CBSE 2023 · Region 3 · Set 2 · Q29 · 3 marks
A rectangular loop of sides $\displaystyle 25$ cm and $\displaystyle 20$ cm is lying in $\displaystyle \mathrm{x}-\mathrm{y}$ plane. It is subjected to a magnetic field $\displaystyle \overrightarrow{\mathrm{B}}=\left(5 \mathrm{t}^{2}+2 \mathrm{t}+10\right) \hat{\mathrm{k}}$, where B is in Tesla and t is in seconds. If the resistance of the loop is $\displaystyle 4 \Omega$, find the emf induced and the induced current in the loop at $\displaystyle \mathrm{t}=5 \mathrm{~s}$.
Marking-scheme solution
Induced Emf Induced Current \[\begin{aligned}
& \phi_{\mathrm{B}}=\vec{\mathrm{B}} \cdot \vec{A}=\mathrm{B} A \cos 0^{o} \\
& =\left(5 \mathrm{t}^{2}+2 \mathrm{t}+10\right) \times\left(25 \times 10^{-2} \times 20 \times 10^{-2}\right) \\
& \quad|\varepsilon|=\left|-\frac{d \phi_{\mathrm{B}}}{d \mathrm{t}}\right|
\end{aligned}
\] \[=5 \times 10^{-2} \times(10 \mathrm{t}+2)
\]
At \(\displaystyle \mathrm{t}=5 \mathrm{sec}\);
\[\varepsilon=\left(5 \times 10^{-2}\right) \times(52)
\]
\[\varepsilon=2.6 \mathrm{~V}
\]
\[I=\frac{\varepsilon}{R}
\]
\[=\frac{2.6}{4}
\]
\[I=0.65 \mathrm{~A}
\]
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CBSE Class 12 Physics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.