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Physics · 2026 · 3 marks
CBSE 2026 · Region 4 · Set 1 · Q24
The figure given below shows a square-shaped loop MNOP of side 25 cm placed horizontally in a uniform magnetic field B directed vertically downward.…
The figure given below shows a square-shaped loop MNOP of side $\displaystyle 25$ cm placed horizontally in a uniform magnetic field $\displaystyle \overrightarrow{\mathrm{B}}$ directed vertically downward. The position of the loop at $\displaystyle \mathrm{t}=0 \mathrm{~s}$ is as shown in figure. The loop is pulled with a constant velocity of $\displaystyle 25$ cm/s till it goes out of the field.
(a)
What will be the direction of the induced current in the loop as it goes out of the field? For how long would the current in the loop persist ?
(b)
Plot graphs showing the variation of magnetic flux and magnitude of induced emf as a function of time.
Marking-scheme solution
(a)
As the magnetic flux linked with the loop is decreasing, hence in accordance with Lenz's law, the induced current will be in the clockwise direction.
$\displaystyle \mathrm{t}=\dfrac{25 \mathrm{~cm}}{25 \mathrm{~cm} / \mathrm{s}}=1 \mathrm{~s}$
(b)
Plot of magnetic flux $\displaystyle \phi$ against time: constant while the loop is fully inside the field, then falling linearly to zero between $\displaystyle t=3 \mathrm{~s}$ and $\displaystyle t=4 \mathrm{~s}$ as the loop leaves the field.
Plot of the magnitude of induced emf $\displaystyle |\varepsilon|$ against time: zero while $\displaystyle \phi$ is constant, then a constant non-zero value between $\displaystyle t=3 \mathrm{~s}$ and $\displaystyle t=4 \mathrm{~s}$, and zero again afterwards.
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CBSE Class 12 Physics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.