. Write the expression for the magnetic force F⃗ acting on it. Using this expression, show that the force on the conductor is zero when it is placed parallel to the field B⃗.

Accepted Answer

The magnetic force on a current-carrying conductor of length L in a uniform magnetic field B⃗ is given by:

F⃗ = I (L⃗ × B⃗)

where L⃗ is a vector of magnitude L directed along the direction of current flow.

The magnitude of this force is:

|F⃗| = BIL sinθ

where θ is the angle between the direction of current (L⃗) and the magnetic field B⃗.

When the conductor is placed parallel to the field B⃗, the angle between L⃗ and B⃗ is θ = 0° (or 180°).

→ |F⃗| = BIL sin 0° = BIL × 0

∴ F⃗ = 0

Hence, the force on the conductor is zero when it is placed parallel to the magnetic field.

Question 10

A galvanometer has a resistance G = 50 Ω and gives full-scale deflection for a current Ig = 2 mA. How will you convert it into an ammeter capable of measuring currents up to 4 A? Calculate the value of the shunt resistance required.

Accepted Answer

To convert a galvanometer into an ammeter, a low resistance called a shunt (S) is connected in parallel with the galvanometer.

The shunt resistance is given by:

S = IgG / (I − Ig)

Substituting the given values (Ig = 2 mA = 2 × 10⁻³ A, G = 50 Ω, I = 4 A):

S = (2 × 10⁻³ × 50) / (4 − 2 × 10⁻³)
    S = 0.1 / 3.998

∴ S ≈ 0.025 Ω

Question 11

A science exhibit at a school fair demonstrates a simple velocity selector. A beam of singly charged ions (charge +e) is directed along the +x axis. A uniform electric field E⃗ = E ĵ (pointing in the +y direction, E = 4.8 × 10³ V/m) and a uniform magnetic field B⃗ = B k̂ (pointing in the +z direction, B = 0.12 T) are applied simultaneously in the same region.

(i) Write the condition on the velocity of an ion so that it passes through the selector undeflected.

(ii) Calculate the speed of ions that pass through undeflected.

(iii) An ion with twice this speed enters the same crossed-field region. State (with reason) the direction in which it deflects — towards +y or −y.

(iv) If the magnetic field is now doubled (B' = 0.24 T) while E is kept the same, what must be the new speed of ions for them to pass through undeflected? How does this new speed compare with the original?

Accepted Answer

(i) Condition for undeflected motion:

For an ion to pass through undeflected, the net force on it must be zero.

Electric force: F⃗_E = qE⃗ = eE ĵ (in +y direction)
Magnetic force: F⃗_B = q(v⃗ × B⃗) = e(v î × B k̂) = evB (î × k̂) = evB(−ĵ) (in −y direction)

∴ Condition: eE = evB

⟹ v = E/B

(ii) Calculation of speed:

Using v = E/B,

v = (4.8 × 10³) / (0.12)

∴ v = 4.0 × 10⁴ m/s

(iii) Direction of deflection for an ion with speed 2v:

At speed 2v, the magnetic force becomes:
F_B = e(2v)B = 2evB (in −y direction)

The electric force remains: F_E = eE = evB (in +y direction)

Since F_B > F_E, the net force is in the −y direction.

∴ The ion deflects towards −y direction (i.e., in the direction of the magnetic force, away from the electric force).

(iv) New speed with B' = 0.24 T:

Using the same condition v' = E/B',

v' = (4.8 × 10³) / (0.24)

∴ v' = 2.0 × 10⁴ m/s

Comparison: v' = v/2. When the magnetic field is doubled, the selected speed is halved. The velocity selector selects a speed inversely proportional to B (for fixed E).

Question 12

A hospital uses a galvanometer with coil resistance G = 50 Ω and full-scale deflection current Ig = 1 mA. A technician needs to:
(i) Convert it into an ammeter to measure currents up to 5 A.
(ii) Convert it into a voltmeter to measure voltages up to 10 V.

(a) Calculate the value of the shunt resistance S required to convert it into an ammeter. [2 marks]
(b) Calculate the value of the series resistance R required to convert it into a voltmeter. [2 marks]

Accepted Answer

Part (a): Converting galvanometer into an Ammeter [2 marks]

To convert a galvanometer into an ammeter, a low resistance shunt S is connected in parallel with it.

Since the galvanometer and shunt are in parallel, the potential difference across both is equal:

Ig × G = (I − Ig) × S

∴ S = (Ig × G) / (I − Ig)

Substituting values: Ig = 1 × 10⁻³ A, G = 50 Ω, I = 5 A

S = (1 × 10⁻³ × 50) / (5 − 1 × 10⁻³)

S = (50 × 10⁻³) / (4.999)

∴ S ≈ 0.01 Ω

(Value: S ≈ 1×10⁻² Ω)

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Part (b): Converting galvanometer into a Voltmeter [2 marks]

To convert a galvanometer into a voltmeter, a high resistance R is connected in series with it.

At full-scale deflection, the total voltage across the combination equals the maximum voltage to be measured:

V = Ig × (G + R)

∴ R = V/Ig − G

Substituting values: V = 10 V, Ig = 1 × 10⁻³ A, G = 50 Ω

R = (10) / (1 × 10⁻³) − 50

R = 10000 − 50

∴ R = 9950 Ω

Question 13

A rectangular coil PQRS of length 8 cm and width 5 cm carries a current of 3 A. It is placed in a uniform magnetic field of magnitude 0.4 T such that the plane of the coil makes an angle of 30° with the direction of the field. The coil has 50 turns.

(i) What is the angle between the magnetic moment vector of the coil and the magnetic field direction?
(ii) Calculate the torque acting on the coil.
(iii) A student suggests that if the plane of the coil is made perpendicular to the field, the torque will become maximum. Is the student correct? Justify your answer.
(iv) If this coil is used as a galvanometer and a shunt of 4 Ω is connected in parallel, and the galvanometer resistance is 36 Ω, what fraction of the total current passes through the galvanometer?

Accepted Answer

(i) The plane of the coil makes an angle of 30° with B⃗.
The magnetic moment vector m⃗ is perpendicular to the plane of the coil.
∴ Angle between m⃗ and B⃗ = 90° − 30° = 60°

(ii) The torque on a current-carrying coil in a magnetic field is given by:
τ = NIBA sinθ
where θ is the angle between m⃗ and B⃗, N = number of turns, I = current, B = magnetic field, A = area of coil.

Area A = length × width = 8 × 10⁻² × 5 × 10⁻² = 4 × 10⁻³ m²

Substituting values:
τ = 50 × 3 × 0.4 × 4 × 10⁻³ × sin 60°
τ = 50 × 3 × 0.4 × 4 × 10⁻³ × (√3/2)
τ = 50 × 3 × 0.4 × 4 × 10⁻³ × 0.866
τ = 0.24 × 0.866
∴ τ = 0.2078 ≈ 0.208 N m

(iii) No, the student is incorrect.
τ = NIBA sinθ, where θ is the angle between m⃗ and B⃗.
Torque is maximum when sinθ = 1, i.e., θ = 90°, meaning m⃗ ⊥ B⃗.
Since m⃗ is perpendicular to the plane of the coil, m⃗ ⊥ B⃗ corresponds to the plane of the coil being parallel to B⃗ (i.e., the plane of the coil contains the field direction).
When the plane of the coil is perpendicular to B⃗, θ = 0°, sinθ = 0, and torque = 0 (minimum, not maximum).
∴ The student's suggestion is incorrect. Maximum torque occurs when the plane of the coil is parallel to the magnetic field.

(iv) In a galvanometer converted to an ammeter, the shunt S is connected in parallel with galvanometer resistance G.
The potential difference across both is the same:
I_g × G = I_s × S
where I_g = current through galvanometer, I_s = current through shunt.
Total current: I = I_g + I_s

Given G = 36 Ω, S = 4 Ω:
I_g × 36 = I_s × 4
I_g/I_s = 4/36 = 1/9

So I_s = 9 I_g
Total current I = I_g + 9I_g = 10 I_g

∴ Fraction of total current through galvanometer = I_g/I = I_g/(10 I_g) = 1/10

Question 14

A space probe carries two charged-particle detectors. Detector 1 registers an alpha particle (charge 2e, mass 4u) moving with speed v, and Detector 2 registers a proton (charge e, mass 1u) moving with the same speed v. Both particles enter a region of uniform magnetic field B⃗ directed perpendicular to their velocities and travel in circular paths.

(i) Write the expression for the radius of a circular path followed by a charged particle of charge q, mass m, moving with speed v in a magnetic field B.

(ii) Calculate the ratio of the radius of the circular path of the alpha particle to that of the proton.

(iii) If the proton completes one full circle in time T_p, find the time T_α taken by the alpha particle to complete one full circle. Express your answer in terms of T_p.

(iv) The probe now enters a region where the magnetic field is doubled (2B) but the speed of the proton is also doubled (2v). How does the new radius of the proton's circular path compare with the original radius? Justify your answer.

Accepted Answer

(i) When a charged particle of charge q and mass m moves with speed v perpendicular to a uniform magnetic field B, the magnetic Lorentz force provides the centripetal force:

qvB = mv²/r

∴ r = mv / qB

(ii) Using r = mv / qB, the radius depends on m and q (since v and B are the same for both particles).

For the alpha particle: charge q_α = 2e, mass m_α = 4u
For the proton: charge q_p = e, mass m_p = u

r_α / r_p = (m_α / q_α) / (m_p / q_p)

r_α / r_p = (4u / 2e) / (u / e) = (2u/e) / (u/e) = 2

∴ r_α : r_p = 2 : 1

The alpha particle travels in a circle of twice the radius of the proton's circle.

(iii) The time period for one complete circle is:

T = 2πr / v = 2π(mv/qB) / v = 2πm / qB

Note: T is independent of speed v.

For the proton: T_p = 2πm_p / (q_p · B) = 2πu / (eB)

For the alpha particle: T_α = 2πm_α / (q_α · B) = 2π(4u) / (2e · B) = 2π(2u) / (eB)

∴ T_α / T_p = [2π(2u)/(eB)] / [2πu/(eB)] = 2

∴ T_α = 2 T_p

(iv) When the magnetic field is doubled to 2B and the proton's speed is doubled to 2v:

New radius r' = m_p(2v) / (e · 2B) = 2m_p v / 2eB = m_p v / eB = r (original radius)

∴ The new radius equals the original radius r_p.

Justification: The radius r = mv/qB. Doubling both v (numerator) and B (denominator) leaves their ratio unchanged, so r' = r. The effects of doubling speed and doubling field exactly cancel each other.

Question 15

A proton and an alpha particle are accelerated through the same potential difference V and then projected perpendicular to a uniform magnetic field B⃗.

(i) Derive the expression for the radius of the circular path of a charged particle moving perpendicular to a uniform magnetic field. (1 mark)

(ii) Find the ratio of the radii of the circular paths of the proton and the alpha particle. (2 marks)

(iii) A scientist wishes to separate these two particles using this magnetic field arrangement. She notes that although their radii differ, both particles complete one full circle in the same time interval. Explain whether this observation is correct or incorrect. (1 mark)

Accepted Answer

(i) Derivation of radius of circular path:

When a charged particle of mass m, charge q, and speed v moves perpendicular to a uniform magnetic field B⃗, the magnetic Lorentz force provides the centripetal force.

By Newton's second law (centripetal condition):

qvB = mv²/r

∴ r = mv / qB

(ii) Ratio of radii of proton and alpha particle:

Both particles are accelerated through the same potential difference V.

By work–energy theorem, the kinetic energy gained:

qV = ½mv² → v = √(2qV/m)

Substituting into r = mv/qB:

r = m·√(2qV/m) / qB = √(2mV/q) / B = (1/B)·√(2mV/q)

So r ∝ √(m/q)

For proton: mass mp = m, charge qp = e

For alpha particle: mass mα = 4m, charge qα = 2e

rp/rα = √(mp/qp) / √(mα/qα)

= √(m/e) / √(4m/2e)

= √(m/e) / √(2m/e)

= √(1/2)

∴ rp : rα = 1 : √2

(iii) Check whether the time period is the same for both particles:

The time period of circular motion in a magnetic field is:

T = 2πr/v = 2π(mv/qB)/v = 2πm/qB

For proton: Tp = 2πm/eB

For alpha particle: Tα = 2π(4m)/(2e)B = 4πm/eB

∴ Tp : Tα = 1 : 2

The observation is INCORRECT. The time period T = 2πm/qB depends on the mass-to-charge ratio (m/q). Since mp/qp = m/e and mα/qα = 4m/2e = 2m/e, the alpha particle has twice the time period of the proton. They do NOT complete one full circle in the same time.

Question 16

A rectangular current-carrying loop PQRS has its sides PQ = 8 cm and QR = 6 cm. It carries a steady current of 5 A. The loop is free to rotate about the axis YY′ which passes through the midpoints of sides QR and SP, as shown. A uniform magnetic field B⃗ = 0.25 T acts along the +x direction throughout the region.

(i) Calculate the magnetic moment of the loop.
(ii) What is the magnitude of the torque acting on the loop when the plane of the loop makes an angle of 30° with the direction of B⃗?
(iii) The loop is released from the position described in (ii). As it rotates, at what orientation will the torque be maximum? What is the value of that maximum torque?
(iv) If the current in the loop is doubled and the area is halved (keeping all other parameters the same), how does the maximum torque change? Justify your answer.

Accepted Answer

(i) The magnetic moment of a current-carrying loop is given by:
    m = N I A
where N = 1 (single loop), I = 5 A, A = PQ × QR = 8 × 10⁻² × 6 × 10⁻² = 48 × 10⁻⁴ m²
    m = 1 × 5 × 48 × 10⁻⁴
∴ m = 24 × 10⁻³ A m² = 2.4 × 10⁻² A m²

(ii) The torque on a magnetic dipole in a uniform field is given by:
    τ = m B sin θ
where θ is the angle between the magnetic moment m⃗ (normal to the loop) and B⃗.

Since the plane of the loop makes 30° with B⃗, the normal to the loop makes (90° − 30°) = 60° with B⃗.
Therefore θ = 60°.

τ = m B sin 60°
    τ = 2.4 × 10⁻² × 0.25 × (√3/2)
    τ = 2.4 × 10⁻² × 0.25 × 0.866
    τ = 2.4 × 10⁻² × 0.2165
∴ τ ≈ 5.2 × 10⁻³ N m

(iii) The torque τ = m B sin θ is maximum when sin θ = 1, i.e., when θ = 90°.
This occurs when the plane of the loop is parallel to B⃗ (i.e., the normal to the loop is perpendicular to B⃗).

Maximum torque:
    τ_max = m B sin 90° = m B
    τ_max = 2.4 × 10⁻² × 0.25
∴ τ_max = 6.0 × 10⁻³ N m

(iv) The maximum torque is τ_max = N I A B.
New current I′ = 2I, new area A′ = A/2.
New magnetic moment m′ = N I′ A′ = N (2I)(A/2) = N I A = m.

Since m′ = m, the maximum torque τ_max = m′ B = m B remains unchanged.
∴ The maximum torque does not change, because doubling the current and halving the area leave the magnetic moment (m = NIA) unchanged.

Question 17

A proton and an alpha particle are accelerated from rest through the same potential difference V and then enter a region of uniform magnetic field B⃗ directed perpendicular to their velocities.

(i) Write the expression for the radius of the circular path of a charged particle of mass m, charge q moving with speed v in a magnetic field B.

(ii) Using the expression above, find the ratio of the radii of the circular paths of the proton and the alpha particle.

(iii) If the same proton and alpha particle enter the magnetic field with the same kinetic energy (instead of the same potential difference), what would be the ratio of their radii? Give a reason for any difference in the two results.

Accepted Answer

(i) Radius of circular path in a magnetic field:

When a charged particle of mass m, charge q moves with speed v perpendicular to a uniform magnetic field B, the magnetic force provides the centripetal force:

qvB = mv²/r

∴  r = mv / qB

(1 mark)

(ii) Ratio of radii when accelerated through the same potential difference V:

For a particle accelerated from rest through potential difference V:

Kinetic energy = work done by electric field

½mv² = qV  →  v = √(2qV/m)

Substituting in r = mv/qB:

r = m√(2qV/m) / qB  =  √(2mV/q) / B

∴  r = (1/B) √(2mV/q)

For a proton:  m_p = m,  q_p = e
For an alpha particle:  m_α = 4m,  q_α = 2e

r_p / r_α = √(m_p / q_p) / √(m_α / q_α)
         = √(m/e) / √(4m/2e)
         = √(m/e) / √(2m/e)
         = √(1/2)

∴  r_p : r_α = 1 : √2  ≈  1 : 1.41

(1 mark)

(iii) Ratio of radii when both enter with the same kinetic energy K:

If kinetic energy K = ½mv² is the same for both, then:

mv = √(2mK)    (since mv = √(2m · ½mv²) = √(2mK))

Substituting in r = mv/qB:

r = √(2mK) / qB

For the same K and B:

r_p / r_α = (√(2m_p K) / q_p B) / (√(2m_α K) / q_α B)
          = (q_α / q_p) × √(m_p / m_α)
          = (2e / e) × √(m / 4m)
          = 2 × (1/2)
          = 1

∴  r_p : r_α = 1 : 1

Reason for the difference: When accelerated through the same potential difference, the alpha particle gains twice the kinetic energy of the proton (since K = qV and q_α = 2e > e = q_p). This extra kinetic energy increases the alpha particle's momentum more than would be expected from its larger mass alone, making r_α > r_p. However, when both particles enter with the same kinetic energy, the greater momentum of the alpha particle (due to larger mass) is exactly offset by its larger charge, so both travel in circles of equal radius.

∴ r_p : r_α = 1 : √2  (same potential difference) and  1 : 1  (same kinetic energy).

(1 mark for part iii result + 1 mark for correct reasoning)

Question 18

A biomedical engineer is designing a velocity selector for a mass spectrometer used to analyse blood-plasma ions. Singly charged calcium ions (Ca²⁺, mass = 6.64×10⁻²⁶ kg, charge = 2×1.6×10⁻¹⁹ C) enter a region where a uniform electric field E⃗ = 4×10⁴ V m⁻¹ (directed upward) and a uniform magnetic field B⃗₁ = 0.2 T (directed into the page) act simultaneously. Only ions moving at the correct speed pass through undeflected and enter the deflecting chamber.

(i) Find the speed of Ca²⁺ ions that pass through the velocity selector undeflected.

(ii) In the deflecting chamber, only the magnetic field B⃗₂ = 0.5 T (directed into the page) is present. Calculate the radius of the circular path traced by the selected Ca²⁺ ions.

(iii) After passing through the deflecting chamber, the same ions are accelerated through a potential difference V. Their speed increases to 4×10⁵ m s⁻¹. Determine the potential difference V applied.

(iv) Give ONE reason why a magnetic field alone cannot be used to select a specific speed in a velocity selector.

Accepted Answer

(i) Speed of undeflected ions:

For a charged particle to pass through the velocity selector undeflected, the electric force must exactly balance the magnetic force:

qE = qv B₁

∴ v = E / B₁

Substituting:
v = (4×10⁴) / (0.2)

∴ v = 2×10⁵ m s⁻¹

(ii) Radius of circular path in the deflecting chamber:

In the deflecting chamber, only B⃗₂ acts. The magnetic force provides the centripetal force:

qvB₂ = mv² / r  →  r = mv / qB₂

Given:
m = 6.64×10⁻²⁶ kg,  q = 2×1.6×10⁻¹⁹ = 3.2×10⁻¹⁹ C
v = 2×10⁵ m s⁻¹,  B₂ = 0.5 T

r = (6.64×10⁻²⁶ × 2×10⁵) / (3.2×10⁻¹⁹ × 0.5)

r = (1.328×10⁻²⁰) / (1.6×10⁻¹⁹)

∴ r = 0.083 m  (≈ 8.3 cm)

(iii) Potential difference for acceleration:

Using the work–energy theorem, the work done by the electric field equals the gain in kinetic energy:

qV = ½mv²_f − ½mv²_i

qV = ½m(v²_f − v²_i)

Substituting:
v_f = 4×10⁵ m s⁻¹,  v_i = 2×10⁵ m s⁻¹

qV = ½ × 6.64×10⁻²⁶ × [(4×10⁵)² − (2×10⁵)²]

qV = ½ × 6.64×10⁻²⁶ × [16×10¹⁰ − 4×10¹⁰]

qV = ½ × 6.64×10⁻²⁶ × 12×10¹⁰

qV = 3.984×10⁻¹⁵ J

V = (3.984×10⁻¹⁵) / (3.2×10⁻¹⁹)

∴ V = 1.245×10⁴ V  ≈ 1.25×10⁴ V

(iv) Reason why magnetic field alone cannot select speed:

A magnetic force on a moving charge is always perpendicular to the velocity (F⃗ = qv⃗×B⃗). It deflects the particle into a circular/curved path but does NO work and cannot balance any force along the direction of motion. Since it cannot exert a restoring or opposing force along the particle's direction of travel, it cannot single out one specific speed — it only changes direction, not speed. A velocity selector requires two forces (electric and magnetic) acting in opposite directions along the same line so that exact cancellation occurs only at one particular speed.

Question 19

A research team is designing a velocity selector for a mass spectrometer. Singly charged ions (charge +e) are accelerated through a potential difference V and then enter a region where a uniform electric field E⃗ (directed upward, magnitude E = 4.8 × 10⁴ V/m) and a uniform magnetic field B⃗ (directed into the page, magnitude B = 0.12 T) are applied simultaneously, perpendicular to the ion beam.

(i) Derive the condition for an ion to pass undeflected through the velocity selector. Hence write the expression for the selected velocity.

(ii) After exiting the velocity selector, the ions enter a second region containing only the magnetic field B⃗ (same magnitude, same direction — into the page) and follow a semicircular path of radius R = 0.25 m. Calculate the mass of the ion.

(iii) The team now replaces these ions with doubly charged ions (charge +2e) of the same mass accelerated through the same potential difference V. State with justification whether the radius of the semicircular path in the second region (magnetic field only) will increase, decrease, or remain the same.

(iv) A technician mistakenly reverses the direction of E⃗ (now directed downward) but keeps B⃗ unchanged (into the page). State the net direction of the resultant force on a positive ion moving to the right with the selected speed, and hence explain whether the ion will still pass undeflected.

Accepted Answer

(i) Condition for undeflected passage and selected velocity:

For an ion moving horizontally (say, to the right) in crossed E⃗ and B⃗ fields, the electric force F⃗_E = qE⃗ acts upward and the magnetic force F⃗_B = qv⃗ × B⃗ acts downward (by right-hand rule, with v⃗ rightward and B⃗ into the page).

For the ion to pass undeflected, the net force must be zero:

qE = qvB

∴ Selected velocity:  v = E/B

Substituting:  v = (4.8 × 10⁴) / (0.12)

∴ v = 4.0 × 10⁵ m/s

(ii) Mass of the ion:

In the second region, the ion (charge +e, speed v) moves in a circle of radius R under the magnetic force alone, which provides the centripetal force:

qvB = mv²/R

→  m = qBR/v

Substituting: q = 1.6 × 10⁻¹⁹ C, B = 0.12 T, R = 0.25 m, v = 4.0 × 10⁵ m/s:

m = (1.6 × 10⁻¹⁹ × 0.12 × 0.25) / (4.0 × 10⁵)

m = (4.8 × 10⁻²¹) / (4.0 × 10⁵)

∴ m = 1.2 × 10⁻²⁶ kg

(iii) Effect of doubling the charge on the radius:

In the velocity selector, the selected speed is v = E/B. This depends only on E and B, not on the charge. So the doubly charged ion exits the selector with the same speed v = 4.0 × 10⁵ m/s.

In the second region, the radius is:

R = mv / (qB)

For the doubly charged ion: q becomes 2e (doubles), m and v remain the same.

→  R' = mv / (2eB) = R/2

∴ The radius will decrease (to half the original value), because the magnetic force on the ion is doubled (F = qvB), providing a larger centripetal force for the same mass and speed, resulting in a smaller circular orbit.

(iv) Effect of reversing E⃗:

With E⃗ now directed downward: the electric force on the positive ion (F⃗_E = qE⃗) acts downward.

With B⃗ into the page and v⃗ to the right: by F⃗_B = qv⃗ × B⃗, using the right-hand rule (v⃗ = +x̂, B⃗ = −ẑ), F⃗_B = q(v x̂ × (−B ẑ)) = q v B (x̂ × (−ẑ)) = q v B ŷ — i.e., F⃗_B acts upward.

Both F⃗_E (downward) and F⃗_B (upward) now act in opposite directions but, at the selected speed v = E/B, their magnitudes are equal:

|F⃗_E| = qE  and  |F⃗_B| = qvB = qE

∴ The net force is ZERO, and the ion still passes undeflected.

However, this is because the forces still balance in magnitude — but both forces have simply swapped roles compared to the original configuration. The ion continues in a straight line with zero net force.

Question 20

A charged particle of charge q and mass m enters a region of uniform magnetic field B⃗ (directed into the page) with velocity v⃗ perpendicular to B⃗. 
(a) Name the shape of the path traced by the particle and write the expression for the radius of this path.
(b) If the kinetic energy of the particle is doubled (keeping B and q unchanged), by what factor does the radius of its circular path change?
(c) Two particles — a proton and an α-particle — enter the same uniform magnetic field B with the same kinetic energy. Find the ratio of the radii of their circular paths (r_p : r_α).

Accepted Answer

(a) When a charged particle enters a uniform magnetic field with velocity perpendicular to B⃗, the magnetic force F⃗ = q(v⃗ × B⃗) acts as the centripetal force — always perpendicular to v⃗ — so the particle traces a CIRCULAR path.

By Newton's second law (centripetal condition):

qvB = mv²/r

∴  r = mv / qB

[½ mark: name of path; ½ mark: correct expression for r]

(b) Express r in terms of kinetic energy K:

K = ½mv²  →  mv = √(2mK)

∴  r = mv/qB = √(2mK) / qB

So r ∝ √K  (for fixed m, q, B).

When kinetic energy is doubled: K′ = 2K

r′/r = √(K′/K) = √(2K/K) = √2

∴ The radius increases by a factor of √2.                [1 mark]

(c) Using r = √(2mK) / qB, and same K and same B for both particles:

r ∝ √m / q

For a proton:  m_p = m,  q_p = e
For an α-particle:  m_α = 4m,  q_α = 2e

r_p / r_α = (√m_p / q_p) / (√m_α / q_α)
             = (√m / e) / (√(4m) / 2e)
             = (√m / e) × (2e / 2√m)
             = (2e√m) / (2e√m)
             ... simplify step by step:

r_p / r_α = (√m_p × q_α) / (√m_α × q_p)
              = (√m × 2e) / (√(4m) × e)
              = (2e√m) / (2e√m)
              = 1 / 1

∴  r_p : r_α = 1 : 1                                   [1 mark]

The proton and the α-particle trace circular paths of EQUAL radius when entering the same magnetic field with the same kinetic energy.

Moving Charges and Magnetism: Class 12 Physics Practice Questions

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