Advertisements
Advertisements
प्रश्न
In a given coil of self-inductance of 5 mH, current changes from 4 A to 1 A in 30 ms. Calculate the emf induced in the coil.
Advertisements
उत्तर
Self inductance, L=5 mH=5×10−3 HChange in current,
dI=(1−4)=−3 A
Change in time, `dt=30 ms=30×10−3 s`
The emf induced in the coil is given by ` e=−L(dI)/(dt)`
`⇒e=((−5xx10^(−3) )xx(−3))/(30×10^(−3))`
`⇒e=0.5 V`
APPEARS IN
संबंधित प्रश्न
Define self-inductance. Write its SI units.
A plot of magnetic flux (Φ) versus current (I) is shown in the figure for two inductors A and Β. Which of the two has larger value of self inductance?

A coil having inductance 2.0 H and resistance 20 Ω is connected to a battery of emf 4.0 V. Find (a) the current at the instant 0.20 s after the connection is made and (b) the magnetic field energy at this instant.
What are the values of the self-induced emf in the circuit of the previous problem at the times indicated therein?
A long wire carries a current of 4.00 A. Find the energy stored in the magnetic field inside a volume of 1.00 mm3 at a distance of 10.0 cm from the wire.
Obtain an expression for the self-inductance of a solenoid.
A coil of self-inductance L is connected in series with a bulb and an a. c. source. Brightness of the bulb decreases when ______.
The current flowing through an inductor of self inductance ‘L’ is continuously increasing at constant rate. The variation of induced e.m.f. (e) versus `((dI)/(dt))`is shown graphically by:
What is the magnetic field inside a long air-core solenoid?




