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CBSE Class 12 Physics 2023 Outside Delhi Set 1 Paper

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Question : 32 of 35
Marks: +1, -0
(a) (i) Define coefficient of self-induction. Obtain an expression for self-inductance of a long solenoid of length ll , area of cross- section A having NN turns.
(ii) Calculate the self-inductance of a coil using the following data of obtained when an AC source of frequency (200π) Hz\left(\frac{200}{\pi}\right)\,\text{Hz} and a DC source is applied across the coil.
 AC Source
 S.No.  V(Volts)  I(A)
 1  3.0  0.5
 2  6.0  1.0
 3  9.0  1.5
 DC Source
 S.No.  V(Volts)  I(A)
 1  4.0  1.0
 2  6.0  1.5
 3  8.0  2.0
OR
(b) (i) With the help of a labelled diagram, describe the principle and working of an ac generator. Hence, obtain an expression for the instantaneous value of the emf generated.
(ii) The coil of an ac generator consists of 100 turns of wire, each of area 0.5 m20.5\,\text{m}^2. The resistance of the wire is 100 Ω100\,\Omega. The coil is rotating in a magnetic field of 0.8 T0.8\,\text{T} perpendicular to its axis of rotation, at a constant angular speed of 60 radian per second. Calculate the maximum emf generated and power dissipated in the coil.
Solution:     👈: Video Solution
(a) (i) Coefficient of self induction: Coefficient of self induction of a coil is defined as the emf induced in the coil per unit rate of change of current in the same coil.
Self-inductance of a long solenoid:
Length of the solenoid =l=l
Total number of turns =N=N
Area of cross section =A=A
Current flowing through the solenoid =i=i
Magnetic flux associated with each turn =μ0μrni A=\mu_0 \mu_r n i\,\text{A}
Magnetic flux at any point on the axis
=μ0μr(Nl)i=\mu_0 \mu_r \left(\frac{N}{l}\right) i
Magnetic flux associated with each turn
=μ0μr(Nl)i A=\mu_0 \mu_r \left(\frac{N}{l}\right) i\,\text{A}
Total magnetic flux associated with the solenoid =Ï•=\phi
=μ0μr(Nl)iAN=\mu_0 \mu_{r} \left(\frac{N}{l}\right) i A N
=μ0μr(N2l)iA=\mu_0 \mu_r \left(\frac{N^2}{l}\right) i A
Self inductance of the solenoid =L=Ï•i=L=\frac{\phi}{i}
=μ0μr(N2l)A Henry=\mu_0 \mu_r \left(\frac{N^2}{l}\right) A\,\text{Henry}
(ii) Calculation of self inductance:
 DC SOURCE
 S. No.  V(Volts)  I(Ampere)  Resistance (Ohms)  Average resistance value (R)
 1  4.0  1.0  4.0   4.0 Ω4.0\,\Omega
 2  6.0  2.0  4.0
 3  8.0  2.0  4.0
 AC SOURCE
 S. No.  V(Volts)  I(Ampere)  Impedance (Ohms)  Average Impedance value (Z)
 1  3.0  0.5  6.0   6.0 Ω6.0\,\Omega
 2  6.0  1.0  6.0
 3  9.0  1.5  6.0
XL=Z2−R2X_L=\sqrt{Z^2-R^2}
Or,  2πfL=62−42\text{Or,}\;2\pi f L=\sqrt{6^2-4^2}
Or,  2πfL=20\text{Or,}\;2\pi f L=\sqrt{20}
Or,  2π×200π×L=20\text{Or,}\;2\pi \times \frac{200}{\pi} \times L=\sqrt{20}
∴L=1405 Henry\therefore L=\frac{1}{40}\sqrt{5}\,\text{Henry}
OR
(b) (i) Refer the answer of Q28 of CBSE 12 DELHI, Set-1.
(ii) Given,\text{Given,}N=100N=100
A=0.5 m2A=0.5\,\text{m}^2
ω=60 radians/second\omega=60\,\text{radians/second}
B=0.8 TB=0.8\,\text{T}
r=100 Ωr=100\,\Omega
Maximum emf generated=NBAω\text{Maximum emf generated}=NBA\omega
Or, Maximum emf generated =100×0.8×0.5×60=100 \times 0.8 \times 0.5 \times 60
∴\therefore Maximum emf generated =E0=2400 V=E_0=2400\,\text{V}
Maximum current =I0=E0R=2400100=24 A=I_0=\frac{E_0}{R}=\frac{2400}{100}=24\,\text{A}
Power dissipation =E0I0=2400×24=57600=E_0 I_0=2400 \times 24=57600 Watt
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