NIOS Class 12 Physics Chapter 19 Electromagnetic Induction and Alternating Current Solutions English Medium As Per New Syllabus to each chapter is provided in the list so that you can easily browse throughout different chapters NIOS Class 12 Physics Chapter 19 Electromagnetic Induction and Alternating Current Notes in English and select need one. NIOS Class 12 Physics Solutions English Medium Download PDF. NIOS Study Material of Class 12 Physics Notes Paper Code: 312.
NIOS Class 12 Physics Chapter 19 Electromagnetic Induction and Alternating Current
Also, you can read the NIOS book online in these sections Solutions by Expert Teachers as per National Institute of Open Schooling (NIOS) Book guidelines. These solutions are part of NIOS All Subject Solutions. Here we have given NIOS Class 12 Physics Notes, NIOS Senior Secondary Course Physics Solutions in English for All Chapter, You can practice these here.
Electromagnetic Induction and Alternating Current
Chapter: 19
| Module-V: Electricity and Magnetism |
INTEXT QUESTIONS 19.1
1. A 1000 turn coil has a radius of 5 cm. Calculate the emf developed across the coil if the magnetic field through the coil is reduced from 10 T to 0 in (a) 1s (b) 1ms.
Ans: Given: Number of turns, N = 1000
Radius, r = 5 cm
= 0.05m
Initial magnetic field,
B₁ = 10 T
Final magnetic field, B₂ = 0
Area, A = πr²
(a) Time, t = 1s
(b) Time, t = 1ms = 0.001s
Formula:
Induced emf, |e| = N × (ΔΦ)/Δt,
where Φ = BA
So,
Change in flux, ΔΦ = (B₂ – B₁)A = (0 – 10)π(0.05)² = -10 × π × 0.0025 = -0.025π
(a) For t = 1s:
|e| = (1000) × (0.025π) / 1
|e| = 25π V
π ≈ 3.14
|e| ≈ 78.5 V
(b) For t = 0.001s:
|e| = (1000) × (0.025π) / 0.001
|e| = 25,000π V
≈ 78,500 V
2. The magnetic flux linking each loop of a 250-turn coil is given by φB (t) = A + Dt², where A = 3 Wb and D = 15 Wb/s² are constants. Show that (a) the magnitude of the induced emf in the coil is given by e = (2ND)t, and (b) evaluate the emf induced in the coil at t = 0s and t = 3.0s.
Ans: (a) Flux in one loop, φ = A + Dt²
= N × 2D × t
So, e = (2ND) t
(b) Given: N = 250,
D = 15
At t = 0:
e = 2 × 250 × 15 × 0
= 0 V
At t = 3s:
e = 2 × 250 × 15 × 3
= 2 × 250 × 45
= 500 × 45
= 22,500 V
3. The perpendicular to the plane of a conducting loop makes a fixed angle θ with a spatially uniform magnetic field. If the loop has area S and the magnitude of the field changes at a rate dB/dt, show that the magnitude of the induced emf in the loop is given by e = (dB/dt) S cosθ. For what orientation(s) of the loop will e be (a) maximum and (b) minimum?
Ans: Magnetic flux, Φ = B × S × cosθ
So, e = (dB/dt) × S × cosθ
Maximum when cosθ = 1
⇒ θ = 0°
i.e., loop perpendicular to B
(b) Minimum when cosθ = 0
⇒ θ = 90°, i.e.
loop parallel to B
Intext Questions 19.2
1. The bar magnet in Fig.19.6 moves to the right. What is the sense of the induced current in the stationary loop A? In loop B?
Ans: Loop A: As the magnet approaches, flux increases; current is induced to oppose the increase (Lenz’s law). So, current direction is such that it creates a magnetic field opposing the motion (anticlockwise when viewed from the side where magnet approaches).
Loop B: As the magnet moves away, flux decreases; induced current supports the receding field (clockwise when viewed from that side).
2. A cross-section of an ideal solenoid is shown in Fig.19.7. The magnitude of a uniform magnetic field is increasing inside the solenoid and B = 0 outside the solenoid. In which conducting loops is there an induced current? What is the sense of the current in each case?
Ans: To answer this question, let’s first interpret the situation:
Given: The cross-section of an ideal solenoid is shown (in Fig. 19.7 — typically a circular cross-section with magnetic field lines pointing into or out of the page).
Magnetic field inside the solenoid is increasing in magnitude.
Magnetic field outside the solenoid is zero (ideal solenoid).
There are conducting loops placed both inside and outside the solenoid.
Faraday’s Law of Electromagnetic Induction.
An emf is induced in a conducting loop if the magnetic flux through the loop changes with time:
Where:
- ΦB = B.A is the magnetic flux through the loop.
- The direction of induced current is given by Lenz’s Law, which states that the current flows in such a way that it opposes the change in magnetic flux.
Analysis of Each Loop
1. Loop Inside the Solenoid:
- The magnetic field inside the solenoid is increasing.
- Therefore, the magnetic flux through the loop increases.
- By Lenz’s Law, the induced current will oppose this increase in flux.
- Hence, the loop produces a magnetic field opposite to the solenoid’s field.
Sense of Induced Current
- If the solenoid’s magnetic field is into the page,
→ the loop must create a field out of the page,
→ therefore, the induced current is clockwise. - If the solenoid’s magnetic field is out of the page,
→ the loop must create a field into the page,
→ therefore, the induced current is anticlockwise.
Conclusion: An induced current flows in the loop inside the solenoid.
2. Loop Outside the Solenoid:
- The magnetic field outside an ideal solenoid is zero and remains zero, even when the internal field increases.
- Thus, the magnetic flux through the outer loop is zero and does not change.
No change in flux⇒No induced current
Conclusion: No current is induced in the loop outside the solenoid.
Hi! my Name is Parimal Roy. I have completed my Bachelor’s degree in Philosophy (B.A.) from Silapathar General College. Currently, I am working as an HR Manager at Dev Library. It is a website that provides study materials for students from Class 3 to 12, including SCERT and NCERT notes. It also offers resources for BA, B.Com, B.Sc, and Computer Science, along with postgraduate notes. Besides study materials, the website has novels, eBooks, health and finance articles, biographies, quotes, and more.
Buy Now