Self-Induction Numerical Class-12 Nootan ISC Physics Solution Ch-11 Electromagnetic Induction. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Self-Induction Numerical Class-12 Nootan ISC Physics Solution Ch-11 Electromagnetic Induction
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-11 | Electromagnetic Induction. |
| Topics | Numericals on Self-Induction |
| Academic Session | 2025-2026 |
Numericals on Self-Induction
Class-12 Nootan ISC Physics Solution Ch-11 Electromagnetic Induction.
Que-21: A magnetic flux of 5 µWb is linked with a coil when a current of 1 mA flows through it. Calculate self-inductance of the coil.
Ans- Q = Li => L = Q/i
=> 5 x 10^-6 / 1 x 10^-3
=> 5 x 10^-3 H = 5mH
Que-22: If a rate of change of current of 2 As^-1 induces an emf of 10 mV in a solenoid, what is the self-inductance of the solenoid?
Ans- e = IL(di/dt)I
=> 10 x 10^-3 = L x 2
=> L = 5 x 10^-3 H = 5mH
Que-23: The current in a coil increases from zero to 2 A in 0.05 s, due to which an induced emf of 80 V is developed. Calculate the coefficient of self-induction of the coil.
Ans- e = L(di/dt)
=> 80 = L x (2-0 / 0.05)
=> L = 80 x 0.05/2 = 2H
Que-24: The self-inductance of a coil is 10 H. Find the rate of change of current in the coil, if an induced emf of 120 V is developed in it.
Ans- e = L(di/dt)
=> 10 x (di/dt) = 120
=> (di/dt) = 120/10 = 12 A s^-1
Que-25: A plot of magnetic flux (φ) versus current (I) is shown in the figure for two inductors A and B. Find the ratio of the coefficient of self-inductance of A and B.
Ans- tan(60) = √3
tan(30) = 1/√3
Therefore the ratio of the slopes is √3 / (1/√3) = 3/1.
Therefore, the ratio of the self-inductance of A to B is 3:1
Que-26: A coil has an inductance of 4.0 H. Find the emf induced, if a current of 2.7 A in it is stopped within 1.35 s.
Ans- e = L(di/dt)
=> 4 x (2.7-0/1.35) = 8 volt
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