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Physics · 2025 · 3 marks
CBSE 2025 · Region 2 · Set 1 · Q23
(a) In a region of a uniform electric field E, a negatively charged particle is moving with a constant velocity v =-v_0 i near a long straight…
In a region of a uniform electric field $\displaystyle \overrightarrow{\mathrm{E}}$, a negatively charged particle is moving with a constant velocity $\displaystyle \vec{v}=-v_{0} \hat{i}$ near a long straight conductor coinciding with $\displaystyle \mathrm{XX}^{\prime}$ axis and carrying current I towards -X axis. The particle remains at a distance d from the conductor.
(i)
Draw diagram showing direction of electric and magnetic fields.
(ii)
What are the various forces acting on the charged particle?
(iii)
Find the value of $\displaystyle v_{0}$ in terms of $\displaystyle \mathrm{E}, \mathrm{d}$ and I.
Two infinitely long conductors kept along $\displaystyle \mathrm{X} \mathrm{X}^{\prime}$ and $\displaystyle \mathrm{Y} \mathrm{Y}^{\prime}$ axes are carrying current $\displaystyle \mathrm{I}_{1}$ and $\displaystyle \mathrm{I}_{2}$ along -X axis and -Y axis respectively. Find the magnitude and direction of the net magnetic field produced at point $\displaystyle \mathrm{P}(\mathrm{X}, \mathrm{Y})$.
Marking-scheme solution
(ii)
Electric force
Magnetic force
$\displaystyle F_{E}=e E$
$\displaystyle F_{B}=e v B$
(iii)
$\displaystyle e v_{0} B=e E$
$\displaystyle v_{0} \times \frac{\mu_{0} I}{2 \pi d}=E$
$\displaystyle v_{0}=\frac{2 \pi d E}{\mu_{0} I}$
Magnetic field due to conductor carrying current $\displaystyle I_{1}$: $\displaystyle \vec{B}_{1}=\frac{\mu_{0} I_{1}}{2 \pi Y}(-\hat{k})$
Magnetic field due to conductor carrying current $\displaystyle I_{2}$: $\displaystyle \vec{B}_{2}=\frac{\mu_{0} I_{2}}{2 \pi X}(\hat{k})$
$\displaystyle \vec{B}_{P}=\vec{B}_{1}+\vec{B}_{2}$
$\displaystyle \vec{B}_{P}=\frac{\mu_{0}}{2 \pi}\left(\frac{I_{2}}{X}-\frac{I_{1}}{Y}\right) \hat{k}$
Direction will be along the Z-axis.
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CBSE Class 12 Physics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.