Given the force, find the field. The questions she left blank on the ILTS — and they are easier than they look.
Part 1 — The concept, explained simply
Most questions give you v and B and ask for the force. These give you the force and ask you to work backwards — and she left both of them blank on the ILTS.
1. Forwards is easy, backwards feels impossible
Going forwards, F = q(v × B) is just a cross product. Going backwards, you know F and v and want B. That sounds like dividing by a cross product, which you cannot do — so it looks impossible.
It is not. The trick is to stop thinking of it as one equation and treat it as three ordinary equations, one for each of î, ĵ and k̂.
Picture it: a locked door with three bolts. You cannot open it in one move, but you can slide each bolt one at a time.
2. Expanding a cross product
If v = vxî + vyĵ + vzk̂ and B = Bxî + Byĵ + Bzk̂, then:
v × B = (vyBz − vzBy)î + (vzBx − vxBz)ĵ + (vxBy − vyBx)k̂
You do not need to memorise this. Write the 3×3 determinant with î ĵ k̂ on the top row, v underneath, B at the bottom, and expand. Or simply multiply out using the cyclic rule — every term is one of î×ĵ = k̂ and its relatives.
Splitting a vector equation into components is the whole technique. Once split, there is no vector algebra left — only ordinary arithmetic.
3. The two shortcuts that skip the algebra
Often you do not need to solve anything, because two facts are always true:
Shortcut 1 — the force is perpendicular to B. So F · B = 0. If you are told the acceleration and asked for an unknown component of B, just set the dot product to zero.
Shortcut 2 — the force is perpendicular to v. So F · v = 0. Useful for checking your answer, or for finding an unknown component of the velocity.
A dot product is far easier than a cross product: multiply matching components and add.
A · B = AxBx + AyBy + AzBz
4. Undeviated motion — the easiest case of all
If a charge passes through undeviated with only a magnetic field present, then v × B = 0, which means v is parallel to B. Parallel vectors have proportional components:
vx/Bx = vy/By = vz/Bz
Two ratios, two unknowns — solved in a line, with no cross product at all.
5. Why the answer is not always unique
Worth understanding, not just accepting. Any component of B that lies along v contributes nothing to the force, because v × v = 0. So from F and v alone you can never pin down that part of B. Exam questions get round this by giving you partial information about B — for instance telling you its k̂ component — and asking for the rest.
Part 2 — Formula sheet
Tool
Formula
When to use it
Cross product, expanded
(vyBz−vzBy)î + (vzBx−vxBz)ĵ + (vxBy−vyBx)k̂
When you must solve for components.
Determinant form
| î ĵ k̂ ; vx vy vz ; Bx By Bz |
Safer than memorising the expansion.
Dot product
A·B = AxBx+AyBy+AzBz
For the two perpendicularity shortcuts.
Force ⊥ field
F · B = 0 (also a · B = 0)
Find an unknown component of B or of a.
Force ⊥ velocity
F · v = 0
Check your answer, or find a v component.
Undeviated motion
v ∥ B ⇒ vx/Bx = vy/By = vz/Bz
Whenever "moves undeviated" appears.
Magnitude of a vector
|A| = √(Ax²+Ay²+Az²)
Final step of most such questions.
Force magnitude
|F| = q|v||B| sinθ
When only magnitudes are wanted.
Acceleration
a = F/m = (q/m)(v × B)
Specific charge questions.
Decision rule — which tool?
Question says "undeviated" → use v ∥ B, compare ratios.
Question gives a or F and asks for a missing component → use the dot product = 0.
Question gives full v and full F and asks for full B → expand and match components.
Question asks only for a magnitude → use qvB sinθ, skip the components.
Part 3 — 20 questions with step-by-step solutions
Attempt each on paper first. Every solution follows the same five steps: Given → Asked → Concept → Formula → Solution.
Q1Back-solve
In F = q(v × B) with q = 1, v = 2î + 4ĵ + 6k̂ and F = 4î − 20ĵ + 12k̂, and B of the form Bî + Bĵ + B0k̂, the complete expression for B is:
(a) 8î + 8ĵ − 6k̂
(b) 6î + 6ĵ − 8k̂
(c) −8î − 8ĵ − 6k̂
(d) −6î − 6ĵ − 8k̂
Show step-by-step solution
Givenq = 1, v = 2î+4ĵ+6k̂, F = 4î−20ĵ+12k̂, B = Bî+Bĵ+B0k̂
AskedThe vector B
ConceptExpand the cross product and match components. Note B has equal î and ĵ parts, leaving only two unknowns.
FormulaF = v × B, expanded component-wise
Solutionî: vyBz − vzBy = 4B0 − 6B = 4 ĵ: vzBx − vxBz = 6B − 2B0 = −20 Adding twice the first to the second: from 4B0−6B = 4 and 6B−2B0 = −20, solve to get B = −6, B0 = −8. So B = −6î − 6ĵ − 8k̂. Check the k̂ component: 2(−6) − 4(−6) = 12 ✓
Answer: −6î − 6ĵ − 8k̂
Q2Dot product
A charged particle has acceleration a = 2î + xĵ in a magnetic field B = −3î + 2ĵ − 4k̂. The value of x is:
(a) 2
(b) 3
(c) 4
(d) −3
Show step-by-step solution
Givena = 2î + xĵ, B = −3î + 2ĵ − 4k̂
AskedThe value of x
ConceptThe magnetic force, and therefore the acceleration, is always perpendicular to B. So their dot product is zero.
Formulaa · B = 0
Solutiona · B = (2)(−3) + (x)(2) + (0)(−4) = −6 + 2x = 0 x = 3.
Answer: 3
Q3Undeviated
A charge of 4 mC enters a field B = 3î + 6ĵ + 6k̂ T with velocity v = 4î − xĵ + yk̂ and moves undeviated. The magnitude of its velocity is:
(a) 10 m/s
(b) 15 m/s
(c) 12 m/s
(d) 8 m/s
Show step-by-step solution
GivenB = 3î+6ĵ+6k̂, v = 4î−xĵ+yk̂, motion undeviated
Asked|v|
ConceptUndeviated with only a magnetic field means v is parallel to B, so the components are proportional.
Formulavx/Bx = vy/By = vz/Bz
Solution4/3 = −x/6 = y/6 From 4/3 = −x/6: x = −8, so the ĵ component is +8. From 4/3 = y/6: y = 8. |v| = √(16 + 64 + 64) = √144 = 12 m/s.
Answer: 12 m/s
Q4Magnitude
A wire carrying current I along the positive x-axis has length L and is kept in a field B = (2î + 3ĵ − 4k̂) T. The magnitude of the magnetic force on it is:
(a) √3 IL
(b) 3 IL
(c) √5 IL
(d) 5 IL
Show step-by-step solution
Givenl = Lî, B = 2î + 3ĵ − 4k̂
Asked|F|
ConceptOnly the components of B perpendicular to the wire produce a force. The î part of B is parallel to l and contributes nothing.
FormulaF = I(l × B)
Solutionl × B = Lî × (2î + 3ĵ − 4k̂) î × î = 0; î × ĵ = k̂; î × k̂ = −ĵ = L(3k̂ + 4ĵ) |F| = IL√(9+16) = 5IL.
Answer: 5 IL
Q5Perpendicular
For a charged particle moving in a magnetic field, which of the following is always zero?
(a) F · B only
(b) F · v only
(c) both F · B and F · v
(d) neither
Show step-by-step solution
GivenMagnetic force on a moving charge
AskedWhich dot products vanish
ConceptA cross product is perpendicular to both of its inputs.
FormulaF = q(v × B)
SolutionF is the cross product of v and B. A cross product is perpendicular to both vectors it is built from. So F ⊥ v and F ⊥ B, making both dot products zero.
Answer: both F · B and F · v
Q6Back-solve
A proton of velocity (3î + 2ĵ) m/s enters a field (2ĵ + 3k̂) T. Given q/m = 0.96 × 108 C/kg, the acceleration produced is:
(a) 2.8 × 108(2î − 3ĵ)
(b) 2.88 × 108(2î − 3ĵ + 2k̂)
(c) 2.8 × 108(2î + 3k̂)
(d) 2.88 × 108(î − 3ĵ + 2k̂)
Show step-by-step solution
Givenv = 3î+2ĵ, B = 2ĵ+3k̂, q/m = 0.96×108
AskedAcceleration a
ConceptCompute v × B first, then multiply by q/m.
Formulaa = (q/m)(v × B)
Solutionv × B with v = (3,2,0), B = (0,2,3): î: (2)(3) − (0)(2) = 6 ĵ: (0)(0) − (3)(3) = −9 k̂: (3)(2) − (2)(0) = 6 v × B = 6î − 9ĵ + 6k̂ = 3(2î − 3ĵ + 2k̂) a = 0.96×108 × 3 × (2î−3ĵ+2k̂) = 2.88×108(2î−3ĵ+2k̂)
Answer: 2.88 × 108(2î − 3ĵ + 2k̂)
Q7Cyclic
The value of ĵ × î is:
(a) k̂
(b) −k̂
(c) î
(d) 0
Show step-by-step solution
GivenCross product of unit vectors
AskedResult
ConceptGoing backwards round the cyclic circle picks up a minus sign.
Formulaî × ĵ = k̂
SolutionForwards: î × ĵ = k̂. Reversing the order reverses the sign. So ĵ × î = −k̂.
Answer: −k̂
Q8Components
A charge q moves with v = vî in a field B = Bxî + Byĵ. The force on it is:
(a) qvBxk̂
(b) qvByk̂
(c) qv(Bx+By)k̂
(d) zero
Show step-by-step solution
Givenv = vî, B = Bxî + Byĵ
AskedForce F
ConceptThe part of B parallel to v does nothing; only the perpendicular part counts.
FormulaF = q(v × B)
Solutionv × B = vî × (Bxî + Byĵ) î × î = 0, so the Bx term vanishes. î × ĵ = k̂, so we are left with vByk̂. F = qvByk̂.
Answer: qvByk̂
Q9Undeviated
A charged particle moves undeviated through a region containing only a magnetic field. This means:
(a) the field is zero
(b) the velocity is parallel or antiparallel to B
(c) the charge is zero
(d) the speed is very high
Show step-by-step solution
GivenUndeviated motion, magnetic field only
AskedWhat must be true
ConceptThe only way for the magnetic force to vanish with a moving charge is v ∥ B.
Formulav × B = 0
SolutionNo deflection means no force. F = q(v × B) = 0 with q ≠ 0 and v ≠ 0. So v × B = 0, meaning v is parallel or antiparallel to B.
Answer: the velocity is parallel or antiparallel to B
Q10Numerical
A charge of 2 C moves with v = 3î m/s in B = 4ĵ T. The force on it is:
(a) 24k̂ N
(b) 12k̂ N
(c) 24î N
(d) 6k̂ N
Show step-by-step solution
Givenq = 2 C, v = 3î, B = 4ĵ
AskedForce F
ConceptStraight cross product, then multiply by the charge.
FormulaF = q(v × B)
Solutionv × B = 3î × 4ĵ = 12(î × ĵ) = 12k̂ F = 2 × 12k̂ = 24k̂ N.
Answer: 24k̂ N
Q11Dot product
A charged particle experiences force F = 3î + 4ĵ in a field B = aî + 2ĵ. The value of a is:
(a) 8/3
(b) −8/3
(c) 3/8
(d) −3/8
Show step-by-step solution
GivenF = 3î + 4ĵ, B = aî + 2ĵ
AskedThe value of a
ConceptThe force is perpendicular to B, so their dot product vanishes.
FormulaF · B = 0
SolutionF · B = 3a + 8 = 0 3a = −8 a = −8/3.
Answer: −8/3
Q12Magnitude
A charge q moves at speed v at 30° to a field B. The magnitude of the force is:
(a) qvB
(b) qvB/2
(c) qvB√3/2
(d) zero
Show step-by-step solution
GivenAngle between v and B is 30°
Asked|F|
ConceptOnly the magnitude is wanted, so use the sine form and skip the components entirely.
Formula|F| = qvB sinθ
Solutionsin30° = 1/2. |F| = qvB × 1/2 = qvB/2.
Answer: qvB/2
Q13Determinant
The safest way to compute v × B without memorising the expansion is to:
(a) use the dot product
(b) write a 3×3 determinant with î ĵ k̂ on the top row
(c) add the components
(d) take the magnitudes and multiply
Show step-by-step solution
GivenComputing a cross product
AskedRecommended method
ConceptThe determinant layout builds in all the signs automatically.
Formula| î ĵ k̂ ; vx vy vz ; Bx By Bz |
SolutionPut î, ĵ, k̂ in the first row. Put the components of v in the second row and of B in the third. Expand along the top row; the alternating signs come out automatically.
Answer: write a 3×3 determinant with î ĵ k̂ on the top row
Q14Uniqueness
From F and v alone, the component of B along v:
(a) can always be found
(b) can never be found
(c) is always zero
(d) equals |B|
Show step-by-step solution
GivenGiven F and v, solving for B
AskedWhether B is fully determined
ConceptAny part of B parallel to v produces no force, so it leaves no trace in F.
Formulav × v = 0
SolutionWrite B as a part along v plus a part perpendicular to v. The parallel part crossed with v gives zero, so it does not appear in F. Hence it cannot be recovered from F and v alone.
Answer: can never be found
Q15Application
A particle of charge q and mass m has velocity v = v0ĵ in a field B = B0k̂. Its acceleration is:
(a) (qv0B0/m)î
(b) (qv0B0/m)ĵ
(c) (qv0B0/m)k̂
(d) zero
Show step-by-step solution
Givenv = v0ĵ, B = B0k̂
AskedAcceleration a
ConceptCross product then divide by mass.
Formulaa = (q/m)(v × B)
Solutionĵ × k̂ = î. v × B = v0B0î a = (q/m)v0B0î.
Answer: (qv0B0/m)î
Q16Check
A student computes F = 5î + 2ĵ for a particle moving with v = 2î + 3ĵ. Without knowing B, this answer is:
(a) certainly correct
(b) certainly wrong
(c) possibly correct
(d) impossible to judge
Show step-by-step solution
GivenF = 5î + 2ĵ, v = 2î + 3ĵ
AskedIs the answer consistent?
ConceptThe force must be perpendicular to the velocity, so their dot product must be zero. Use this to check any answer.
FormulaF · v = 0
SolutionF · v = (5)(2) + (2)(3) = 10 + 6 = 16. This is not zero, so F is not perpendicular to v. The answer must therefore be wrong. This check takes five seconds and catches most slips.
Answer: certainly wrong
Q17Numerical
A charge of 1 C moving with v = î + ĵ m/s in a field B = k̂ T experiences a force of magnitude:
(a) 1 N
(b) √2 N
(c) 2 N
(d) zero
Show step-by-step solution
Givenq = 1 C, v = î+ĵ, B = k̂
Asked|F|
ConceptCross product, then take the magnitude.
FormulaF = q(v × B)
Solutionî × k̂ = −ĵ and ĵ × k̂ = î. v × B = −ĵ + î |F| = √(1+1) = √2 N.
Answer: √2 N
Q18Undeviated
A particle moves undeviated through crossed electric and magnetic fields. If E = 100î V/m and B = 0.5k̂ T, the speed is:
(a) 50 m/s
(b) 200 m/s
(c) 100 m/s
(d) 500 m/s
Show step-by-step solution
GivenE = 100î V/m, B = 0.5k̂ T, undeviated
AskedSpeed v
ConceptHere both fields are present, so the velocity-selector condition applies rather than v ∥ B.
Formulav = E/B
SolutionFor no deflection the electric and magnetic forces must cancel. qE = qvB ⇒ v = E/B v = 100/0.5 = 200 m/s.
Answer: 200 m/s
Q19Components
If v = 2î and B = 3ĵ + 4k̂, then v × B is:
(a) 8ĵ − 6k̂
(b) −8ĵ + 6k̂
(c) 6ĵ + 8k̂
(d) 6k̂ − 8ĵ
Show step-by-step solution
Givenv = 2î, B = 3ĵ + 4k̂
Askedv × B
ConceptDistribute over the two terms of B, using the cyclic rule for each.
A question gives v and F and asks for B, and also tells you the k̂ component of B. The best method is:
(a) guess and check
(b) expand the cross product and match components
(c) use v = E/B
(d) take magnitudes only
Show step-by-step solution
GivenPartial information about B supplied
AskedBest approach
ConceptWith one component known, the remaining two follow from matching the î and ĵ equations.
FormulaExpand F = q(v × B) component-wise
SolutionExpand the cross product to get three scalar equations. Substitute the known k̂ component. Solve the remaining two equations for the two unknowns. The extra information is given precisely because B is otherwise not unique.
Answer: expand the cross product and match components