One strange fact runs through this whole topic: a magnetic field can steer a moving charge, but it can never make it go faster or slower.
1. The Lorentz force — the total push on a charge
Put a charge q moving with velocity v into a region that has both an electric field and a magnetic field. The total push it feels is:
F = q [ E + v × B ] = Felectric + Fmagnetic
Read it as: total push = electric push + magnetic push. The electric part is the old friend from Chapter 1. The magnetic part is the new one, and notice what sits inside it — v. No movement, no magnetic push.
Two pushes on the same particle. Notice the sideways one is always at right angles to the motion, however fast the charge goes — that single fact runs through this whole topic.
2. Three rules about the magnetic part
Rule 1 — It depends on q, v and B, and it flips for a negative charge.
An electron and a proton flying side by side through the same field bend apart, one up and one down.
Rule 2 — It vanishes when the charge moves along the field.
If the charge flies straight along the field lines (θ = 0° or 180°), it feels nothing at all. The field only pushes when the charge cuts across it.
Picture it: swimming in a river. Swim straight downstream with the current and the water doesn't shove you sideways. Only if you cut across does it push you.
Rule 3 — There is no force at all on a stationary charge.
Park the charge and the magnet ignores it completely. This is the deepest difference from electricity: an electric field pushes anything charged, moving or not; a magnetic field only bothers with things that are moving.
Three ways the magnetic force behaves. The middle and right panels are the ones NEET dresses up — "projected along the axis of a solenoid", "a charge at rest in a field". Both mean zero force.
Putting the size together:
F = q v B sin θ
The sin θ is the "cutting across" idea in maths form. Moving along the field: θ = 0, sin 0 = 0, no force. Moving straight across: θ = 90°, sin 90° = 1, maximum force.
3. The direction — sideways to everything
The strangest rule of all: the push is never forwards, never backwards, never along the field. It is sideways to both the velocity and the field at once.
How to find it with your hand: point your right fingers along v, curl them towards B, and your thumb sticks out along the force. For a negative charge such as an electron, work it out for a positive charge first and then reverse the answer.
The force is never forwards and never backwards. It is always exactly sideways — which is why it can steer the particle but can never change how fast it goes.
4. The big consequence — no work, ever
A force can only change an object's speed if some part of it points forwards or backwards along the motion. The magnetic force is always exactly sideways. Therefore:
A magnetic field does no work on a charged particle. Its speed never changes. Its kinetic energy never changes. Its momentum's size never changes. Only the direction changes.
This is asked in almost every paper in some form, so it must be automatic.
Compare with the electric force, which can point along the motion and so can change energy. That is the contrast NEET tests: "does a magnetic field change the KE of a charged particle?" The answer is always no.
The force is perpendicular to the motion at every instant, so it does no work. Careful though: the momentum vector does change, because its direction changes — only its size is fixed.
5. The circle
Now suppose the charge cuts straight across the field (θ = 90°). The push is always at right angles to its motion — and a force that is always at right angles, always pointing at one fixed spot, is exactly what makes a circle. It acts as the centripetal force.
Picture it: whirling a stone on a string around your head. The string pulls the stone sideways, towards your hand, all the time. The stone never gets faster or slower — it just keeps turning. Let go and it flies off in a straight line at the same speed it always had.
Setting the magnetic force equal to the centripetal force:
m v² / r = q v B ⇒ r = m v / q B
Since mv is momentum, this also reads r = p / qB. Bigger momentum → wider circle. Stronger field → tighter circle. Just like a car: go faster and you can't take the corner as sharply.
Same job, different agent. A force that is always sideways and always aimed at one fixed point is a centripetal force — so the path has to be a circle.
6. The surprise — time is independent of speed
From v = ωr and r = mv/qB:
ω = q B / m T = 2π m / q B ν = q B / 2π m
Look at what is missing from all three: there is no v in them.
Speed up the particle and it travels a bigger circle — but it covers that bigger circle in exactly the same time. The two effects cancel perfectly. Double the speed and the time period does not budge. This is the single most repeated conceptual question from section 4.3.
Watch the two dots stay perfectly in step, lap after lap. Doubling the speed doubles the radius, so the extra distance is covered at exactly the extra speed — and the two effects cancel.
7. The helix
What if the charge enters at a slant rather than straight across? Split its velocity into two parts:
The part across the field, v sin θ, gets bent into a circle.
The part along the field, v cos θ, is completely ignored by the magnetic force and just keeps going steadily.
Do both at once and you get a spiral — going round and round while drifting steadily forward. Like a screw being turned into wood: rotating and advancing at the same time.
The pitch is how far it advances in one full turn — the gap between the threads of the screw:
p = v∥ T = 2π m v cos θ / q B rhelix = m v sin θ / q B
A screw being turned into wood: rotating and advancing at the same time. Split the velocity into the part across the field and the part along it, and the helix is just those two motions happening together.
The whole of section 4.3 in one table. Everything depends on the angle at which the charge enters.
Entry angle θ
Across the field
Along the field
Path
0° or 180°
zero
full v
Straight line — no force at all
90°
full v
zero
Circle
anything else
v sin θ
v cos θ
Helix
8. The three ratio problems — read the question carefully
NEET's favourite trick here is to give two particles and ask for the ratio of their radii. The formula changes depending on which quantity is held fixed, and the three versions look almost identical on the page.
What is the same for both
Use
So r depends on
Same speed v
r = mv / qB
r ∝ m / q
Same momentum p
r = p / qB
r ∝ 1 / q
Same kinetic energy K
r = √(2mK) / qB
r ∝ √m / q
Same accelerating voltage V
r = (1/B)√(2mV/q)
r ∝ √(m/q)
Before writing anything, underline the word that says what is fixed — "same velocity", "equal momenta", "equal kinetic energy", "accelerated through the same potential". That one word decides which of the four rows you are in. Choosing the wrong row is the commonest way to lose this mark.
Nothing about the particles changes; only the phrase in the question changes. Underline the word that says what is fixed before writing anything down.
9. Bonus — crossed fields (NEET has asked this)
Send a charge through a region where E and B are at right angles to each other and to the motion. The electric push and the magnetic push can be made to cancel exactly, so the particle sails through undeflected:
q E = q v B ⇒ v = E / B
This arrangement is called a velocity selector: only particles travelling at exactly E/B get through in a straight line, whatever their charge or mass. NEET 2025 used exactly this idea.
A different case worth knowing: if E and B point in the same direction and the charge is projected along them, then v × B = 0 so there is no magnetic force at all — only the electric one, which speeds the particle up or slows it down along a straight line.
Go faster and the magnetic push wins, bending you one way. Go slower and the electric push wins, bending you the other. Only at v = E/B do they balance exactly.
10. Units of magnetic field
The force formula also defines how we measure field strength. One tesla is the field in which a charge of 1 C moving at 1 m/s perpendicular to the field feels a force of 1 N. Tesla is a big unit, so a smaller one is often used: 1 gauss = 10⁻⁴ T. For scale, the Earth's field is only about 3.6 × 10⁻⁵ T — less than a gauss.
The whole Earth manages only about 0.000036 T. That is exactly why Oersted needed a large current and a compass held close — his wire had to out-shout the entire planet.
Part 2 — Formula sheet
Everything in sections 4.2.2 and 4.3, on one page.
Quantity
Formula
Watch out for
Lorentz force (full)
F = q(E + v × B)
Electric part + magnetic part.
Magnetic force (magnitude)
F = q v B sin θ
θ is between v and B.
Maximum force
F = q v B
When v ⊥ B (θ = 90°).
Zero force
F = 0
If v = 0, or v ∥ B, or v antiparallel to B.
Work done by magnetic force
W = 0
Always. Speed and KE never change.
Radius of the circular path
r = m v / q B
The single most-used formula here.
Radius from momentum
r = p / q B
Use when "equal momenta" is stated.
Radius from kinetic energy
r = √(2 m K) / q B
Use when "equal kinetic energy" is stated.
Radius after accelerating through V
r = (1/B) √(2 m V / q)
Since K = qV.
Angular frequency
ω = q B / m
No v in it.
Time period
T = 2π m / q B
Independent of speed and of radius.
Frequency
ν = q B / 2π m
Also independent of speed.
Momentum from the path
p = q B r
Rearranged radius formula.
Kinetic energy from the path
K = q²B²r² / 2m
From K = p²/2m.
Pitch of the helix
p = 2π m v cos θ / q B
Uses only the along-field part of v.
Radius of the helix
r = m v sin θ / q B
Uses only the across-field part of v.
Velocity selector
v = E / B
Undeflected motion in crossed fields.
Unit conversion
1 gauss = 10⁻⁴ T
Earth's field ≈ 3.6 × 10⁻⁵ T.
The four facts that answer most conceptual questions in this topic
A magnetic force does no work; speed and KE are unchanged.
A stationary charge feels no magnetic force.
A charge moving along the field feels no magnetic force.
T and ν do not depend on speed — only on q, B and m.
Part 3 — 50 questions with step-by-step solutions
Attempt each one on paper first, then open the solution. Every solution follows the same five steps: Given → Asked → Concept → Formula → Solution.
Q1Lorentz force · numeric
An electron moving with a speed of 10⁷ m/s enters a magnetic field of 2 × 10⁻⁴ T at right angles to it. The force on the electron is:
(a) 1.6 × 10⁻¹⁶ N
(b) 3.2 × 10⁻¹⁶ N
(c) 3.2 × 10⁻¹⁵ N
(d) 1.6 × 10⁻¹⁵ N
Show step-by-step solution
Givenq = 1.6 × 10⁻¹⁹ C, v = 10⁷ m/s, B = 2 × 10⁻⁴ T, θ = 90°
AskedMagnetic force F
ConceptAt right angles sin θ = 1, so the force is simply qvB.
A charge of 1 μC moves with a speed of 10⁶ m/s at 30° to a magnetic field of 0.5 T. The force on it is:
(a) 0.125 N
(b) 0.25 N
(c) 0.5 N
(d) 1 N
Show step-by-step solution
Givenq = 1 μC = 10⁻⁶ C, v = 10⁶ m/s, B = 0.5 T, θ = 30°
AskedMagnetic force F
ConceptThe angle is not 90°, so the sin θ factor must be included.
FormulaF = q v B sin θ
Solutionsin 30° = 0.5 F = 10⁻⁶ × 10⁶ × 0.5 × 0.5 = 1 × 0.25 = 0.25 N
Answer: 0.25 N
Q3Lorentz force · concept
A charged particle experiences no magnetic force while moving through a magnetic field. This is because:
(a) its charge is zero
(b) it is moving parallel to the field
(c) the field is zero
(d) it is moving perpendicular to the field
Show step-by-step solution
GivenCharged particle in a magnetic field feels no force
AskedThe reason
ConceptCheck the sin θ factor. It vanishes when the velocity lies along the field.
FormulaF = q v B sin θ
SolutionIf v is parallel (or antiparallel) to B, then θ = 0° or 180°. sin 0° = sin 180° = 0 So F = 0 even though q, v and B are all non-zero.
Answer: it is moving parallel to the field
Q4Lorentz force · concept
A stationary charge is placed in a uniform magnetic field. The force acting on it is:
(a) qB
(b) qB/2
(c) zero
(d) infinite
Show step-by-step solution
GivenCharge at rest, v = 0, in a field B
AskedMagnetic force on it
ConceptThe velocity appears as a factor in the magnetic force, so no motion means no force.
FormulaF = q v B sin θ
Solutionv = 0 F = q × 0 × B × sin θ = 0 Only a MOVING charge feels a magnetic force — unlike an electric field, which pushes it either way.
Answer: zero
Q5Lorentz force · concept
A charged particle moves in a uniform magnetic field. Which of the following remains unchanged?
(a) its velocity
(b) its momentum
(c) its kinetic energy
(d) the direction of its motion
Show step-by-step solution
GivenCharged particle moving in a magnetic field
AskedThe quantity that stays constant
ConceptThe magnetic force is always perpendicular to the velocity, so it does no work.
FormulaW = 0 ⇒ KE constant
SolutionThe force is always perpendicular to v, so no work is done. No work means the kinetic energy cannot change. Speed is constant, but velocity and momentum change because their DIRECTION changes. So the kinetic energy is the unchanged quantity.
Answer: its kinetic energy
Q6Lorentz force · direction
A proton moves along the +x axis in a magnetic field directed along the +y axis. The force on the proton is along:
(a) +z
(b) −z
(c) +y
(d) −x
Show step-by-step solution
Givenv along +x, B along +y, positive charge
AskedDirection of the force
ConceptUse the cross product v × B with the standard cyclic rule, then apply the sign of the charge.
FormulaF = q (v × B); î × ĵ = k̂
Solutionv × B is along î × ĵ = k̂, i.e. the +z direction. The proton is positive, so F is in the same direction as v × B. F is along +z.
Answer: +z
Q7Lorentz force · direction
An electron moves along the +x axis in a magnetic field directed along the +y axis. The force on it is along:
(a) +z
(b) −z
(c) +x
(d) −y
Show step-by-step solution
Givenv along +x, B along +y, negative charge
AskedDirection of the force
ConceptWork it out for a positive charge first, then reverse it because the electron is negative.
FormulaF = q (v × B), with q negative
Solutionv × B = î × ĵ = k̂, i.e. +z. The electron carries a negative charge, so the force is opposite to v × B. F is along −z.
Answer: −z
Q8Lorentz force · concept
Two particles, a proton and an electron, enter the same magnetic field with the same velocity, perpendicular to the field. They will:
(a) both bend the same way
(b) bend in opposite directions
(c) travel in straight lines
(d) come to rest
Show step-by-step solution
GivenProton and electron, same v, same B, both perpendicular
AskedDirections of bending
ConceptThe sign of the charge decides the sign of the force, so opposite charges bend opposite ways.
FormulaF = q (v × B)
SolutionFor the proton (positive), F is along v × B. For the electron (negative), F is opposite to v × B. So they bend in opposite directions — one curves up, the other down.
Answer: bend in opposite directions
Q9Units
One gauss is equal to:
(a) 10⁻² T
(b) 10⁻⁴ T
(c) 10⁴ T
(d) 10⁻⁷ T
Show step-by-step solution
GivenRelation between gauss and tesla
AskedValue of 1 gauss in tesla
ConceptTesla is a large unit, so gauss is used for weak fields such as the Earth's.
Formula1 gauss = 10⁻⁴ T
Solution1 gauss = 10⁻⁴ tesla The Earth's field is about 3.6 × 10⁻⁵ T, which is less than 1 gauss.
Answer: 10⁻⁴ T
Q10Crossed fields
An electron travelling at 3 × 10⁶ m/s enters a region where a magnetic field of 9 × 10⁻⁴ T acts perpendicular to its motion. The electric field that must be applied so that the electron passes through undeflected is:
ConceptFor undeflected motion the electric force must exactly balance the magnetic force. This is the velocity-selector condition.
Formulaq E = q v B ⇒ E = v B
SolutionThe charge q cancels from both sides. E = v B = 3 × 10⁶ × 9 × 10⁻⁴ = 27 × 10² = 2.7 × 10³ V/m
Answer: 2.7 × 10³ V/m
Q11Crossed fields
In a velocity selector, the electric field is 3.2 × 10⁵ V/m and the magnetic field is 0.4 T. The speed of the particles that pass through undeflected is:
(a) 8 × 10⁴ m/s
(b) 8 × 10⁵ m/s
(c) 1.28 × 10⁵ m/s
(d) 1.25 × 10⁶ m/s
Show step-by-step solution
GivenE = 3.2 × 10⁵ V/m, B = 0.4 T
AskedSelected speed v
ConceptOnly particles with v = E/B travel straight through, whatever their charge or mass.
Formulav = E / B
Solutionv = 3.2 × 10⁵ / 0.4 = 8 × 10⁵ m/s
Answer: 8 × 10⁵ m/s
Q12Crossed fields · concept
An electric field and a magnetic field point in the SAME direction, and an electron is projected along that same direction. The electron will:
(a) move in a circle
(b) move in a helix
(c) slow down along a straight line
(d) continue with constant speed
Show step-by-step solution
GivenE parallel to B, electron projected along both
AskedResulting motion
ConceptIf v is parallel to B then v × B = 0, so there is no magnetic force at all — only the electric one acts.
FormulaF = q(E + v × B), with v × B = 0
Solutionv is parallel to B ⇒ v × B = 0 ⇒ no magnetic force. Only the electric force qE remains, and it acts along the line of motion. For an electron (negative charge) this force opposes the motion, so the electron slows down while moving in a straight line.
Answer: slow down along a straight line
Q13Circular motion · numeric
An electron (mass 9 × 10⁻³¹ kg, charge 1.6 × 10⁻¹⁹ C) moves at 3 × 10⁷ m/s perpendicular to a magnetic field of 6 × 10⁻⁴ T. The radius of its path is about:
(a) 14 cm
(b) 28 cm
(c) 56 cm
(d) 2.8 cm
Show step-by-step solution
Givenm = 9 × 10⁻³¹ kg, v = 3 × 10⁷ m/s, q = 1.6 × 10⁻¹⁹ C, B = 6 × 10⁻⁴ T
AskedRadius r of the circular path
ConceptThe magnetic force provides the centripetal force, giving r = mv/qB.
A charged particle moving perpendicular to a magnetic field has time period T. If its speed is doubled, the new time period is:
(a) T / 2
(b) T
(c) 2T
(d) 4T
Show step-by-step solution
GivenSpeed doubled, same q, m and B
AskedNew time period
ConceptLook at the formula for T — there is no v in it at all. This is the most-repeated conceptual question in the topic.
FormulaT = 2π m / q B
SolutionT depends only on m, q and B. Doubling the speed doubles the radius, but the particle covers the bigger circle at twice the speed. The two effects cancel exactly, so T is unchanged.
Answer: T
Q17Circular motion · ratio
A proton and an alpha particle enter a magnetic field perpendicularly with the SAME velocity. The ratio of the radii of their paths (proton : alpha) is:
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 4
Show step-by-step solution
GivenSame speed v; proton (m, q), alpha (4m, 2q)
AskedRatio r_proton : r_alpha
ConceptWith the speed fixed, r depends on m/q. An alpha particle has 4 times the mass and 2 times the charge.
Ionised hydrogen and an alpha particle with EQUAL MOMENTA enter a magnetic field perpendicularly. The ratio of the radii of their circular paths is:
(a) 1 : 4
(b) 1 : 2
(c) 2 : 1
(d) 4 : 1
Show step-by-step solution
GivenEqual momenta; hydrogen ion charge q, alpha charge 2q
AskedRatio of radii
ConceptWith momentum fixed, use r = p/qB — the mass drops out entirely and only the charge matters.
Formular = p / q B ⇒ r ∝ 1 / q
Solutionp and B are the same for both. r_H / r_alpha = q_alpha / q_H = 2q / q = 2 Ratio = 2 : 1
Answer: 2 : 1
Q19Circular motion · ratio
A proton and an alpha particle having the SAME KINETIC ENERGY enter a magnetic field perpendicularly. The ratio of their radii is:
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 4
Show step-by-step solution
GivenSame kinetic energy K; proton (m, q), alpha (4m, 2q)
AskedRatio of radii
ConceptWith kinetic energy fixed, momentum is √(2mK), so r depends on √m / q.
Formular = √(2 m K) / q B ⇒ r ∝ √m / q
SolutionProton: √m / q Alpha: √(4m) / 2q = 2√m / 2q = √m / q The two are identical, so the ratio is 1 : 1.
Answer: 1 : 1
Q20Circular motion · ratio
Two particles of the same charge but masses in the ratio 1 : 2 enter a magnetic field perpendicularly with the same speed. The ratio of their radii is:
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 4
Show step-by-step solution
GivenSame q, same v, same B; m₁ : m₂ = 1 : 2
AskedRatio r₁ : r₂
ConceptWith everything else fixed, the radius is directly proportional to mass.
Formular = m v / q B ⇒ r ∝ m
Solutionr₁ / r₂ = m₁ / m₂ = 1 / 2 Ratio = 1 : 2
Answer: 1 : 2
Q21Circular motion · numeric
The frequency of revolution of an electron in a magnetic field of 3.57 × 10⁻² T, given e/m = 1.76 × 10¹¹ C/kg, is:
(a) 1 MHz
(b) 100 MHz
(c) 1 GHz
(d) 10 GHz
Show step-by-step solution
GivenB = 3.57 × 10⁻² T, e/m = 1.76 × 10¹¹ C/kg
AskedFrequency ν
ConceptThe specific charge e/m is given directly, so use the frequency formula written as (e/m)B/2π.
If the magnetic field acting on a charged particle in a circular path is doubled while its speed stays the same, the radius becomes:
(a) double
(b) half
(c) four times
(d) unchanged
Show step-by-step solution
GivenB → 2B, same v, q and m
AskedNew radius
ConceptRead the dependence from the formula: r is inversely proportional to B.
Formular = m v / q B ⇒ r ∝ 1 / B
SolutionDoubling B halves r. r′ = r / 2 — a stronger field bends the particle more tightly.
Answer: half
Q23Circular motion · concept
The time taken by a charged particle to complete a HALF circle in a magnetic field is:
(a) 2πm/qB
(b) πm/qB
(c) πm/2qB
(d) 4πm/qB
Show step-by-step solution
GivenHalf a revolution in a magnetic field
AskedTime taken
ConceptHalf a circle takes half the time period, and T is independent of speed.
FormulaT = 2πm / qB; t = T/2
SolutionT = 2πm / qB t = T / 2 = πm / qB Note this does not depend on the particle's speed.
Answer: πm/qB
Q24Circular motion · derived
A charged particle moves in a circle of radius r in a magnetic field B. Its momentum is:
(a) qBr
(b) qB/r
(c) qr/B
(d) Br/q
Show step-by-step solution
GivenCircular path of radius r in field B, charge q
AskedMomentum p
ConceptRearrange the radius formula to make momentum the subject.
Formular = p / qB
SolutionFrom r = p/qB, multiply both sides by qB. p = q B r
Answer: qBr
Q25Circular motion · derived
A particle of charge q and mass m moves in a circle of radius r in a field B. Its kinetic energy is:
(a) q²B²r²/2m
(b) qBr/2m
(c) q²B²r/2m
(d) qB²r²/2m
Show step-by-step solution
GivenCharge q, mass m, radius r, field B
AskedKinetic energy K
ConceptGet the momentum first from p = qBr, then use K = p²/2m.
Formulap = qBr; K = p² / 2m
Solutionp = qBr K = p²/2m = (qBr)² / 2m K = q²B²r² / 2m
Answer: q²B²r²/2m
Q26Circular motion · accelerated
A charged particle is accelerated from rest through a potential difference V and then enters a magnetic field B perpendicularly. The radius of its path is:
(a) (1/B)√(2mV/q)
(b) (1/B)√(2qV/m)
(c) B√(2mV/q)
(d) (1/B)√(mV/2q)
Show step-by-step solution
GivenAccelerated through potential V, then enters field B perpendicularly
AskedRadius r
ConceptThe accelerating voltage gives the kinetic energy, K = qV. Then use r = √(2mK)/qB.
FormulaK = qV; r = √(2mK) / qB
SolutionK = qV, so p = √(2mK) = √(2mqV) r = p / qB = √(2mqV) / qB = (1/B) √(2mV/q)
Answer: (1/B)√(2mV/q)
Q27Circular motion · concept
The work done by the magnetic force on a charged particle completing one full circular revolution is:
(a) 2πrqvB
(b) qvB
(c) zero
(d) ½mv²
Show step-by-step solution
GivenOne complete revolution in a magnetic field
AskedWork done by the magnetic force
ConceptThe force is perpendicular to the displacement at every instant, so no work is done at any point of the path.
FormulaW = F·d = F d cos 90° = 0
SolutionAt every point the magnetic force is perpendicular to the velocity. So the work done in each small step is zero. Total work over the whole revolution = zero. (This is why the speed never changes.)
Answer: zero
Q28Circular motion · graph
For a charged particle moving perpendicular to a fixed magnetic field, the graph of radius r against speed v is:
(a) a horizontal straight line
(b) a straight line through the origin
(c) a rectangular hyperbola
(d) a parabola
Show step-by-step solution
GivenFixed q, m and B; varying speed v
AskedShape of the r versus v graph
ConceptRead the relationship from the formula: r is directly proportional to v.
Formular = m v / q B ⇒ r ∝ v
SolutionWith m, q and B fixed, r = (m/qB) × v. This is of the form y = kx. So the graph is a straight line passing through the origin. (By contrast, the T versus v graph is a horizontal line.)
Answer: a straight line through the origin
Q29Circular motion · geometry
A charged particle enters a region of magnetic field of width d, perpendicular to the field. If the radius of its path is r, the angle through which it is deflected is given by:
(a) sin θ = d / r
(b) cos θ = d / r
(c) tan θ = d / r
(d) θ = d / r
Show step-by-step solution
GivenField region of width d, path radius r
AskedDeflection angle θ
ConceptDraw the arc inside the strip: the width d is the perpendicular from the entry point, forming a right triangle with the radius as hypotenuse.
FormulaGeometry of the circular arc
SolutionThe particle travels along an arc of radius r inside the strip. The horizontal distance covered is d, and the radius r is the hypotenuse. So sin θ = d / r.
Answer: sin θ = d / r
Q30Helix · concept
A charged particle enters a uniform magnetic field at an angle of 60° to the field. Its path will be:
(a) a straight line
(b) a circle
(c) a helix
(d) a parabola
Show step-by-step solution
GivenEntry angle 60° to the field
AskedShape of the path
ConceptSplit the velocity into a part across the field (which curves) and a part along the field (which does not).
Formulav⊥ = v sin θ, v∥ = v cos θ
SolutionThe component v sin 60° across the field bends into a circle. The component v cos 60° along the field is unaffected and carries the particle forward steadily. Circular motion plus steady forward drift = a helix.
Answer: a helix
Q31Helix · formula
The pitch of the helical path of a charged particle entering a magnetic field at an angle θ is:
(a) 2πmv sin θ / qB
(b) 2πmv cos θ / qB
(c) 2πm / qB
(d) mv sin θ / qB
Show step-by-step solution
GivenCharge q, mass m, speed v, entry angle θ, field B
AskedPitch of the helix
ConceptThe pitch is the distance travelled ALONG the field in one full revolution, so it uses v cos θ.
Formulapitch = v∥ T = (v cos θ)(2πm/qB)
SolutionTime for one revolution: T = 2πm/qB Forward speed along the field: v∥ = v cos θ pitch = v cos θ × 2πm/qB = 2πmv cos θ / qB
Answer: 2πmv cos θ / qB
Q32Helix · formula
The radius of the helical path of a charged particle entering a field at angle θ is:
(a) mv / qB
(b) mv cos θ / qB
(c) mv sin θ / qB
(d) mv tan θ / qB
Show step-by-step solution
GivenCharge q, mass m, speed v, entry angle θ, field B
AskedRadius of the helix
ConceptOnly the component of velocity ACROSS the field takes part in the circular motion.
Formular = m v⊥ / qB with v⊥ = v sin θ
SolutionThe across-field component is v sin θ. r = m (v sin θ) / qB = mv sin θ / qB
Answer: mv sin θ / qB
Q33Helix · ratio
A charged particle enters a magnetic field at 45° to it. The ratio of the pitch of its helical path to the radius of the helix is:
(a) π
(b) 2π
(c) π/2
(d) 4π
Show step-by-step solution
GivenEntry angle θ = 45°
AskedRatio pitch : radius
ConceptWrite both expressions and divide — most factors cancel, leaving a neat trigonometric result.
Formulapitch = 2πmv cos θ/qB; r = mv sin θ/qB
Solutionpitch / r = (2πmv cos θ / qB) ÷ (mv sin θ / qB) = 2π cos θ / sin θ = 2π cot θ At θ = 45°, cot 45° = 1 So the ratio is 2π.
Answer: 2π
Q34Helix · concept
A charged particle is projected along the direction of a uniform magnetic field. Its path is:
(a) a circle
(b) a helix
(c) a straight line
(d) a parabola
Show step-by-step solution
GivenVelocity parallel to the field, θ = 0°
AskedShape of the path
ConceptWith θ = 0 there is no across-field component, so no bending occurs at all.
FormulaF = q v B sin θ, with θ = 0
Solutionsin 0° = 0, so the magnetic force is zero. With no force acting, the particle continues in a straight line at constant speed.
Answer: a straight line
Q35Helix · concept
Which component of the velocity is responsible for the circular part of a helical path?
(a) v cos θ
(b) v sin θ
(c) the full v
(d) neither component
Show step-by-step solution
GivenCharged particle entering at angle θ
AskedComponent causing the circular motion
ConceptThe magnetic force acts only on the part of the velocity that cuts across the field.
Formulav⊥ = v sin θ
SolutionThe component along the field, v cos θ, feels no force. The component across the field, v sin θ, is bent into a circle. So v sin θ is responsible for the circular part.
Answer: v sin θ
Q36Helix · numeric
A charged particle takes time T to complete one revolution in a magnetic field. In a time interval of 5T, the number of revolutions it completes is:
(a) 2.5
(b) 5
(c) 10
(d) depends on its speed
Show step-by-step solution
GivenTime period T, total time 5T
AskedNumber of revolutions
ConceptDivide the total time by the time for one revolution. Note that the speed is irrelevant, since T does not depend on it.
Formulan = total time / T
Solutionn = 5T / T = 5 revolutions The answer does not depend on the speed, because T = 2πm/qB has no v in it.
Answer: 5
Q37Comparison
Which of the following statements correctly distinguishes the electric and magnetic forces on a charged particle?
(a) Both can change the particle's kinetic energy
(b) Neither can change the kinetic energy
(c) The electric force can change the kinetic energy but the magnetic force cannot
(d) The magnetic force can change the kinetic energy but the electric force cannot
Show step-by-step solution
GivenCharged particle in electric and magnetic fields
AskedThe correct distinction
ConceptAn electric force can have a component along the motion; a magnetic force never can.
FormulaF_electric = qE; F_magnetic = q(v × B)
SolutionThe electric force qE can point along or against v, so it does work and changes the energy. The magnetic force is always perpendicular to v, so it does no work. Hence only the electric force can change the kinetic energy.
Answer: The electric force can change the kinetic energy but the magnetic force cannot
Q38Concept · specific charge
Two particles with the same charge but different masses enter the same magnetic field with the same speed. The particle that takes LONGER to complete one revolution is:
(a) the lighter one
(b) the heavier one
(c) both take the same time
(d) it depends on the field strength
Show step-by-step solution
GivenSame q, same v, same B; different masses
AskedWhich has the greater time period
ConceptT = 2πm/qB, so the time period is directly proportional to mass.
FormulaT = 2π m / q B
SolutionT ∝ m when q and B are fixed. The heavier particle therefore has the longer time period. (Its circle is also bigger, since r ∝ m.)
Answer: the heavier one
Q39Concept · direction
A positive charge moves vertically downwards in a region where the Earth's horizontal magnetic field points from south to north. The charge is deflected towards the:
(a) north
(b) south
(c) east
(d) west
Show step-by-step solution
Givenv vertically downwards, B horizontal from south to north, positive charge
AskedDirection of deflection
ConceptSet up axes and use the cross product carefully — do not guess.
FormulaF = q (v × B)
SolutionTake x = east, y = north, z = up. v is downwards ⇒ v = −ẑ. B is northwards ⇒ B = ŷ. v × B = (−ẑ) × ŷ = −(ẑ × ŷ) = −(−x̂) = +x̂ The charge is positive, so the force is along +x, i.e. towards the east.
Answer: east
Q40Concept · energy
A charged particle enters a uniform magnetic field with kinetic energy K. After completing a quarter circle, its kinetic energy is:
(a) K/4
(b) K/2
(c) K
(d) 2K
Show step-by-step solution
GivenInitial kinetic energy K, quarter circle completed
AskedFinal kinetic energy
ConceptThe magnetic force does no work, so the kinetic energy cannot change at any point of the path.
FormulaW = 0 ⇒ K constant
SolutionThe magnetic force is always perpendicular to the velocity. So no work is done and the speed stays constant. The kinetic energy is still K.
Answer: K
Q41Concept · combination
A charged particle moving with velocity v in a region passes through undeflected. This is possible if:
(a) only a magnetic field is present, perpendicular to v
(b) only an electric field is present, perpendicular to v
(c) both fields are present and qE = qvB with the forces opposing
(d) it is impossible for both fields to be present
Show step-by-step solution
GivenParticle passes through a field region undeflected
AskedCondition for no deflection
ConceptUndeflected means the net force is zero, so the two forces must be equal in size and opposite in direction.
Formulaq E = q v B
SolutionA magnetic field alone perpendicular to v would bend it into a circle. An electric field alone perpendicular to v would bend it into a parabola. Both together can cancel exactly when qE = qvB and the forces oppose — the velocity selector.
Answer: both fields are present and qE = qvB with the forces opposing
Q42Assertion–Reason
Assertion (A): A magnetic field cannot change the speed of a charged particle. Reason (R): The magnetic force always acts perpendicular to the velocity of the particle.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Show step-by-step solution
GivenCharged particle in a magnetic field
AskedTruth of A and R, and whether R explains A
ConceptCheck each statement, then ask whether the reason genuinely causes the assertion.
FormulaF = q(v × B); W = 0
SolutionA is TRUE: a magnetic field cannot change speed or kinetic energy. R is TRUE: a cross product is always perpendicular to both its inputs, including v. And that perpendicularity is exactly WHY no work is done and the speed cannot change. So R correctly explains A.
Answer: Both A and R are true, and R is the correct explanation of A
Q43Assertion–Reason
Assertion (A): The time period of a charged particle moving in a circular path in a magnetic field depends on its speed. Reason (R): A faster particle moves in a circle of larger radius.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Show step-by-step solution
GivenCharged particle in circular motion in a magnetic field
AskedTruth of A and R
ConceptThe second statement is correct but the first is not — a classic pairing where a true reason supports a false assertion.
FormulaT = 2πm/qB; r = mv/qB
SolutionA is FALSE: T = 2πm/qB contains no v, so the period does not depend on speed. R is TRUE: r = mv/qB, so a faster particle does travel a larger circle. The larger circle is covered at proportionally higher speed, so the time stays the same. Hence A is false but R is true.
Answer: A is false but R is true
Q44Assertion–Reason
Assertion (A): A charged particle at rest in a magnetic field experiences no force. Reason (R): The magnetic force is proportional to the velocity of the charge.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Show step-by-step solution
GivenStationary charge in a magnetic field
AskedTruth of A and R
ConceptBoth statements follow directly from the force formula.
FormulaF = q v B sin θ
SolutionA is TRUE: with v = 0 the force is zero. R is TRUE: v appears as a multiplying factor in the force expression. And that is precisely why the force vanishes at rest, so R explains A.
Answer: Both A and R are true, and R is the correct explanation of A
Q45Match the following
Match each situation with the resulting path of the charged particle:
(a) v parallel to B (b) v perpendicular to B (c) v at 30° to B (d) v = 0
(i) Circle (ii) Helix (iii) Remains at rest (iv) Straight line
(a) a–iv, b–i, c–ii, d–iii
(b) a–i, b–iv, c–ii, d–iii
(c) a–iv, b–ii, c–i, d–iii
(d) a–ii, b–i, c–iv, d–iii
Show step-by-step solution
GivenFour different entry conditions
AskedCorrect matching
ConceptSplit the velocity into across-field and along-field parts in each case.
Formulav⊥ = v sin θ, v∥ = v cos θ
Solution(a) v ∥ B ⇒ no force ⇒ straight line ⇒ (iv) (b) v ⊥ B ⇒ all of v bends ⇒ circle ⇒ (i) (c) v at 30° ⇒ both components present ⇒ helix ⇒ (ii) (d) v = 0 ⇒ no magnetic force ⇒ remains at rest ⇒ (iii)
Answer: a–iv, b–i, c–ii, d–iii
Q46Match the following
Match each quantity with its correct expression for a charge q of mass m moving perpendicular to a field B:
(a) Radius (b) Time period (c) Angular frequency (d) Momentum
(i) qB/m (ii) mv/qB (iii) qBr (iv) 2πm/qB
(a) a–ii, b–iv, c–i, d–iii
(b) a–ii, b–i, c–iv, d–iii
(c) a–iii, b–iv, c–i, d–ii
(d) a–ii, b–iv, c–iii, d–i
Show step-by-step solution
GivenStandard results for circular motion in a magnetic field
AskedCorrect matching
ConceptThese four are the core formulas of section 4.3 — worth being able to write down instantly.
Formular = mv/qB; T = 2πm/qB; ω = qB/m; p = qBr
Solution(a) Radius = mv/qB ⇒ (ii) (b) Time period = 2πm/qB ⇒ (iv) (c) Angular frequency = qB/m ⇒ (i) (d) Momentum = qBr ⇒ (iii)
Answer: a–ii, b–iv, c–i, d–iii
Q47Match the following
Match each condition with the correct dependence of the radius:
(a) Same speed (b) Same momentum (c) Same kinetic energy (d) Same accelerating potential
(i) r ∝ 1/q (ii) r ∝ m/q (iii) r ∝ √(m/q) (iv) r ∝ √m/q
(a) a–ii, b–i, c–iv, d–iii
(b) a–i, b–ii, c–iii, d–iv
(c) a–ii, b–iv, c–i, d–iii
(d) a–iii, b–i, c–iv, d–ii
Show step-by-step solution
GivenFour versions of the radius formula
AskedCorrect matching
ConceptThis is the ratio-problem table in question form — the version NEET rotates between year after year.
Solution(a) Same v: r = mv/qB ⇒ r ∝ m/q ⇒ (ii) (b) Same p: r = p/qB ⇒ r ∝ 1/q ⇒ (i) (c) Same K: r = √(2mK)/qB ⇒ r ∝ √m/q ⇒ (iv) (d) Same V: r = (1/B)√(2mV/q) ⇒ r ∝ √(m/q) ⇒ (iii)
Answer: a–ii, b–i, c–iv, d–iii
Q48Recognition
A question states: 'Two ions with equal kinetic energies enter the same magnetic field perpendicularly. Find the ratio of the radii of their paths.' Which formula should be used?
(a) r = mv/qB
(b) r = p/qB
(c) r = √(2mK)/qB
(d) r = (1/B)√(2mV/q)
Show step-by-step solution
GivenTwo ions with EQUAL KINETIC ENERGIES
AskedThe correct form of the radius formula
ConceptThe phrase that says what is held constant decides the formula. Here it is kinetic energy.
Formular = √(2 m K) / q B
SolutionThe fixed quantity is kinetic energy K, not speed, momentum or voltage. Momentum in terms of K is p = √(2mK). So r = √(2mK)/qB, giving r ∝ √m / q.
Answer: r = √(2mK)/qB
Q49Concept · summary
Which of the following statements about a charged particle moving in a uniform magnetic field is INCORRECT?
(a) Its speed remains constant
(b) Its kinetic energy remains constant
(c) Its momentum remains constant
(d) Its time period is independent of its speed
Show step-by-step solution
GivenCharged particle in a uniform magnetic field
AskedThe incorrect statement
ConceptCheck each one. The trap is the difference between the SIZE of the momentum and the momentum itself, which is a vector.
FormulaW = 0; T = 2πm/qB; p = mv (a vector)
Solution'Speed constant' — correct, no work is done. 'Kinetic energy constant' — correct, for the same reason. 'Momentum constant' — INCORRECT. Its magnitude is constant, but its DIRECTION keeps changing, so the momentum vector is not constant. 'Time period independent of speed' — correct, T = 2πm/qB.
Answer: Its momentum remains constant
Q50Circular motion · numeric
A proton (mass 1.67 × 10⁻²⁷ kg, charge 1.6 × 10⁻¹⁹ C) moves at 10⁶ m/s perpendicular to a magnetic field of 0.5 T. The radius of its path is about:
(a) 1.0 cm
(b) 2.1 cm
(c) 4.2 cm
(d) 8.4 cm
Show step-by-step solution
Givenm = 1.67 × 10⁻²⁷ kg, v = 10⁶ m/s, q = 1.6 × 10⁻¹⁹ C, B = 0.5 T
AskedRadius r
ConceptStraight substitution into r = mv/qB. Keep the powers of ten organised.