Advertisements
Advertisements
प्रश्न
Obtain an expression for the energy stored in a solenoid of self-inductance ‘L’ when the current through it grows from zero to ‘I’.
उत्तर
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`
संबंधित प्रश्न
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.
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 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 solenoid of radius a and length l has n turns per unit length. It carries an electric current i. Consider a length dx of the solenoid at a distance x from one end. This contains n dx turns and may be approximated as a circular current i n dx. (a) Write the magnetic field at the centre of the solenoid due to this circular current. Integrate this expression under proper limits to find the magnetic field at the centre of the solenoid. (b) verify that if l >> a, the field tends to B = µ0ni and if a >> l, the field tends to `B =(mu_0nil)/(2a)` . Interpret these results.
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 current of 1.0 A is established in a tightly wound solenoid of radius 2 cm having 1000 turns/metre. Find the magnetic energy stored in each metre of the solenoid.
A long solenoid carrying a current produces a magnetic field B along its axis. If the current is doubled and the number of turns per cm is halved, the new value of magnetic field will be equal to ______.
