Advertisements
Advertisements
Question
Obtain an expression for the energy stored in a solenoid of self-inductance ‘L’ when the current through it grows from zero to ‘I’.
Advertisements
Solution
When the current is varied, the flux linked with the coil changes and an e.m.f. is induced in the coil. It is given as
`in=-(d(Nphi_E))/dt=-L(dI)/dt`
The self-induced e.m.f. is also called back e.m.f. as it opposes any change in current in the circuit. So, work needs to be done against back e.m.f. in establishing current.
This work done is stored as magnetic potential energy.
The rate of doing work is given as
`(dW)/(dt)=|in|I=LI(dI)/dt " (neglecting negative sign)"`
Thus, the total work done in establishing current from 0 to I is
`w=intdw=int_0^1LIdI=1/2LI^2`
RELATED QUESTIONS
Define the term self-inductance of a solenoid.
Obtain the expression for mutual inductance of a pair of long coaxial solenoids each of length l and radii r1 and r2 (r2 >> r1). Total number of turns in the two solenoids are N1 and N2, respectively.
A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.
A magnetic field of 100 G (1 G = 10−4 T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10−3 m2. The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m−1. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.
Define self-inductance of a coil.
A wire AB is carrying a steady current of 12 A and is lying on the table. Another wire CD carrying 5 A is held directly above AB at a height of 1 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]
How is the magnetic field inside a given solenoid made strong?
The magnetic field inside a tightly wound, long solenoid is B = µ0 ni. It suggests that the field does not depend on the total length of the solenoid, and hence if we add more loops at the ends of a solenoid the field should not increase. Explain qualitatively why the extra-added loops do not have a considerable effect on the field inside the solenoid.
A copper wire having resistance 0.01 ohm in each metre is used to wind a 400-turn solenoid of radius 1.0 cm and length 20 cm. Find the emf of a battery which when connected across the solenoid will cause a magnetic field of 1.0 × 10−2 T near the centre of the solenoid.
