Move a conductor through a magnetic field and it becomes a battery. This file finds its voltage, its plus end, the force needed to keep it moving and where that energy goes.
Charges sorted by motion. The rod carries its free charges through the field. Each feels a magnetic force along the rod. They pile up at the ends until the electric field they create stops any more piling.
Picture it
Imagine a bus driving through a steady side wind. Passengers are pushed towards one side of the aisle. They crowd there until the crowd itself pushes back as hard as the wind. The rod's free charges are the passengers; moving through the field is the wind.
In exam language
A charge q moving with the rod feels F = q(v × B). Charges build up until the electric force balances it: E = vB. Across a rod of length ℓ:
e = (v × B) · ℓ when v, B and ℓ are mutually perpendicular: e = Bℓv
If any two of v, B and ℓ are parallel, the emf is zero.
Trap
Which end is at higher potential? Work out v × B: point your fingers along v and curl them towards B; your thumb points to the end where positive charge collects. That end is the higher-potential end (the + terminal of the rod as a battery). Inside the rod, current flows from the low end to the high end, just as inside a battery.
Tilted rods and curved wires
Effective length. Replace any wire by the straight line joining its ends. Only the part of that line perpendicular to both v and B counts.
The emf in a wire of any shape, moving through a uniform field, equals the emf in an imaginary straight wire joining its two ends (Allen p. 121). For a straight rod at angle θ to its velocity, only ℓ sin θ cuts across the motion:
e = Bv(ℓ sin θ) = B(v sin θ)ℓ
Trap
A semicircular wire whose ends lie along the direction of motion, or an inverted V whose two ends are level with the velocity, has zero emf between its ends, however long the wire (Beginner's Box 7 Q7(iii)).
A rod sliding on rails
Rod on rails. The moving rod is the battery, the resistor is the load. The current in the rod feels a magnetic force backwards, so an external push is needed to keep the speed steady.
emf e = Bℓv
current I = Bℓv/R
backward force F = BIℓ = B²ℓ²v/R
power needed P = Fv = B²ℓ²v²/R
heat per second I²R = B²ℓ²v²/R (equal: energy is conserved)
If the speed v is doubled
Factor
emf, current, force
× 2 (∝ v)
Mechanical power, heat per second
× 4 (∝ v²)
Allen Illustration 27. A square loop of side ℓ, mass m and resistance R falls with its top edge in a horizontal field. It speeds up until the upward magnetic force B²ℓ²v/R equals mg, giving a terminal velocity v = mgR/B²ℓ². Checked: correct.
A loop crossing a field region
Allen Illustration 26. Only the edges that are cutting field lines produce emf. Fully inside, both side edges cut equally and cancel, so the emf is zero even though the flux is largest.
L = 40 mm, b = 10 cm, d = 15 cm, R = 1.6 Ω, B = 2.0 T, v = 1.0 m/s. Maximum flux = BLb = 2 × 0.04 × 0.10 = 8 mWb. Emf while entering or leaving = BLv = 2 × 0.04 × 1 = 80 mV, negative while flux rises and positive while it falls. Checked: correct.
Stage
Flux
Emf
Current (R = 1.6 Ω)
Entering (x = 0 to 10 cm)
Rising
−80 mV
50 mA
Fully inside (10 to 15 cm)
8 mWb, constant
0
0
Leaving (15 to 25 cm)
Falling
+80 mV
50 mA the other way
Formula sheet
Formula
Meaning
When to use
Watch out
e = (v × B) · ℓ
General motional emf
Any straight rod
Zero if any two are parallel
e = Bℓv
All three perpendicular
Rod across rails, wing of a plane
B is the component ⊥ to v and ℓ
e = Bvℓ sin θ
Rod at angle θ to v
Tilted rods
Use the projection ⊥ v
e = Bv × (end-to-end length ⊥ v)
Bent or curved wire
Semicircles, V-shapes
Ends in line with v → 0
I = Bℓv/R
Current on rails
Closed circuit
R is total resistance
F = B²ℓ²v/R
Retarding force
Force to keep v constant
∝ v
P = B²ℓ²v²/R
Power and heat rate
Energy questions
∝ v²
vT = mgR/B²ℓ²
Terminal speed
Falling loop or rod on vertical rails
Set B²ℓ²v/R = mg
Allen pages 120–125 checked
Item
Check
Motional emf from Lorentz force; V = (v × B) · ℓ
Correct
Emf in wire acb equals emf in straight ab
Correct
Rails: I, F, Pmech = Pheat
Correct
Illustration 26: 8 mWb, ±80 mV
Correct
Illustration 27: v = mgR/B²ℓ²
Correct
Beginner's Box 7 answer key
Q
Answer
Working
1 (i)
No emf; neither end is higher
v is parallel to B, so v × B = 0.
1 (ii)
B out of the page
v is down the page and the left end L is higher, so v × B must point left. (−ĵ) × k̂ = −î: B = out of page.
1 (iii)
B into the page
v to the right, top end higher: v × B must point up. î × (−k̂) = +ĵ.
1 (iv)
Current in PQ flows from Q to P
The lower end N is higher, so B at the rod must be out of the page. Below a wire, that needs current from right to left.
1 (v)
Lower end N is the low-potential end
Field from N pole to S pole points right; v is out of the page. v × B = k̂ × î = +ĵ, so the top end L is higher.
1 (vi)
X is N, Y is S
Lower end N higher with v out of page needs B to point left at the rod, beyond end Y on the axis: field points into Y, so Y is S.
1 (vii)
X is S, Y is N
Top end higher with v into page needs B to point left above the magnet's middle. On the equatorial line the field runs from the N end towards the S end, so N is on the right (Y).
2
(1) 3.75 × 10⁻³ N
F = B²ℓ²v/R = 0.15² × 0.5² × 2/3.
3
(1) 2.25 × 10⁻³ N
The 3 cm edge cuts the field: e = 0.5 × 0.03 × 0.01 = 1.5 × 10⁻⁴ V, I = 0.15 A, F = BIℓ = 0.5 × 0.15 × 0.03.
4
(3) 2 × 10⁻² m/s, clockwise
The five 3 Ω resistors form a balanced bridge: 3 Ω. Total R = 4 Ω, so e = 4 mV = Bℓv gives v = 0.004/(2 × 0.1). The loop's inside edge is moving out of the field, so the flux into the page falls and the current is clockwise.
5
(4) 3 × 10⁻⁴ V, zero
At t = 2 s the loop is entering: e = 0.6 × 0.05 × 0.01. At t = 10 s it is fully inside the 20 cm region: zero.
6
(1) Straight line, parabola
F = B²ℓ²v/R ∝ v; P = B²ℓ²v²/R ∝ v².
7 (i)
2BvR, upper end L higher
Chord of the semicircle is 2R, perpendicular to v.
7 (ii)
2Bvℓ(1 + sin θ), upper end L higher
Vertical distance from L to N: ℓ + ℓ sin θ + ℓ sin θ + ℓ, perpendicular to v.
7 (iii) a, b, c
Zero in all three
Both ends lie on a line parallel to v: no perpendicular length.
NEET practice: 35 questions
Motional emf is one of the most frequently tested parts of this chapter. Direction questions state the field and velocity in words so that no drawing can be misread. Every number was recalculated in Python.
Q1Numerical
A rod of length 20 cm moves at 5 m/s perpendicular to its length and to a uniform field of 0.5 T. The emf between its ends is
(A)0.5 V
(B)50 V
(C)0.05 V
(D)2 V
Show the solution
Given
ℓ = 0.2 m, v = 5 m/s, B = 0.5 T, all perpendicular
Asked
e
Concept
Motional emf.
Formula
e = Bℓv
Baby steps
0.5 × 0.2 × 5 = 0.5 V.
Answer
(A) 0.5 V
Why not the others
50 V uses ℓ = 20 m. 0.05 V divides by 10. 2 V divides B by v.
Shortcut
B × ℓ × v.
Where it went wrong
Not converting cm to m.
Q2Numerical
A straight wire 1 m long moves at 8 m/s at right angles to a 2 T magnetic field, with the wire also perpendicular to the field. The emf between its ends is
(A)4 V
(B)16 V
(C)0.25 V
(D)10 V
Show the solution
Given
ℓ = 1 m, v = 8 m/s, B = 2 T
Asked
e
Concept
Motional emf.
Formula
e = Bℓv
Baby steps
2 × 1 × 8 = 16 V.
Answer
(B) 16 V
Why not the others
4 V divides v by B. 0.25 V inverts. 10 V adds.
Shortcut
Multiply all three.
Where it went wrong
Dividing instead of multiplying.
Q3Numerical
A 1 m rod lies in the plane perpendicular to a 0.4 T field and moves at 2 m/s in a direction making 30° with its own length. The emf between its ends is
(A)0.8 V
(B)0.69 V
(C)0.4 V
(D)0.2 V
Show the solution
Given
ℓ = 1 m, v = 2 m/s, angle between ℓ and v = 30°, B ⊥ both
Asked
e
Concept
Only the component of ℓ perpendicular to v counts.
Formula
e = Bvℓ sin θ
Baby steps
ℓ sin 30° = 0.5 m.
e = 0.4 × 2 × 0.5 = 0.4 V.
Answer
(C) 0.4 V
Why not the others
0.69 V uses cos 30°. 0.8 V ignores the angle. 0.2 V halves twice.
Shortcut
Project the rod across the motion.
Where it went wrong
Using cos θ.
Q4Numerical
A semicircular wire of radius 10 cm moves at 2 m/s in its own plane, perpendicular to the diameter joining its ends, in a 0.5 T field perpendicular to the plane. The emf between its ends is
(A)0.1 V
(B)0.314 V
(C)0.2 V
(D)0.628 V
Show the solution
Given
R = 0.1 m, v = 2 m/s, B = 0.5 T
Asked
e
Concept
A curved wire acts like the straight line joining its ends.
Formula
e = Bv(2R)
Baby steps
End-to-end length = 2R = 0.2 m, perpendicular to v.
e = 0.5 × 2 × 0.2 = 0.2 V.
Answer
(C) 0.2 V
Why not the others
0.314 V uses the arc length πR. 0.1 V uses R. 0.628 V uses 2πR.
Shortcut
Chord, not arc.
Where it went wrong
Using the curved length.
Q5Numerical
A rod of length 0.5 m slides at 4 m/s on rails joined by a 2 Ω resistor, in a 0.2 T field perpendicular to the rails' plane. The current is
(A)0.2 A
(B)0.4 A
(C)0.8 A
(D)0.1 A
Show the solution
Given
B = 0.2 T, ℓ = 0.5 m, v = 4 m/s, R = 2 Ω
Asked
I
Concept
Rod as a battery.
Formula
I = Bℓv/R
Baby steps
e = 0.2 × 0.5 × 4 = 0.4 V.
I = 0.4/2 = 0.2 A.
Answer
(A) 0.2 A
Why not the others
0.4 A is the emf in volts. 0.8 A multiplies by R. 0.1 A divides by 4.
Shortcut
Emf ÷ R.
Where it went wrong
Giving the emf as the current.
Q6Numerical
For the rod in the previous question (I = 0.2 A), the external force needed to keep it moving at constant speed is
(A)0.02 N
(B)0.04 N
(C)0.2 N
(D)0.08 N
Show the solution
Given
B = 0.2 T, I = 0.2 A, ℓ = 0.5 m
Asked
F
Concept
The external force balances the magnetic force on the current.
Formula
F = BIℓ
Baby steps
F = 0.2 × 0.2 × 0.5 = 0.02 N.
Answer
(A) 0.02 N
Why not the others
0.04 N forgets ℓ. 0.2 N forgets B. 0.08 N is the power (Fv) in watts.
Shortcut
BIℓ.
Where it went wrong
Leaving out ℓ.
Q7Numerical
For the same rod (F = 0.02 N at v = 4 m/s, R = 2 Ω), the power supplied by the external agent is
(A)0.005 W
(B)0.08 W
(C)0.04 W
(D)0.16 W
Show the solution
Given
F = 0.02 N, v = 4 m/s
Asked
P
Concept
Mechanical power equals heat per second.
Formula
P = Fv = I²R
Baby steps
Fv = 0.02 × 4 = 0.08 W.
Check: I²R = 0.04 × 2 = 0.08 W.
Answer
(B) 0.08 W
Why not the others
0.005 W divides F by v. 0.04 W is I² × 1. 0.16 W doubles.
Shortcut
Two routes agree: Fv and I²R.
Where it went wrong
Using F/v.
Q8Numerical
A 30 cm rod moves at 5 m/s on rails in a 0.2 T field perpendicular to the circuit, which has a total resistance of 2 Ω. The force needed to keep the speed constant is
(A)1.8 × 10⁻² N
(B)3 × 10⁻² N
(C)4.5 × 10⁻³ N
(D)9 × 10⁻³ N
Show the solution
Given
B = 0.2 T, ℓ = 0.3 m, v = 5 m/s, R = 2 Ω
Asked
F
Concept
Retarding force on a rod on rails.
Formula
F = B²ℓ²v/R
Baby steps
B²ℓ² = 0.04 × 0.09 = 3.6 × 10⁻³.
× v/R = × 2.5 → 9 × 10⁻³ N.
Answer
(D) 9 × 10⁻³ N
Why not the others
3 × 10⁻² N is the emf (0.3 V) ÷ 10. 4.5 × 10⁻³ N halves. 1.8 × 10⁻² N forgets to divide by R.
Shortcut
B²ℓ²v/R.
Where it went wrong
Squaring only one of B and ℓ.
Q9Numerical
A rectangular loop of resistance 2 mΩ is pulled at 2 cm/s out of a 0.5 T field. The edge still inside the field is 4 cm long. The force needed to keep the speed constant is
(A)4 × 10⁻⁴ N
(B)8 × 10⁻³ N
(C)2 × 10⁻³ N
(D)4 × 10⁻³ N
Show the solution
Given
B = 0.5 T, ℓ = 0.04 m, v = 0.02 m/s, R = 2 × 10⁻³ Ω
Asked
F
Concept
Only the edge inside the field cuts field lines.
Formula
F = B²ℓ²v/R
Baby steps
e = 0.5 × 0.04 × 0.02 = 4 × 10⁻⁴ V.
I = 0.2 A.
F = BIℓ = 0.5 × 0.2 × 0.04 = 4 × 10⁻³ N.
Answer
(D) 4 × 10⁻³ N
Why not the others
8 × 10⁻³ N doubles. 2 × 10⁻³ N halves. 4 × 10⁻⁴ N is the emf in volts.
Shortcut
Emf, current, force.
Where it went wrong
Using the long side of the loop.
Q10Numerical
A loop 4 cm wide enters a 2 T field region at 1 m/s, its leading edge (4 cm) perpendicular to the motion. The emf while it enters is
(A)8 mV
(B)80 mV
(C)800 mV
(D)40 mV
Show the solution
Given
B = 2 T, ℓ = 0.04 m, v = 1 m/s
Asked
|e|
Concept
Entering: only the leading edge is in the field.
Formula
e = Bℓv
Baby steps
2 × 0.04 × 1 = 0.08 V = 80 mV.
Answer
(B) 80 mV
Why not the others
8 mV is the maximum flux in mWb for a 10 cm loop. 800 mV uses ℓ = 0.4 m. 40 mV halves.
Shortcut
Same as Allen Illustration 26.
Where it went wrong
Mixing up flux and emf.
Q11Concept
A square loop of side 5 cm moves at 1 cm/s into a 20 cm wide region of 0.4 T field; its front edge enters at t = 0. The emf at t = 12 s and at t = 22 s is
(A)2 × 10⁻⁴ V both times
(B)2 × 10⁻⁴ V, and zero
(C)zero, and 2 × 10⁻⁴ V
(D)zero both times
Show the solution
Given
Side 5 cm, v = 1 cm/s, region 20 cm, B = 0.4 T
Asked
e(12 s), e(22 s)
Concept
Emf only while an edge is crossing a boundary.
Formula
e = Bℓv
Baby steps
t = 12 s: front edge at 12 cm, back edge at 7 cm, both inside → 0.
t = 22 s: front edge at 22 cm (outside), back edge at 17 cm (inside) → leaving.
e = 0.4 × 0.05 × 0.01 = 2 × 10⁻⁴ V.
Answer
(C) zero, and 2 × 10⁻⁴ V
Why not the others
The swapped option places 12 s at entry. Both non-zero forgets the fully-inside stage. Both zero forgets the leaving stage.
Shortcut
Track both edges.
Where it went wrong
Tracking only the front edge.
Q12Numerical
A square wire loop of mass 10 g, side 10 cm and resistance 0.1 Ω falls with its top edge in a horizontal 0.5 T field (g = 10 m/s²). Its terminal velocity is
(A)2 m/s
(B)0.4 m/s
(C)40 m/s
(D)4 m/s
Show the solution
Given
m = 0.01 kg, ℓ = 0.1 m, R = 0.1 Ω, B = 0.5 T
Asked
vT
Concept
Terminal velocity when magnetic force equals weight (Allen Illustration 27).
Formula
vT = mgR/B²ℓ²
Baby steps
mgR = 0.01 × 10 × 0.1 = 0.01.
B²ℓ² = 0.25 × 0.01 = 2.5 × 10⁻³.
vT = 4 m/s.
Answer
(D) 4 m/s
Why not the others
0.4 m/s and 40 m/s shift a power of ten. 2 m/s halves.
Shortcut
mgR over B²ℓ².
Where it went wrong
Forgetting to square ℓ.
Q13Numerical
A rod of mass 20 g and length 50 cm slides without friction down two vertical rails joined at the top by a 2 Ω resistor, in a horizontal 0.4 T field perpendicular to the rails (g = 10 m/s²). Its terminal speed is
(A)2.5 m/s
(B)5 m/s
(C)10 m/s
(D)20 m/s
Show the solution
Given
m = 0.02 kg, ℓ = 0.5 m, R = 2 Ω, B = 0.4 T
Asked
vT
Concept
Magnetic force balances the weight.
Formula
vT = mgR/B²ℓ²
Baby steps
mgR = 0.4.
B²ℓ² = 0.16 × 0.25 = 0.04.
vT = 10 m/s.
Answer
(C) 10 m/s
Why not the others
5 m/s forgets the 2 Ω. 2.5 m/s squares R. 20 m/s doubles.
Shortcut
Same formula as a falling loop.
Where it went wrong
Using ℓ instead of ℓ².
Q14Numerical
A loop of resistance 2 Ω is connected to a balanced bridge whose equivalent resistance is 4 Ω. One 20 cm edge of the loop moves out of a 1 T field. For a steady current of 2 mA, the speed must be
(A)6 cm/s
(B)2 cm/s
(C)4 cm/s
(D)12 cm/s
Show the solution
Given
Rtotal = 6 Ω, I = 2 × 10⁻³ A, B = 1 T, ℓ = 0.2 m
Asked
v
Concept
Total emf = I × total resistance.
Formula
v = IR/(Bℓ)
Baby steps
e = 2 × 10⁻³ × 6 = 0.012 V.
v = 0.012/(1 × 0.2) = 0.06 m/s = 6 cm/s.
Answer
(A) 6 cm/s
Why not the others
2 cm/s uses only the loop's 2 Ω. 4 cm/s uses only the bridge's 4 Ω. 12 cm/s doubles.
Shortcut
Add the resistances first.
Where it went wrong
Leaving out part of the circuit (Beginner's Box 7 Q4 pattern).
Q15Numerical
An aircraft with a wingspan of 20 m flies horizontally at 250 m/s where the vertical component of the earth's field is 5 × 10⁻⁵ T. The emf between its wing tips is
(A)0.025 V
(B)2.5 V
(C)0.25 V
(D)1.0 V
Show the solution
Given
ℓ = 20 m, v = 250 m/s, BV = 5 × 10⁻⁵ T
Asked
e
Concept
Only the vertical component is perpendicular to both the wings and the velocity.
Formula
e = BVℓv
Baby steps
20 × 250 = 5000.
× 5 × 10⁻⁵ = 0.25 V.
Answer
(C) 0.25 V
Why not the others
2.5 V and 0.025 V shift powers of ten. 1.0 V has no basis.
Shortcut
Horizontal flight → vertical component.
Where it went wrong
Using the horizontal component.
Q16Concept
The speed of a rod sliding on rails in a steady field (with a resistor) is doubled. The mechanical power needed to keep it moving becomes
(A)four times
(B)twice
(C)half
(D)unchanged
Show the solution
Given
v → 2v
Asked
New P
Concept
P = B²ℓ²v²/R.
Formula
P ∝ v²
Baby steps
Emf and current double; force doubles.
P = Fv: 2 × 2 = 4.
Answer
(A) four times
Why not the others
Twice is the force. Half and unchanged have no basis.
Shortcut
Force ∝ v, power ∝ v².
Where it went wrong
Treating power like force.
Q17Concept
A straight rod moves along its own length through a uniform magnetic field perpendicular to it. The emf between its ends is
(A)Bℓv
(B)zero
(C)Bℓv/2
(D)maximum
Show the solution
Given
v parallel to ℓ
Asked
e
Concept
v × B is perpendicular to v, so it has no component along ℓ.
Formula
e = (v × B) · ℓ
Baby steps
v ∥ ℓ.
v × B ⊥ v, so ⊥ ℓ.
Dot product = 0.
Answer
(B) zero
Why not the others
Each non-zero option needs v × B to have a component along the rod.
Shortcut
Sliding along itself: no cutting.
Where it went wrong
Assuming any motion in a field gives Bℓv.
Q18Concept
A rod moves through a uniform magnetic field with its velocity parallel to the field. The emf between its ends is
(A)Bℓv
(B)zero
(C)depends on the rod's orientation
(D)Bℓv sin 45°
Show the solution
Given
v ∥ B
Asked
e
Concept
v × B = 0 when v ∥ B.
Formula
e = (v × B) · ℓ
Baby steps
v × B = 0.
So e = 0 whatever the rod's orientation.
Answer
(B) zero
Why not the others
Bℓv needs v ⊥ B. Orientation cannot matter when v × B is zero.
Shortcut
Moving along field lines cuts none.
Where it went wrong
Thinking orientation can rescue it.
Q19Direction
A vertical rod moves to the right through a magnetic field pointing into the page. Which end is at higher potential?
(A)Both ends are at the same potential
(B)The bottom end
(C)The top end
(D)The end nearer the reader
Show the solution
Given
v = right, B = into page, rod vertical
Asked
Higher-potential end
Concept
Positive charges collect in the direction of v × B.
Formula
F = q(v × B)
Baby steps
v = î, B = −k̂.
v × B = î × (−k̂) = +ĵ.
Positive charges pushed up: top end higher.
Answer
(C) The top end
Why not the others
Bottom end is the answer for a field out of the page. Equal potentials need v ∥ B or v ∥ ℓ. The rod has no front/back ends.
Shortcut
Fingers along v, curl to B, thumb → + end.
Where it went wrong
Getting î × k̂ backwards.
Q20Direction
A vertical rod moves to the right through a magnetic field pointing out of the page. Which end is at higher potential?
(A)The top end
(B)The bottom end
(C)Both ends are at the same potential
(D)The middle of the rod
Show the solution
Given
v = right, B = out of page
Asked
Higher-potential end
Concept
Direction of v × B.
Formula
F = q(v × B)
Baby steps
v × B = î × k̂ = −ĵ.
Positive charges pushed down: bottom end higher.
Answer
(B) The bottom end
Why not the others
Top end is the into-page answer. Equal potentials need v ∥ B or ℓ. Ends, not the middle, collect charge.
Shortcut
Reversing B reverses the answer.
Where it went wrong
Using the into-page answer.
Q21Direction
A horizontal rod moves up the page through a magnetic field pointing into the page. Which end is at higher potential?
(A)It depends on the rod's length
(B)The right end
(C)Both ends are at the same potential
(D)The left end
Show the solution
Given
v = up, B = into page, rod horizontal
Asked
Higher-potential end
Concept
Direction of v × B.
Formula
F = q(v × B)
Baby steps
v × B = ĵ × (−k̂) = −î.
Positive charges pushed left: left end higher.
Answer
(D) The left end
Why not the others
Right end reverses the cross product. Equal potentials need parallel vectors. Length changes the size, not the direction.
Shortcut
ĵ × k̂ = î, so ĵ × (−k̂) = −î.
Where it went wrong
Cross-product order error.
Q22Concept
A semicircular wire moves in its plane, perpendicular to a uniform field, with its velocity parallel to the diameter joining its two ends. The emf between the ends is
(A)BvπR
(B)Bv(2R)
(C)zero
(D)BvR
Show the solution
Given
Ends on a line parallel to v
Asked
e
Concept
Effective length is the component of the end-to-end line perpendicular to v.
Formula
e = Bv × (length ⊥ v)
Baby steps
The line joining the ends is parallel to v.
Its perpendicular component is zero.
e = 0.
Answer
(C) zero
Why not the others
2R applies when the diameter is perpendicular to v. πR is the arc. R has no basis.
Shortcut
Ends level with v → zero.
Where it went wrong
Using the curved length.
Q23Concept
A rod is kept moving at constant velocity on rails with a resistor in a magnetic field. The work done by the external agent is converted into
(A)magnetic field energy that keeps growing
(B)kinetic energy of the rod
(C)heat in the resistance
(D)potential energy of the rod
Show the solution
Given
Constant velocity
Asked
Where the work goes
Concept
Energy conservation: Pmech = I²R.
Formula
Fv = I²R
Baby steps
Speed is constant: no gain in kinetic energy.
The field is steady: no growing field energy.
All the work becomes Joule heat.
Answer
(C) heat in the resistance
Why not the others
Kinetic energy is constant. The external field is not growing. Horizontal rails: no potential energy change.
Shortcut
Mechanical → electrical → heat.
Where it went wrong
Thinking the rod gains energy.
Q24Direction
A rod slides to the right on rails in a magnetic field, with the circuit closed through a resistor. The magnetic force on the rod due to its induced current points
(A)to the right, along the velocity
(B)to the left, opposite to the velocity
(C)up along the rod
(D)into the page
Show the solution
Given
Rod moving right
Asked
Direction of F = IL × B
Concept
Lenz's law: the force opposes the motion.
Formula
F = BIℓ
Baby steps
The induced current makes the rod feel a force.
By Lenz's law, it opposes the cause (the motion).
So it points left.
Answer
(B) to the left, opposite to the velocity
Why not the others
Along v would create energy. Up the rod or into the page are not perpendicular to both I and B correctly.
Shortcut
Induced force always brakes.
Where it went wrong
Applying the force rule with the wrong current direction.
Q25Concept
A thin rod bent into an inverted V has its two ends level with each other. It moves horizontally (in the direction of the line joining its ends) through a field perpendicular to its plane. The emf between the ends is
(A)zero
(B)2Bvℓ sin θ
(C)2Bvℓ
(D)Bvℓ cos θ
Show the solution
Given
End-to-end line parallel to v
Asked
e
Concept
Only the perpendicular component of the end-to-end line counts.
Formula
e = Bv × (length ⊥ v)
Baby steps
The ends are level, and v is horizontal.
The end-to-end line has no component perpendicular to v.
e = 0 (Beginner's Box 7 Q7(iii)).
Answer
(A) zero
Why not the others
The non-zero options use the arm lengths, which cancel between the two arms.
Shortcut
Ends in line with v → zero.
Where it went wrong
Adding the emfs of the two arms instead of noticing they cancel.
Q26Graph
A rod moves at constant velocity v on rails with a resistor. Which graph shows the power supplied by the external agent against v?
(A)
(B)
(C)
(D)
Show the solution
Given
P against v
Asked
Shape
Concept
P = B²ℓ²v²/R.
Formula
P ∝ v²
Baby steps
Square law → parabola through the origin.
Answer
(B) the graph in option B
Why not the others
The straight line is the force. The root curve and flat line have no basis.
Shortcut
Force: line. Power: parabola.
Where it went wrong
Choosing the force graph.
Q27Graph
A rectangular loop of length b moves at constant speed through a field region of width d > b. Which graph shows the emf against the position of its front edge?
(A)
(B)
(C)
(D)
Show the solution
Given
Loop crossing a wider region
Asked
e against x
Concept
Emf only while an edge crosses a boundary; sign flips between entering and leaving.
Formula
e = −dφ/dt
Baby steps
Entering (width b): constant emf of one sign.
Fully inside (width d − b): zero.
Leaving (width b): constant emf of the opposite sign.
Answer
(D) the graph in option D
Why not the others
The trapezoid is the flux graph. The graph with no zero stretch ignores the fully-inside stage. A single block ignores leaving.
Shortcut
Block, gap, opposite block.
Where it went wrong
Drawing the flux instead of the emf.
Q28Graph
A rod is released from rest on frictionless vertical rails joined by a resistor, in a horizontal magnetic field. Which graph shows its speed against time?
(A)
(B)
(C)
(D)
Show the solution
Given
Released from rest
Asked
v against t
Concept
Magnetic braking grows with speed until it balances gravity.
Formula
m dv/dt = mg − B²ℓ²v/R
Baby steps
At first a ≈ g.
Braking force grows with v.
Speed levels off at vT = mgR/B²ℓ².
Answer
(B) the graph in option B
Why not the others
The straight line is free fall. The decaying curve starts from rest wrongly. Rise-then-fall would need braking to exceed gravity.
Shortcut
Terminal velocity: rises and levels.
Where it went wrong
Ignoring the braking force.
Q29Graph
A rod on horizontal rails with a resistor is given a push and then left to slide freely (no friction) in a magnetic field. Which graph shows its speed against time?
(A)
(B)
(C)
(D)
Show the solution
Given
Only magnetic braking
Asked
v against t
Concept
Braking force ∝ v gives exponential decay.
Formula
m dv/dt = −B²ℓ²v/R
Baby steps
Force proportional to speed.
Speed falls by the same fraction in equal times.
Exponential decay towards zero.
Answer
(D) the graph in option D
Why not the others
A straight fall to zero needs a constant force. Constant speed ignores braking. The rising curve needs a push.
Shortcut
Same shape as F = −bv in Topic 03.
Where it went wrong
Drawing constant deceleration.
Q30Assertion–reason
Assertion (A): A rod moving along its own length in a magnetic field has no emf between its ends. Reason (R): The magnetic force on its charges, q(v × B), has no component along the rod.
(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 the solution
Given
A: no emf. R: force has no component along the rod.
Asked
Truth and link
Concept
e = (v × B) · ℓ.
Formula
—
Baby steps
A is true.
R is true.
R is exactly why no charge separates along the rod.
Answer
(A) Both A and R are true, and R is the correct explanation of A.
Why not the others
(B) denies a direct link. (C) and (D) need a false statement.
Shortcut
The mechanism explains the result.
Where it went wrong
Choosing (B) by habit.
Q31Assertion–reason
Assertion (A): An external force is needed to keep a rod moving at constant velocity on rails joined by a resistor in a magnetic field. Reason (R): The induced current in the rod experiences a magnetic force opposing the motion.
(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 the solution
Given
A: external force needed. R: opposing magnetic force.
Asked
Truth and link
Concept
Newton's first law plus Lenz's law.
Formula
Fext = B²ℓ²v/R
Baby steps
A is true.
R is true.
Without the push, the opposing force would slow the rod: R explains A.
Answer
(A) Both A and R are true, and R is the correct explanation of A.
Why not the others
(B) denies a direct link. (C) and (D) need a false statement.
Shortcut
Opposing force → need a balancing push.
Where it went wrong
Picking (B) out of caution.
Q32Assertion–reason
Assertion (A): The force needed to move a rod at constant speed on rails is proportional to the square of the speed. Reason (R): The retarding force is B²ℓ²v/R.
(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 the solution
Given
A: F ∝ v². R: F = B²ℓ²v/R.
Asked
Truth and link
Concept
F ∝ v; power ∝ v².
Formula
F = B²ℓ²v/R
Baby steps
R is true.
By R, F ∝ v, so A is false.
A false, R true.
Answer
(D) A is false, but R is true.
Why not the others
(A), (B) and (C) need A true.
Shortcut
Use R to test A.
Where it went wrong
Confusing force with power.
Q33Assertion–reason
Assertion (A): A semicircular wire of radius R moving perpendicular to its diameter has an emf of 2BvR between its ends. Reason (R): The emf in a curved wire depends on its arc length πR.
(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 the solution
Given
A: 2BvR. R: arc length decides.
Asked
Truth and link
Concept
Only the end-to-end line counts.
Formula
e = Bv(2R)
Baby steps
A is true.
R is false.
A true, R false.
Answer
(C) A is true, but R is false.
Why not the others
(A) and (B) need R true. (D) needs A false.
Shortcut
Chord, not arc.
Where it went wrong
Using the arc length.
Q34Two statements
Statement I: Motional emf arises from the magnetic Lorentz force on the free charges in a moving conductor. Statement II: The end of the rod towards which positive charges are pushed is at the higher potential.
(A)Both Statement I and Statement II are true.
(B)Both Statement I and Statement II are false.
(C)Statement I is true, but Statement II is false.
(D)Statement I is false, but Statement II is true.
Show the solution
Given
Origin of motional emf; higher-potential end
Asked
Which are true
Concept
F = q(v × B); charges separate.
Formula
—
Baby steps
Statement I is true.
Statement II is true.
Answer
(A) Both Statement I and Statement II are true.
Why not the others
Every other option calls a true statement false.
Shortcut
Both are the basic mechanism.
Where it went wrong
Doubting correct statements.
Q35Two statements
Statement I: The mechanical power needed to move a rod at constant speed on rails is proportional to its speed. Statement II: The external force needed is proportional to its speed.
(A)Both Statement I and Statement II are true.
(B)Both Statement I and Statement II are false.
(C)Statement I is true, but Statement II is false.
(D)Statement I is false, but Statement II is true.
Show the solution
Given
Power and force scaling
Asked
Which are true
Concept
F = B²ℓ²v/R; P = Fv.
Formula
F ∝ v, P ∝ v²
Baby steps
Statement I is false: P ∝ v².
Statement II is true.
Answer
(D) Statement I is false, but Statement II is true.
Why not the others
Both-true fails on I. Both-false fails on II. I-true-II-false reverses them.
Shortcut
Force line, power parabola.
Where it went wrong
Mixing up the two.
Answer key
1 A
2 B
3 C
4 C
5 A
6 A
7 B
8 D
9 D
10 B
11 C
12 D
13 C
14 A
15 C
16 A
17 B
18 B
19 C
20 B
21 D
22 C
23 C
24 B
25 A
26 B
27 D
28 B
29 D
30 A
31 A
32 D
33 C
34 A
35 D
Spread across letters: A 9, B 9, C 9, D 8. No letter repeats more than twice in a row. Question mix: Numerical 14, Concept 7, Direction 4, Graph 4, Assertion–reason 4, Two statements 2. Balancing seed 0.