Advertisements
Advertisements
Question
Obtain the expression for the magnetic energy stored in an inductor of self-inductance L to build up a current I through it.
Advertisements
Solution
Consider the circuit shown above consisting of an inductor L and a resistor R, connected to a source of emf E. As the connections are made, the current grows in the circuit and the magnetic field increases in the inductor. Part of the work done by the battery during the process is stored in the inductor as magnetic field energy and the rest appears as thermal energy in the resistor. After sufficient time, the current, and hence the magnetic field, becomes constant and further work done by the battery appears completely as thermal energy. If i be the current in the circuit at time t, we have
`E-L(di)/dt=iR`
`=>Eidt=i^2Rdt+Lidi`
`=>int_0^tEidt=int_0^ti^2Rdt+int_0^iLidi`
`=>int_0^t Eidt=int_0^ti^2Rdt+1/2Li^2`
Now (idt) is the charge flowing through the circuit during the time t to t+dt. Thus (Eidt) is the work done by the battery in this period. The quantity on the left-hand side of the equation (i) is, therefore, the total work done by the battery in time 0 to t. Similarly, the first term on the right-hand side of equation (i) is the total thermal energy developed in the resistor at time t. Thus
`1/2Li^2`is the energy stored in the inductor as the current in it increases from 0 to i. As the energy is zero when the current is zero, the energy in an inductor carrying a current i, is `U=1/2Li^2`
APPEARS IN
RELATED QUESTIONS
Derive the expression for the magnetic field due to a solenoid of length ‘2l’, radius ‘a’ having ’n’ number of turns per unit length and carrying a steady current ‘I’ at a point
on the axial line, distance ‘r’ from the centre of the solenoid. How does this expression compare with the axial magnetic field due to a bar magnet of magnetic moment ‘m’?
An observer to the left of a solenoid of N turns each of cross section area 'A' observes that a steady current I in it flows in the clockwise direction. Depict the magnetic field lines due to the solenoid specifying its polarity and show that it acts as a bar magnet of magnetic moment m = NIA.
Derive an expression for the mutual inductance of two long co-axial solenoids of same length wound one over the other,
Define self-inductance of a coil.
A wire AB is carrying a steady current of 10 A and is lying on the table. Another wire CD carrying 6 A is held directly above AB at a height of 2 mm. Find the mass per unit length of the wire CD so that it remains suspended at its position when left free. Give the direction of the current flowing in CD with respect to that in AB. [Take the value of g = 10 ms−2]
Define mutual inductance between two long coaxial solenoids. Find out the expression for the mutual inductance of inner solenoid of length l having the radius r1 and the number of turns n1 per unit length due to the second outer solenoid of same length and r2 number of turns per unit length.
Draw and compare the pattern of the magnetic field lines in the two cases ?
A long solenoid of radius 2 cm has 100 turns/cm and carries a current of 5 A. A coil of radius 1 cm having 100 turns and a total resistance of 20 Ω is placed inside the solenoid coaxially. The coil is connected to a galvanometer. If the current in the solenoid is reversed in direction, find the charge flown through the galvanometer.
A tightly-wound, long solenoid is kept with its axis parallel to a large metal sheet carrying a surface current. The surface current through a width dl of the sheet is Kdl and the number of turns per unit length of the solenoid is n. The magnetic field near the centre of the solenoid is found to be zero. (a) Find the current in the solenoid. (b) If the solenoid is rotated to make its axis perpendicular to the metal sheet, what would be the magnitude of the magnetic field near its centre?
A capacitor of capacitance 100 µF is connected to a battery of 20 volts for a long time and then disconnected from it. It is now connected across a long solenoid having 4000 turns per metre. It is found that the potential difference across the capacitor drops to 90% of its maximum value in 2.0 seconds. Estimate the average magnetic field produced at the centre of the solenoid during this period.
