CBSE physics

Question
CBSEENPH12039233

A circular coil of N turns and radius R carries a current I. It is unwound and rewound to make another coil of radius R/2, current I remaining the same. Calculate the ratio of the magnetic moments of the new coil and the original coil.

Solution

Length of the wire remains same,
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Magnetic moment of a coil, m = NAI

For the coil of radius R, magnetic moment, 
straight m subscript 1 space equals space straight N subscript 1 IA subscript 1 space equals space straight N subscript 1 Iπ space straight R squared
Magnetic moment for coil of radius R/2,
straight m subscript 2 space equals space straight N subscript 2 space IA subscript 2 space equals space fraction numerator 2 straight N subscript 1 IπR squared over denominator 4 end fraction equals fraction numerator straight N subscript 1 IπR squared over denominator 2 end fraction
Therefore, 
straight m subscript 2 over straight m subscript 1 space equals space 1 colon thin space 2

Question
CBSEENPH12039234

Deduce the expression for the electrostatic energy stored in a capacitor of capacitance ‘C’ and having charge ‘Q’.

How will the (i) energy stored and (ii) the electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant ‘K’? 

Solution

When a capacitor is charged by a battery, work is done by the charging battery at the expense of its chemical energy. This work is stored in the capacitor in the form of electrostatic potential energy.

Consider a capacitor of capacitance C. Initial charge on capacitor is zero. Initial potential difference between capacitor plates =0. Let a charge Q be given to it in small steps. When charge is given to capacitor, the potential difference between its plates increases.

 

Potential difference between its plates, V= q/C

Work done to give an infinitesimal charge dq to the capacitor is given by, 
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If V is the final potential difference between capacitor plates, then Q = CV
therefore space straight W space equals space fraction numerator left parenthesis CV right parenthesis squared over denominator 2 straight C end fraction space equals 1 half CV squared space equals space 1 half QV

Work is stored in the form of electrostatic potential energy.
Electrostatic potential energy, U = fraction numerator straight Q squared over denominator 2 straight C end fraction space equals space 1 half C V squared space equals space 1 half Q V

When battery is disconnected,

i) Energy stored will decrease.

Energy becomes, U = fraction numerator straight Q subscript straight o squared over denominator 2 straight C end fraction space equals space fraction numerator Q subscript o squared over denominator 2 K C subscript o end fraction space equals space U subscript o over K

So, energy is reduced to 1/K times its initial energy.

 

i) In the presence of dielectric, electric field becomes, E = straight E subscript straight o over straight K

Question
CBSEENPH12039235

Calculate the value of the resistance R in the circuit shown in the figure so that the current in the circuit is 0.2 A. What would be the potential difference between points B and E? 


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Solution

We have,

straight R subscript BCD space equals space 5 space straight capital omega space plus space 10 space straight capital omega space equals space 15 space straight capital omega

Effective resistance between B and E is,
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On applying Kirchoff's law, we have
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Hence,
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Question
CBSEENPH12039238

A series LCR circuit is connected to an ac source. Using the phasor diagram, derive the expression for the impedance of the circuit. Plot a graph to show the variation of current with frequency of the source, explaining the nature of its variation.

Solution

Suppose a resistance R, inductance L and capacitance C connected in series.
An alternating source of voltage V = Vo sin straight omegat is applied across it. Since all the components are connected in series, the current flowing through all is same.

Voltage across resistance R is VR, voltage across inductance L is VL and voltage across capacitance C is VC.

VR and (VC -VL) are mutually perpendicular and the phase difference between them is 90°.

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From the figure above, we have 
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and
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The phase difference between current and voltage is given by, 
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From the graph, we can see that with increase in frequency, current first increases and then decreases. At resonant frequency, current amplitude is maximum.

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Question
CBSEENPH12039243

Define relaxation time of the free electrons drifting in a conductor. How is it related to the drift velocity of free electrons? Use this relation to deduce the expression for the electrical resistivity of the material.            

Solution

The average time elapsed between two successive collisions is known as the relaxation time of free electrons drifting in a conductor.
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Relation between straight tau and vd is given by, 
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Consider a conductor of length ‘l’, area of cross-section A and current density n.

Current flowing through the conductor is given by, 
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Electric field applied across the ends is given by, E = V/l

So current flowing through the conductor becomes,
                       WiredFaculty
Then, 
straight V over straight I space equals space fraction numerator ml over denominator ne squared straight tau space straight A end fraction
Using ohm's law, we get
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Therefore, 
                                            straight rho space equals space fraction numerator straight m over denominator ne squared straight tau end fraction