Maharashtra Board Class 12 Physics Chapter 12 Electromagnetic Induction PDF Download

Electromagnetic Induction Problems

7. A long solenoid consisting of 1.5×10³ turns/m has an area of cross-section of 25 cm². A coil C, consisting of 150 turns (Nc) is wound tightly around the centre of the solenoid. Calculate for a current of 3.0 A in the solenoid

(a) the magnetic flux density at the centre of the solenoid,

(b) the flux linkage in the coil C,

(c) the average emf induced in coil C if the current in the solenoid is reversed in direction in a time of 0.5 s.

(\(\mu_0 = 4\pi \times 10^{-7}\) T.m/A)

[Ans: (a) 5.66×10⁻³ T, (b) 2.12×10⁻³ Wb, (c) 8.48×10⁻³ V]

Teacher's Note

This problem teaches us about solenoids and how they create magnetic fields. In India, solenoids are used in many electrical devices like doorbells and electric locks.

Exam Trick

Remember the formula for magnetic field in a solenoid: B = μ₀nI where n is turns per meter. Write down the given values clearly before solving.

Points to Remember

Magnetic flux density B depends on current I and number of turns per metre n.
Flux linkage means how many magnetic field lines pass through the coil.
When current reverses direction, the change in flux is twice the original flux.
Always convert area to m² from cm² before using in formulas.

8. A search coil having 2000 turns with area 1.5 cm² is placed in a magnetic field of 0.60T. The coil is moved rapidly out of the field in a time of 0.2 second. Calculate the induced emf across the search coil.

[Ans: 0.9 V]

Teacher's Note

When you move a coil out of a magnetic field, emf is induced in it. This is how metal detectors at airports work.

Exam Trick

Use Faraday's law: emf = -N × (change in flux)/(change in time). Remember to multiply by number of turns.

Points to Remember

Change in flux = B × A when coil moves completely out of field.
Number of turns multiplies the induced emf.
Time matters - faster movement means larger emf.
Convert area from cm² to m² always.

9. An aircraft of wing span of 50 m flies horizontally in earth's magnetic field of 6×10⁻⁵ T at a speed of 400 m/s. Calculate the emf generated between the tips of the wings of the aircraft.

[Ans: 1.2 V]

Teacher's Note

When an aircraft flies through Earth's magnetic field, the wings cut magnetic field lines and generate emf. This is motional emf.

Exam Trick

Use formula: emf = B × L × v where B is field, L is length of wing, and v is velocity. This is called motional emf.

Points to Remember

Motional emf occurs when a conductor moves through a magnetic field.
The emf = B × L × v only when motion is perpendicular to the field.
Longer the conductor, more the emf generated.
Faster the motion, more the emf.

10. A stiff semi-circular wire of radius R is rotated in a uniform magnetic field B about an axis passing through its ends. If the frequency of rotation of wire be f, calculate the amplitude of alternating emf induced in the wire.

[Ans: \(e_0 = \pi BR^2f\)]

Teacher's Note

This is how generators work in power plants in India. Rotating coils in magnetic fields produce the electricity we use at home.

Exam Trick

The amplitude of emf in a rotating coil is maximum when using the full area. Remember: e₀ = 2πfNBA where N=1 for a single loop.

Points to Remember

Rotating coils generate alternating emf.
Amplitude depends on area, magnetic field, and frequency.
Larger radius means larger area and more emf.
Higher frequency means faster changes and more emf.

11. Calculate the value of induced emf between the ends of an axle of a railway carriage 1.75 m long travelling on level ground with a uniform velocity of 50 km per hour. The vertical component of Earth's magnetic field (Bv) is given to be 5×10⁻⁵T.

[Ans: 0.122 mV]

Teacher's Note

When a train moves through Earth's magnetic field, small emf is induced in the axles. This is why trains need special care in very magnetic areas.

Exam Trick

Convert speed from km/h to m/s by dividing by 3.6. Use only the vertical component of Earth's field for this calculation.

Points to Remember

Use motional emf formula: emf = B × L × v.
Always convert velocity to m/s before calculating.
Only vertical component of Earth's field matters here.
The axle acts as a conductor moving through the magnetic field.

12. The value of mutual inductance of two coils is 10 mH. If the current in one of the coil changes from 5A to 1A in 0.2 s, calculate the value of emf induced in the other coil. Also calculate the value of induced charge passing through the coil if its resistance is 5 ohm.

[Ans: 8 m C]

Teacher's Note

Mutual inductance is how two coils affect each other. Power transformers in your home use this principle to change voltage.

Exam Trick

Induced emf = M × (change in current)/(change in time). For charge, use: charge = (change in flux)/resistance.

Points to Remember

Mutual inductance M tells how much one coil affects another coil.
Change in current in first coil induces emf in second coil.
Induced charge depends on total change in flux and resistance.
Unit of mutual inductance is Henry (H) or millihenry (mH).

13. An emf of 96.0 mV is induced in the windings of a coil when the current in a nearby coil is increasing at the rate of 1.20 A/s. What is the mutual inductance (M) of the two coils?

[Ans: 80 mH]

Teacher's Note

This shows how we measure mutual inductance between coils. Transformers work on this same principle in India's electricity supply system.

Exam Trick

Use formula: emf = M × (dI/dt). Rearrange to find M = emf ÷ (dI/dt). This is the direct method.

Points to Remember

Mutual inductance M = emf ÷ (rate of current change).
Larger mutual inductance means stronger connection between coils.
Unit is Henry (H) or millihenry (mH).
Two coils must be close together for good mutual inductance.

14. A long solenoid of length l, cross-sectional area A and having N₁ turns (primary coil), has a small coil of N₂ turns (secondary coil) wound about its centre. Determine the Mutual inductance (M) of the two coils.

[Ans: \(M = \mu_0 \frac{N_1 N_2 A}{l}\)]

Teacher's Note

This formula shows the relationship between mutual inductance and the physical properties of the coils. It helps us design transformers used in Indian homes.

Exam Trick

Remember: M depends on permeability (μ₀), both numbers of turns (N₁ and N₂), area (A), and inversely on length (l).

Points to Remember

Mutual inductance increases with more turns on both coils.
Larger cross-sectional area means larger mutual inductance.
Shorter solenoid means larger mutual inductance.
Permeability μ₀ is a constant for free space.

15. The primary and secondary coil of a transformer each have an inductance of 200 ×10⁻⁶H. The mutual inductance (M) between the windings is 4×10⁻⁶ H. What percentage of the flux from one coil reaches the other?

[Ans: 2%]

Teacher's Note

This tells us how efficient a transformer is. In real transformers in India, the efficiency is usually 95% or more due to good design.

Exam Trick

Use the formula: coupling factor k = M / √(L₁ × L₂). Percentage = k² × 100%. This tells how much flux is shared.

Points to Remember

Coupling factor k shows how well two coils are coupled.
k = M ÷ √(L₁ × L₂) where L₁ and L₂ are self-inductances.
Percentage of flux = k² × 100%.
Perfect coupling has k = 1 (100% flux transfer).

16. A toroidal ring, made from a bar of length (l) 1 m and diameter (d) 1 cm, is bent into a circle. It is wound tightly with 100 turns per cm. If the permeability of bar is equal to that of free space (μ₀), calculate the magnetic field inside the bar (B) when the current (i) circulating through the turns is 100 A. Also determine the self-inductance (L) of the coil.

[Ans: 1.256T, 10mH]

Teacher's Note

Toroidal coils are used in electrical equipment in India because they keep magnetic fields contained inside the ring and don't disturb other devices.

Exam Trick

For a toroid: B = μ₀ × n × i where n is turns per meter. Self-inductance L = μ₀ × N² × A / length of ring.

Points to Remember

Toroid is a doughnut-shaped coil wound with wire.
Magnetic field only exists inside the toroid, not outside.
Total number of turns = turns per cm × length in cm.
Self-inductance depends on N², area, and length of ring.

17. A uniform magnetic field B(t), pointing upward fills a circular region of radius, s in horizontal plane. If B is changing with time, find the induced electric field.

[Ans: \(E = \frac{s}{2}\frac{dB}{dt}\)]

Teacher's Note

This shows how changing magnetic fields create electric fields. This is the basis of how induction furnaces work to heat metals in Indian industries.

Exam Trick

Remember Faraday's law in circular form: induced electric field is circular around the changing magnetic field. At distance s from centre: E = (s/2) × (dB/dt).

Points to Remember

Changing magnetic field creates circular electric field around it.
Electric field increases with distance from centre inside the region.
Rate of change of B matters, not B itself.
This is electromagnetic induction in its most basic form.

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