CBSE Class 12 Physics 2020 Delhi Set 1 Paper

Question : 34 of 37

Marks: +1, -0

When a conducting loop of resistance

10 Ω

and area

10 {\cm}^2

is removed from external magnetic field acting normally, the variation of induced curren in the loop with time is as shown in the figure -

Find the -
(a) total charges passed through the loop.
(b) change in magnetic flux through the loop.
(c) magnitude of the magnetic field applied.

Solution:

Given,

\;\text " Area "\;=A=10 {\cm}^2=\;{10}/{10000} {\m}^2

Resistance

= {\R}=10 Ω

From the graph, change in current

=0.4 {\A}

Change in time

=1.0 {\ s}

(i) Total charge passed through the loop

=\;\text " current "\; × \;\text " time "\;=0.4 × 1.0

=0.4 {\C}

(ii)

|e|=\;{∆ ϕ}/{∆ t}

Or,

i {\R}=\;{∆ ϕ}/{∆ t}

Or,

0.4 × 10=\;{∆ ϕ}/{1.0}

∴

Change in flux

=∆ ϕ=4 {\Wb}

(iii)

\;{|e|}= {\A} \;{∆ B}/{∆ t}

\;\;\text " Or, "\; i {\R}= {\A} \;{∆ B}/{∆ t}

\;\;\text " Or, "\; 0.4 × 10=\;{10}/{10000} \;{∆ B}/{1.0}

\; ∴ ∆ {\B}=4000 {\T}

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