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Physics · 2024 · 3 marks
CBSE 2024 · Region 1 · Set 1 · Q24
A proton with kinetic energy 1.3384 × 10^-14 J moving horizontally from north to south, enters a uniform magnetic field B of 2 · 0 mT directed…
A proton with kinetic energy $\displaystyle 1.3384 \times 10^{-14} \mathrm{~J}$ moving horizontally from north to south, enters a uniform magnetic field B of $\displaystyle 2 \cdot 0 \mathrm{mT}$ directed eastward. Calculate :
(a)
the speed of the proton
(b)
the magnitude of acceleration of the proton
(c)
the radius of the path traced by the proton [Take ( $\displaystyle \mathrm{q} / \mathrm{m}$ ) for proton $\displaystyle =1.0 \times 10^{8} \mathrm{C} / \mathrm{kg}$ ]
Marking-scheme solution
(a)
$\displaystyle v = \sqrt{\left(\dfrac{2 \times \text{K.E.}}{m}\right)}$
$\displaystyle = 4 \times 10^{6}$ m/s
(b)
acceleration $\displaystyle = qvB / m$
$\displaystyle = 8 \times 10^{11}$ m/s$\displaystyle ^2$
(c)
$\displaystyle r = mv / Bq$
$\displaystyle = 20$ m
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