CBSE Class 12 Physics 2023 Delhi Set 1 Paper

Question : 25 of 35

Marks: +1, -0

Two coils

C_1

and

C_2

are placed close to each other. The magnetic flux

ϕ_2

linked with the coil

C_2

varies with the current

I_1

flowing in coil

C_1

, as shown in the figure. Find

(i) the mutual inductance of the arrangement, and
(ii) the rate of change of current

(\;{d {\I}_1}/{d t})

that will induce an emf of

100 {\V}

in coil

C_2

Solution:

\;\;\text " (i) Since, "\; \;\; {\N}_2 ϕ_2=M I_1

\;\;\text " From graph, "\; ϕ_2=10 {\Wb} \;\text " corresponding to "\; I_1=4 {\A}

\;\;\text " and "\; ϕ_2=10 {\Wb}

\; ∴ \;\; {\N}_2 × 10= {\M} × 4

\;\;\text " Considering "\; N_2=1

\;M=\;{10}/{4}=2.5 {\H}

\;\;\text " (ii) Again, "\; \;\; {\N}_2 ϕ_2= {\MI}_1

\;\;\text " Or, "\; \;\; \;{d}/{d t} (N_2 ϕ_2) =\;{d}/{d t} (M I_1)

\;\;\text " Or, "\; \;\; N_2 \;{d ϕ_2}/{d t}=M \;{d I_1}/{d t}

\;\;\text " Considering "\; {\N}_2=1

\;ε=M \;{d I_1}/{d t}

\;\;\text " or, "\; \;\; 100=2.5 × \;{d I_1}/{d t}

\; ∴ \;\; \;{d I_1}/{d t}=40 {\A} \/ {\ s}

\;

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