In the following part of a circuit i 1 = 5 A , d l 1 d t = 3 A s - 1 , i 2 = 2 A , d i 2 d t = - 4 A / s . Find V A - V B .

Given, i 1 = 5 A , d l 1 d t = 3 A s - 1 , i 2 = 2 A , d i 2 d t = - 4 A / s Potential difference in the circuit is given by V 0 - V A = - 2 + 2 d i 1 d t + 2 i ⇒ - 2 + 2 × 3 + 2 × 5 ⇒ 14 V V 0 - V B = - 3 + 3 d i 2 d t + 3 i 2 ⇒ - 3 + 3 × - 4 + 3 × 2 ⇒ - 9 V V 0 - V B - V 0 - V A = - 9 - 14 ⇒ - 23 V

Find mutual inductance of (a) (b) (c) (d)

Case1) Magnetic field inside the smaller coil due to large coil is uniform and equal to B = μ i 2 R Thus, flux through the smaller coil due to this magnetic field =. μ i 2 R π r 2 Mutual inductance, M = ϕ i M = μ π r 2 2 R 2 Case 2) Magnetic field inside smaller rectangle due to large rectangle is uniform and equal to μ 0 i a 2 The magnetic flux through this rectangle is ∮ d ϕ = ∮ B . d A ϕ = μ 0 i a 2 . b 2 M = μ 0 a 2 . b 2 M = μ 0 a 2 . b 2 Mutual inductance M = 0 Case (3) Magnetic field inside the smaller coil due to rectangle is B = μ 0 i a b Mutual inductance is given by M = ϕ i M = B A i = μ 0 i a b i π r 2 M = μ 0 π r 2 a b (4) Magnetic field on the axis of a circular coil is B = μ 0 i R 2 2 x 2 + R 2 2 S i n c e , x > > R T h e n B = μ 0 i π R 2 r 2 2 x 3 Magnetic flux linked with the coil is ϕ = B . π r 2 M = ϕ i M = μ 0 i π 2 R 2 r 2 x 3 i

Find time constant for following circuits. (a) (b) (c) (d)

(a) Time constant is given by τ = L R e q In the above circuit resistance are in parallel combination R e q = R 1 R 2 R 1 + R 2 τ = L R 1 + R 2 R 1 R 2 (b) In this circuit i R e q = R 1 R 2 R 3 R 1 + R 2 + R 3 Now, time constant τ = L R 1 R 2 R 3 R 1 + R 2 + R 3 (c) In this circuit equivalent resistance is R p = 1 R + 1 2 R = 2 R 3 R e q = 2 R 3 + 2 R 3 R e q = 4 R 3 Time constant τ = 3 L 4 R (d) IN this circuit equivalent resistance is R p = 1 3 R + 1 6 R = 2 R R e q = 2 R + 2 R R e q = 4 R Time constant is τ = L 4 R

E M B I B E

Gamma Question Bank for Engineering Physics>Chapter 20 - Electromagnetic Induction>EXERCISE 6.4>Q 28

EASY

JEE Main/Advance

IMPORTANT

Earn 100

Network shown in the diagram is part of an electrical circuit. The current is $2 A$ and is decreasing at the rate of $2 \frac{A}{s}$ then $V_{P} - V_{Q}$ equals

50% studentsanswered this correctly

Important Questions on Electromagnetic Induction

EASY

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Physics>Chapter 20 - Electromagnetic Induction>EXERCISE 6.4>Q 29

There is a uniform magnetic field directed perpendicular and into the plane of the paper. An irregular shaped conducting loop is slowly changing into a circular loop in the plane of paper, then

EASY

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Physics>Chapter 20 - Electromagnetic Induction>EXERCISE 6.4>Q 30

Two coils of self inductance

2 mH

and

8 mH

are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coil is

MEDIUM

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Physics>Chapter 20 - Electromagnetic Induction>EXERCISE 6.4>Q 31

Find potential difference between $A$ and $C$ if potential difference between $A$ and $B$ at a certain instant is $12 V$

Question Image

MEDIUM

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Physics>Chapter 20 - Electromagnetic Induction>EXERCISE 6.4>Q 32

In the following part of a circuit $i_{1} = 5 A, \frac{d l_{1}}{d t} = 3 A s^{- 1}, i_{2} = 2 A, \frac{d i_{2}}{d t} = - 4 A / s$ . Find $V_{A} - V_{B} .$

Question Image

MEDIUM

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Physics>Chapter 20 - Electromagnetic Induction>EXERCISE 6.5>Q 33

Current in the primary coil of a mutual inductor is

2 A

and is increasing at the rate of

3 A s^{- 1}

. If the mutual inductance is

4 H

, find flux and emf in the secondary coil.

MEDIUM

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Physics>Chapter 20 - Electromagnetic Induction>EXERCISE 6.5>Q 34

Find mutual inductance of

(a) Question Image (b)

(c) Question Image (d)

MEDIUM

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Physics>Chapter 20 - Electromagnetic Induction>EXERCISE 6.5>Q 35

An inductor $(L = 100 mH),$ a resistor $(R = 100 Ω)$ and a battery $(E = 100 V)$ are initially connected in series as shown in the figure. After a long time, the battery is disconnected after short-circuiting points $A$ and $B$ . The current in the circuit, $1 ms$ after the short circuit is: