Class 12 Physics, Chapter 6 (NCERT). 15 questions by concept, 1 with pictures. Open “Show answer” for the answer and why.
How to teach this chapter · Common misconceptions & worked visual for this chapter
A flat surface of area 1 m² has its chosen normal along +z. The uniform magnetic field is B=(1 i + 3 j + 5 k) T. What is the signed magnetic flux?
(a) 5 Wb
Only the component along the chosen normal contributes: Phi=Bz times area. The x and y components are tangential.
A flat surface of area 2 m² has signed flux +10 Wb. A uniform field has a tangential component of 1 T and an unknown component Bn along the chosen normal. What is Bn?
(a) 5 T
Phi=Bn times area, so Bn=+10/2 T. The tangential component has zero dot product with the area vector.
The flux linkage of an isolated resistive loop changes monotonically between the same two values in two trials. Its resistance is unchanged, and self-inductance is negligible. Trial 2 takes 3 times as long as trial 1. What is the ratio of total charge magnitudes Q2:Q1?
(c) 1:1
Integrating I=-(1/R)dLambda/dt gives |Q|=|DeltaLambda|/R. Duration changes the current but not total charge for the same endpoints.
A closed coil of resistance 3 Ω has negligible self-inductance. Its flux linkage changes monotonically by 24 Wb turn in 2 s. What total charge magnitude flows?
(a) 8 coulomb
Integrate the induced current over the transition: |Q|=|DeltaLambda|/R=24/3 coulomb. The duration cancels.
A conducting loop with a narrow insulating cut is stationary in a magnetic field whose flux is changing. In the ideal open-circuit limit, which statement is correct?
(a) An induced emf can exist, while no sustained conduction current circulates around the loop
Faraday law permits an induced emf from changing flux without a closed conducting path. The cut prevents sustained circulating conduction current.
An isolated straight rod of length 1 m translates at 4 m/s in a uniform 1 T field. The rod, velocity and magnetic field are mutually perpendicular. What is the magnitude of the open-circuit voltage between its ends?
(a) 4 V
Charge separation balances the magnetic force in the open rod. Its end-to-end voltage is Bℓv even though no closed conduction path exists.
The external magnetic field through the closed conducting loop shown is decreasing in magnitude. Viewed as drawn, what is the direction of induced current?
(d) Counterclockwise
The induced field reinforces the decreasing external field. Apply the right-hand rule: into-page field requires clockwise current; out-of-page field requires counterclockwise current.
A rod is pulled at constant speed along frictionless conducting rails through a uniform perpendicular magnetic field. The rails and resistor form a closed circuit; self-inductance is negligible. Which statement describes the applied force and energy transfer?
(d) The applied force does positive work against magnetic drag; this becomes Joule heat
Magnetic drag opposes the motion. An equal applied force maintains speed, and its mechanical power equals the electrical heating power.
Two equal-length rods move with the same nonzero speed in the same uniform nonzero magnetic field. For each rod, its length, velocity and field are mutually perpendicular. Rod 2 has 3 times the resistance of rod 1. What is the ratio of their motional-emf magnitudes E2:E1?
(d) 1:1
The motional emf Bℓv depends on field, geometry and speed. Resistance can alter a closed-circuit current but does not multiply the motional emf.
Equal-length rods move in the same uniform field. For each rod, its length, velocity and field are mutually perpendicular. Their nonzero motional-emf magnitudes are equal. Rod 2 has 2 times the resistance of rod 1. What is v2:v1?
(c) 1:1
Equal Bℓv with equal B and length implies equal speed, independently of rod resistance.
Two coils have mutual inductance 0.5 H. Current in one increases by 20 A in 500 ms. What is the magnitude of the resulting change in flux linkage of the other?
(c) 10 Wb turn
Linkage change is M Delta I=0.5×20 Wb turn. Dividing by duration would instead calculate an induced-emf magnitude.
Coils of mutual inductance 1.5 H undergo a transition lasting 500 ms. The flux linkage of the secondary increases by 15 Wb turn. What increase in primary current produced this?
(b) 10 ampere
DeltaLambda=M DeltaI gives DeltaI=15/1.5 amperes. The transition time affects emf but does not enter this linkage relation.
A current flows from P to Q through an ideal inductor and is decreasing in magnitude. In which direction does the self-induced emf tend to drive current?
(b) From P to Q
The induced emf opposes the decrease, so it tends to maintain the existing current direction. It opposes the change of current rather than always opposing the current.
An ideal inductor of 2 H carries 4 A. What magnetic energy is stored?
(c) 16 J
Integrate dW=L I dI from zero current: U=LI²/2. The one-half factor is required.
A generator coil rotates uniformly in a uniform nonzero magnetic field and has nonzero peak emf E0. At an instant its normal is antiparallel to the field. What is the instantaneous emf magnitude?
(a) 0 V
At parallel or antiparallel alignment the flux is at an extremum and its instantaneous derivative is zero. Emf peaks a quarter cycle away.
Wrong
A strong magnet held inside a coil keeps inducing a current.
Why it feels right. The magnet keeps producing a field through the coil.
Right
An emf is induced while the magnetic flux through the coil changes.
Worked example
