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CBSE Class 12 Physics 2023 Delhi Set 1 Paper

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Question : 25 of 35
Marks: +1, -0
Two coils C1C_1 and C2C_2 are placed close to each other. The magnetic flux ϕ2\phi_2 linked with the coil C2C_2 varies with the current I1I_1 flowing in coil C1C_1, as shown in the figure. Find
(i) the mutual inductance of the arrangement, and
(ii) the rate of change of current (  dI1dt)\left(\;\frac{d I_1}{d t}\right) that will induce an emf of 100V100 \mathrm{V} in coil C2C_2.
Solution:     👈: Video Solution
     (i) Since,     N2ϕ2=MI1\;\;\text{ (i) Since, }\;\; N_2 \phi_2 = M I_1
     From graph,   ϕ2=10Wb   corresponding to   I1=4A\;\;\text{ From graph, }\; \phi_2 = 10 \mathrm{Wb} \;\text{ corresponding to }\; I_1 = 4 \mathrm{A}
     and   ϕ2=10Wb\;\;\text{ and }\; \phi_2 = 10 \mathrm{Wb}
  ∴    N2×10=M×4\; \therefore \;\; N_2 \times 10 = M \times 4
     Considering   N2=1\;\;\text{ Considering }\; N_2 = 1
  M=  104=2.5H\;M = \;\frac{10}{4} = 2.5 \mathrm{H}
     (ii) Again,     N2ϕ2=MI1\;\;\text{ (ii) Again, }\;\; N_2 \phi_2 = M I_1
     Or,       ddt(N2ϕ2)=  ddt(MI1)\;\;\text{ Or, }\;\;\;\frac{d}{d t} (N_2 \phi_2) = \;\frac{d}{d t} (M I_1)
     Or,     N2  dϕ2dt=M  dI1dt\;\;\text{ Or, }\;\; N_2 \;\frac{d \phi_2}{d t} = M \;\frac{d I_1}{d t}
     Considering   N2=1\;\;\text{ Considering }\; N_2 = 1
  ϵ=M  dI1dt\;\epsilon = M \;\frac{d I_1}{d t}
     or,     100=2.5×  dI1dt\;\;\text{ or, }\;\; 100 = 2.5 \times \;\frac{d I_1}{d t}
  ∴      dI1dt=40A/s\; \therefore \;\; \;\frac{d I_1}{d t} = 40 \mathrm{A} / \mathrm{s}
  \;
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