Question

A solenoid wound with 2020 turns/m is supplied with current that varies in time according to I = (4 A)sin(120πt), where t is in s. A small coaxial circular coil of 40 turns and radius r = 4.1 cm is located inside the solenoid near its center. (a) Derive an expression that describes the manner in which the emf in the small coil varies in time. (Use the following as necessary: t.) (b) At what average rate is energy transformed into internal energy in the small coil if the windings have a total resistance of 8.00 Ω?

Answer 1

magnetic field B inside solenoid is given by, B = n I .......................(1)

where is permeability of free space, n is number of turns per unit length an I is the current passing through the solenoid.

magnetic field B inside solenoid is given by, B = 410^-7 2020 4 sin(120t) = 1.015 10^-2 sin(120t)

induced emf in the coil, E(t) = - d/dt

where is the magnetic flux which is given by, = n B A

where n is number of turns and A is area of coil

Hence, induced emf in the coil, E(t) = - d/dt = - n A ( dB/dt )

E(t) = - 40 (0.0410.041)1.01510^-2120 cos(120t) = -0.8083 cos(120t)

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Energy dissipation p(t) in the coil is given by, p(t) = [ i(t) ]² R

where i(t) is induced current, i(t) = [ E(t) ] / R

Hence, energy dissipation in the coil, p(t) = [ E(t) ]² / R

Average rate of energy transformed to internal energy is average power dissipation in the coil

To get average energy dissipation, we integrate the above energy dissipation , which is a function of time for one cycle of variation

where T is period of time variation of one cycle, T = 1/60 s

to perform integration let us substitute 120t = , limits will change from 0 to 2