CBSE 2024 · Region 5 · Set 3 · Q27 · 3 marks
An electron is moving with a velocity $\displaystyle \vec{v}=\left(3 \times 10^{6} \frac{\mathrm{~m}}{\mathrm{~s}}\right) \hat{\mathrm{i}}$. It enters a region of magnetic field $\displaystyle \overrightarrow{\mathrm{B}}=(91 \mathrm{mT}) \hat{\mathrm{k}}$.(a)Calculate the magnetic force $\displaystyle \overrightarrow{\mathrm{F}}_{\mathrm{B}}$ acting on electron and the radius of its path.(b)Trace the path described by it.
An electron is moving with a velocity $\displaystyle \vec{v}=\left(3 \times 10^{6} \frac{\mathrm{~m}}{\mathrm{~s}}\right) \hat{\mathrm{i}}$. It enters a region of magnetic field $\displaystyle \overrightarrow{\mathrm{B}}=(91 \mathrm{mT}) \hat{\mathrm{k}}$.
(a)
Calculate the magnetic force $\displaystyle \overrightarrow{\mathrm{F}}_{\mathrm{B}}$ acting on electron and the radius of its path.
(b)
Trace the path described by it.
Marking-scheme solution
(a)
$\displaystyle \vec{F}_{B}=q(\vec{v} \times \vec{B})$
$\displaystyle =-1.6 \times 10^{-19}\left[\left(3 \times 10^{6} \hat{i}\right) \times\left(91 \times 10^{-3} \hat{k}\right)\right]$
$\displaystyle =1.6 \times 10^{-19}\left[3 \times 10^{6} \times 91 \times 10^{-3}\right] \hat{j}$
$\displaystyle =4.368 \times 10^{-14} \hat{j} \mathrm{~N}$
$\displaystyle r=\frac{m v}{q B}$
$\displaystyle r=\frac{9.1 \times 10^{-31} \times 3 \times 10^{6}}{1.6 \times 10^{-19} \times 91 \times 10^{-3}} \mathrm{~m}$
$\displaystyle r=1.875 \times 10^{-4} \mathrm{~m}$
(b)
Anticlockwise circular path.
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CBSE Class 12 Physics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.