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4.Moving Charges and Magnetism
medium
A charge particle of $2\,\mu\,C$ accelerated by a potential difference of $100\,V$ enters a region of uniform magnetic field of magnitude $4\,m\,T$ at right angle to the direction of field. The charge particle completes semicircle of radius $3\,cm$ inside magnetic field. The mass of the charge particle is $........\times 10^{-18}\,kg$.
A
$142$
B
$144$
C
$141$
D
$140$
(JEE MAIN-2023)
Solution
$=\frac{m v}{q B}=\frac{\sqrt{2 k m}}{q B}, m=\frac{r^2 q^2 B^2}{2 k}$
$m = \frac{\frac{1}{100} \times \frac{3}{100} \times 2 \times 2 \times 4 \times 10^{-3} \times 4 \times 10^{-3} \times 10^{-12}}{2 \times(100) \times 10^{-6}}$
$=144 \times 10^{-18}\,kg$
Standard 12
Physics
