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In an RLC ac circuit, the values of the inductance and capacitance are both doub

ID: 1411740 • Letter: I

Question

In an RLC ac circuit, the values of the inductance and capacitance are both doubled. As a result of this change, the resonance frequency of the circuit is the same as before. reduced to one-half the original value. reduced to one-fourth the original value. twice the original value. four times the original value. A bar magnet is pushed through a coil of wire of cross-sectional area 0.020 m^2 as shown in the figure. The coil has seven turns, and the rate of change of the strength of the magnetic field in it due to the motion of the bar magnet is 0.040 T/s. What is the magnitude of the induced emf in that coil of wire? 5.6 times 10^-3 V 5.6 times 10^-2 V 5.6 times 10^-1 V 5.6 times 10^-4 V 5.6 times 10^-5 V A conducting rod of length I = 25 cm is placed on a U-shaped metal wire that is connected to a lightbulb having a resistance of 8.0 Ohm, as shown in the figure. The wire and the rod are in the plane of the page. A constant uniform magnetic field of strength 0.40 T is applied perpendicular to and into the paper. An applied external force pulls the rod to the right with a constant speed of 6.0 m/s. What is the magnitude of the emf induced in the rod?

Explanation / Answer

(1) ans

the resonant frequency f=1/{2*pi*[LC]^½}

if L=2L , C=2C

the resonant frequency f'=1/{2*pi*[2L*2C]^½} =1/2{2*pi*[LC]^½}

the resonant frequency is reduced to half of original value

the correct option is B

(2) ans

area of crossection A=0.020 m^2

the rate of change of magnetic field dB/dt =0.040 T/s

number of turns N=7

according to the concept of faraday law of induced emf

the induced emf e=AN dB/dt

=7*0.02*0.04=5.6*10^-3 V

the correct option is A

(3) ans

length L=25 cm=0.25 m

resistance r=8 ohms

magnetic field B=0.40T

speed V=6 m/s

according to the motional emf of a rod

the motional emf e=BLV=0.40*0.25*6=0.6 V

the current passing in the rod i=e/r=0.6/8=0.075 A

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